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bg-background","children":["$","$L39",null,{"backHref":"/a-level/maths/wjec/questions","basePath":"/a-level/maths/wjec/sequences-and-series/questions","formulaUrl":"https://www.wjec.co.uk/media/4jfdytcz/ss-w27-18-e.pdf","initialIndex":145,"initialQuestionWindow":[{"id":"58ab2398-682f-4b01-a663-d719899ca3c9","revisionId":"ebb6a52c-b1d5-484e-a98c-0c247b8ecfc4","questionNumber":null,"questionType":"LA","specification":"The flow rate of a coolant into a chemical reactor, $R_t$, is adjusted every minute such that the sequence of flow rates forms an arithmetic progression, where $t$ is the time in minutes.","difficultyBand":"medium","tier":null,"totalMarks":8,"sourceTitle":null,"calculatorUse":null,"isCalculation":true,"parts":[{"id":"afd4057f-0012-4d53-bb22-1a0415147731","label":"a","content":"In the 4th minute, the flow rate is $40\\text{ L/min}$. The total volume of coolant delivered during the first 10 minutes is $325\\text{ L}$. Determine the initial flow rate, $a$, and the constant change in flow rate per minute, $d$.","metadata":null,"markscheme":"- Uses $u_n = a + (n-1)d$ to state $a + 3d = 40$ M1\n- Uses $S_n = \\frac{n}{2}(2a + (n-1)d)$ to state $\\frac{10}{2}(2a + 9d) = 325$, which simplifies to $2a + 9d = 65$ M1\n- Attempts to solve the simultaneous equations, e.g., by multiplying the first by 2 to get $2a + 6d = 80$ and subtracting M1\n- Obtains $d = -5$ and $a = 55$ A1","marks":4,"order":0,"markingRubric":null,"tableImages":[],"explanation":{"methods":[{"id":"algebraic_simultaneous","label":"Algebraic Simultaneous Equations","steps":[{"type":"text","order":0,"title":"Set up first equation","content":"The question states that the flow rate forms an arithmetic progression. We use the formula for the $n$-th term of an arithmetic sequence from our formula booklet:\n\n$$u_n = a + (n-1)d$$\n\nwhere $a$ is the initial term and $d$ is the common difference.\n\nIn the 4th minute, the flow rate is $40\\text{ L/min}$. Substituting $n = 4$ and $u_4 = 40$ gives:\n\n$$a + (4-1)d = 40$$\n\n$$\\Rightarrow a + 3d = 40\\quad\\text{--- (Equation 1)}$$","highlights":[{"text":"In the 4th minute, the flow rate is 40 L/min","location":"specification"}]},{"type":"text","order":1,"title":"Set up second equation","content":"Next, we are given that the total volume of coolant delivered during the first $10$ minutes is $325\\text{ L}$. The sum of the first $n$ terms of an arithmetic series is given by:\n\n$$S_n = \\frac{n}{2}(2a + (n-1)d)$$\n\nSubstituting $n = 10$ and $S_{10} = 325$:\n\n$$S_{10} = \\frac{10}{2}(2a + (10-1)d) = 325$$\n\n$$5(2a + 9d) = 325$$\n\nDividing both sides by $5$ gives:\n\n$$2a + 9d = 65\\quad\\text{--- (Equation 2)}$$","highlights":[{"text":"The total volume of coolant delivered during the first 10 minutes is 325 L","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$$\\begin{array}{rcc}\\text{Multiply Equation 1 by 2:} & \\htmlData{anchor=eq1-scaled}{\\color{@sky}{2a + 6d = 80}} \\\\ \\text{Equation 2:} & \\htmlData{anchor=eq2}{\\color{@amber}{2a + 9d = 65}} \\end{array}$$"},{"latex":"$$$\\text{Subtracting the first equation from the second yields:}$$"},{"latex":"$$$\\htmlData{anchor=subtraction}{\\color{@purple}{3d = -15}}$$"},{"latex":"$$$\\htmlData{anchor=d-val}{\\color{@purple}{d = -5}}$$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"eq1-scaled"},{"color":"amber","style":"text-color","anchor":"eq2"},{"color":"purple","style":"box","anchor":"d-val"}],"connections":[{"to":"subtraction","from":"eq1-scaled","color":"sky","label":"subtraction","style":"solid"},{"to":"subtraction","from":"eq2","color":"amber","label":"subtraction","style":"solid"},{"to":"d-val","from":"subtraction","color":"purple","label":"solve","style":"solid"}]},"order":2,"title":"Solve the system","description":"We can solve this system of linear equations by elimination. We multiply Equation 1 by $2$ to match the $2a$ term in Equation 2."},{"type":"text","order":3,"title":"Find the initial rate","content":"Now substitute $d = -5$ back into Equation 1 to find the initial flow rate, $a$:\n\n$$a + 3(-5) = 40$$\n\n$$a - 15 = 40$$\n\n$$a = 55$$\n\nThus, the initial flow rate is $a = 55\\text{ L/min}$ and the constant change in flow rate per minute is $d = -5\\text{ L/min}^2$.","highlights":[{"text":"Determine the initial flow rate, a, and the constant change in flow rate per minute, d.","location":"specification"}]}]},{"id":"sequence_properties_mean","label":"Properties of Average Term","steps":[{"type":"text","order":0,"title":"Analyze the given information","content":"Let the sequence of flow rates be an arithmetic progression with first term $a$ and common difference $d$.\n\nWe are given:\n- The flow rate in the 4th minute: $u_4 = 40\\text{ L/min}$\n- The total volume delivered in the first 10 minutes: $S_{10} = 325\\text{ L}$\n\nRecall that the general formula for the $n$-th term of an arithmetic progression is:\n$$u_n = a + (n-1)d$$\n\nSo for the 4th minute, we have:\n$$a + 3d = 40 \\quad \\text{--- (Equation 1)}$$","highlights":[{"text":"4th minute, the flow rate is 40 L/min","location":"specification"},{"text":"total volume of coolant delivered during the first 10 minutes is 325 L","location":"specification"}]},{"type":"text","order":1,"title":"Find the average term","content":"The sum of the first 10 terms, $S_{10}$, is $325\\text{ L}$. \n\nThe average value of these first 10 terms is simply the total sum divided by the number of terms:\n$$\\text{Average} = \\frac{S_{10}}{10} = \\frac{325}{10} = 32.5\\text{ L/min}$$\n\nIn any arithmetic progression with an even number of terms $N$, the average of the first $N$ terms is equal to the average of the two middle terms.\n\nFor $N = 10$, the two middle terms are the 5th term ($u_5$) and the 6th term ($u_6$). Therefore:\n$$\\text{Average} = \\frac{u_5 + u_6}{2} = 32.5$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\text{Since } u_5 = a + 4d \\text{ and } u_6 = a + 5d, \\text{ their average is:}$"},{"latex":"$$\\frac{(a + 4d) + (a + 5d)}{2} = \\htmlData{anchor=avg-form}{a + 4.5d}$"},{"latex":"$$\\htmlData{anchor=avg-val}{a + 4.5d = 32.5} \\quad \\text{--- (Equation 2)}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"avg-form"},{"color":"sky","style":"text-color","anchor":"avg-val"}],"connections":[{"to":"avg-val","from":"avg-form","color":"sky","label":"substitute average","style":"solid"}]},"order":2,"title":"Link to the middle point","description":"The average of the 5th and 6th terms lies exactly at the midpoint, which corresponds to $a + 4.5d$."},{"type":"text","order":3,"title":"Determine the common difference","content":"Now we can compare our two equations to find the common difference $d$:\n1. $a + 3d = 40$\n2. $a + 4.5d = 32.5$\n\nSubtract Equation 1 from Equation 2:\n$$(a + 4.5d) - (a + 3d) = 32.5 - 40$$\n$$1.5d = -7.5$$\n\nSolving for $d$:\n$$d = \\frac{-7.5}{1.5} = -5$$"},{"type":"text","order":4,"title":"Find the initial flow rate","content":"With $d = -5$, we can work backward to find the initial flow rate, $a$, by substituting $d$ back into Equation 1:\n$$a + 3(-5) = 40$$\n$$a - 15 = 40$$\n$$a = 55$$\n\nThus, the initial flow rate is $a = 55\\text{ L/min}$ and the constant change in flow rate per minute is $d = -5\\text{ L/min}^2$."}]},{"id":"gdc_system_solver","label":"Using Graphing Calculator","steps":[{"type":"text","order":0,"title":"Set up the term equation","content":"An arithmetic progression has an $n$-th term formula given by:\n\n$$u_n = a + (n-1)d$$\n\nwhere $a$ is the initial term (initial flow rate) and $d$ is the common difference (constant change per minute).\n\nWe are given that in the 4th minute ($n = 4$), the flow rate is $40\\text{ L/min}$. Substituting these values into the formula gives:\n\n$$a + (4-1)d = 40 \\implies a + 3d = 40$$","highlights":[{"text":"4th minute, the flow rate is 40 L/min","location":"specification"}]},{"type":"text","order":1,"title":"Set up the sum equation","content":"The sum of the first $n$ terms of an arithmetic progression is given by:\n\n$$S_n = \\frac{n}{2}\\big(2a + (n-1)d\\big)$$\n\nWe are given that the total volume delivered over the first 10 minutes ($n = 10$) is $325\\text{ L}$. Substituting $n = 10$ and $S_{10} = 325$:\n\n$$\\frac{10}{2}(2a + 9d) = 325$$\n\n$$5(2a + 9d) = 325$$\n\nDividing both sides by 5 simplifies this to:\n\n$$2a + 9d = 65$$","highlights":[{"text":"total volume of coolant delivered during the first 10 minutes is 325 L","location":"specification"}]},{"type":"text","order":2,"title":"Solve using GDC","content":"We now have a system of two linear simultaneous equations:\n\n$$\\begin{cases} a + 3d = 40 \\\\ 2a + 9d = 65 \\end{cases}$$\n\nWe can use a graphing calculator (GDC) to solve this system directly instead of doing manual algebraic manipulation.","calculator":[{"gdc":{"expression":"x = 55, y = -5","instructions":"1. Go to the **Equation/Solver** menu (often under **MENU -> Equation** or **EQUA**).\n2. Choose **F1: Simultaneous**.\n3. Choose **F1: 2** unknowns.\n4. Input the coefficient matrix:\n - Row 1: `1`, `3`, `40`\n - Row 2: `2`, `9`, `65`\n5. Press **F6: SOLVE** to find the values."},"ti84":{"expression":"x1 = 55, x2 = -5","instructions":"1. Press **[APPS]** and select **PlySmlt2** (Polynomial Root Finder and Simultaneous Equation Solver).\n2. Select **2: Simul Eqn Solver**.\n3. Set **Number of Equations = 2** and **Number of Unknowns = 2**, then press **[NEXT]** (graph key).\n4. Enter the coefficients into the system matrix:\n - Row 1: `1`, `3`, `40`\n - Row 2: `2`, `9`, `65`\n5. Press **[SOLVE]** (graph key)."},"desmos":{"expression":"(a, d) = (55, -5)","instructions":"In the graphing calculator or matrix calculator, define the system of equations as equations:\n1. Type `a + 3d = 40`\n2. Type `2a + 9d = 65`\nDesmos will find the intersection of the two lines."},"description":"Use the built-in system solver or matrix solver to find the values of $a$ (as $X$) and $d$ (as $Y$)."}]},{"type":"text","order":3,"title":"State final answers","content":"From the calculator solver output, we get:\n\n$$a = 55$$\n$$d = -5$$\n\nThus, the initial flow rate is $55\\text{ L/min}$ and the constant change in flow rate is $-5\\text{ L/min}^2$ (decreasing by $5\\text{ L/min}$ every minute)."}]}],"generatedAt":"2026-06-09T05:43:36.819Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]},{"id":"48384b40-dfd0-47ea-bf44-76ae83e6bfdd","label":"b","content":"A secondary auxiliary system has an initial flow rate of $6\\text{ L/min}$, which increases by $2\\text{ L/min}$ each minute. The total volume delivered by this auxiliary system over $n$ minutes is $T_n$. Find the value of $n$ such that the total volume delivered by the primary system, $S_n$, is equal to $T_n$, where $n > 0$.","metadata":null,"markscheme":"- Writes an expression for the sum of the primary series: $S_n = \\frac{n}{2}(2(55) + (n-1)(-5)) = \\frac{n}{2}(115 - 5n)$ M1\n- Writes an expression for the sum of the auxiliary series: $T_n = \\frac{n}{2}(2(6) + (n-1)(2)) = \\frac{n}{2}(10 + 2n)$ M1\n- Sets $S_n = T_n$ and eliminates $n$ (since $n \\neq 0$): $115 - 5n = 10 + 2n$ M1\n- Solves for $n$ to find $n = 15$ A1","marks":4,"order":1,"markingRubric":null,"tableImages":[],"explanation":{"methods":[{"id":"algebraic","label":"Algebraic Approach","steps":[{"type":"text","order":0,"title":"Define primary system sum","content":"To find the total volume delivered by the primary system over $n$ minutes, $S_n$, we use the sum formula for an arithmetic series from the formula booklet:\n\n$$S_n = \\frac{n}{2}[2a + (n-1)d]$$\n\nFor the primary system, the initial flow rate is $a = 55\\text{ L/min}$ and it decreases by $5\\text{ L/min}$ each minute, so $d = -5$.\n\nSubstituting these values into the formula gives:\n\n$$S_n = \\frac{n}{2}[2(55) + (n-1)(-5)]$$\n\n$$S_n = \\frac{n}{2}[110 - 5n + 5]$$\n\n$$S_n = \\frac{n}{2}(115 - 5n)$$\n\nThis earns the first mark **(M1)**.","highlights":[{"text":"total volume delivered by the primary system, S_n","location":"specification"}]},{"type":"text","order":1,"title":"Define auxiliary system sum","content":"For the secondary auxiliary system, we are given that the initial flow rate is $a = 6\\text{ L/min}$ and the common difference is $d = 2\\text{ L/min}$.\n\nUsing the same arithmetic series sum formula, we find the total volume delivered, $T_n$:\n\n$$T_n = \\frac{n}{2}[2(6) + (n-1)(2)]$$\n\n$$T_n = \\frac{n}{2}[12 + 2n - 2]$$\n\n$$T_n = \\frac{n}{2}(10 + 2n)$$\n\nThis earns the second mark **(M1)**.","highlights":[{"text":"initial flow rate of 6 L/min","location":"specification"},{"text":"increases by 2 L/min each minute","location":"specification"},{"text":"total volume delivered by this auxiliary system over n minutes is T_n","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\htmlData{anchor=s-expr}{\\frac{n}{2}(115 - 5n)} = \\htmlData{anchor=t-expr}{\\frac{n}{2}(10 + 2n)}$"},{"text":"Since we are given that $n > 0$, we can divide both sides of the equation by $\\frac{n}{2}$ to simplify it directly:"},{"latex":"$$\\htmlData{anchor=simp-expr}{115 - 5n = 10 + 2n}$"}],"highlights":[{"color":"sky","style":"box","anchor":"s-expr"},{"color":"amber","style":"box","anchor":"t-expr"}],"connections":[{"to":"simp-expr","from":"s-expr","color":"sky","label":"divide by n/2","style":"solid"}]},"order":2,"title":"Equate and simplify","description":"We set the two total volume expressions equal to each other, $S_n = T_n$, and simplify the resulting algebraic equation by eliminating the common factor of $n$."},{"type":"text","order":3,"title":"Solve for n","content":"Now, we solve the simplified linear equation for $n$ by isolating the variable terms:\n\n$$115 - 5n = 10 + 2n$$\n\n$$115 - 10 = 2n + 5n$$\n\n$$105 = 7n$$\n\n$$n = 15$$\n\nSolving this equation yields $n = 15$ minutes, earning the final marks **(M1 A1)**.","highlights":[{"text":"total volume delivered by the primary system, S_n, is equal to T_n, where n > 0","location":"specification"}]}]},{"id":"graphical","label":"Graphical/Calculator Approach","steps":[{"type":"text","order":0,"title":"Define primary system sum","content":"To find when the total volumes are equal, we first need the formula for the sum of an arithmetic series from our formula booklet:\n\n$$S_n = \\frac{n}{2}[2a + (n-1)d]$$\n\nFor the primary system, the initial flow rate is $a = 55\\text{ L/min}$ and it decreases by $d = -5\\text{ L/min}$ each minute. Substituting these into the formula gives:\n\n$$S_n = \\frac{n}{2}[2(55) + (n-1)(-5)] = \\frac{n}{2}(115 - 5n)$$\n\nWe can define this function as $y_1 = \\frac{x}{2}(115 - 5x)$ on our calculator.","highlights":[{"text":"total volume delivered by the primary system, S_n","location":"specification"}]},{"type":"text","order":1,"title":"Define auxiliary system sum","content":"For the auxiliary system, the initial flow rate is $a = 6\\text{ L/min}$ and it increases by $d = 2\\text{ L/min}$ each minute.\n\nUsing the same arithmetic series sum formula, we get:\n\n$$T_n = \\frac{n}{2}[2(6) + (n-1)(2)] = \\frac{n}{2}(10 + 2n)$$\n\nWe can define this second function as $y_2 = \\frac{x}{2}(10 + 2x)$ on our calculator.","highlights":[{"text":"initial flow rate of 6 L/min, which increases by 2 L/min each minute. The total volume delivered by this auxiliary system over n minutes is T_n","location":"specification"}]},{"type":"text","order":2,"title":"Graph and find intersection","content":"We want to find the value of $n > 0$ where $S_n = T_n$.\n\nInstead of solving the quadratic equation algebraically, we can plot both $y_1 = \\frac{x}{2}(115 - 5x)$ and $y_2 = \\frac{x}{2}(10 + 2x)$ on a graphing calculator and find their intersection point in the first quadrant ($x > 0$).","highlights":[{"text":"S_n, is equal to T_n, where n > 0","location":"specification"}]},{"type":"text","order":3,"title":"State the final solution","content":"The calculator displays the intersection point of the two curves as:\n\n$$(15, 300)$$\n\nSince $x$ represents the time in minutes $n$, we have:\n\n$$n = 15$$\n\n*(Note: There is another intersection at $n = 0$, but the question specifies $n > 0$.)*"}]}],"generatedAt":"2026-06-09T05:45:07.693Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"afd4057f-0012-4d53-bb22-1a0415147731:algebraic_simultaneous","label":"(a) Algebraic Simultaneous Equations","steps":[{"type":"text","order":0,"title":"Set up first equation","content":"The question states that the flow rate forms an arithmetic progression. We use the formula for the $n$-th term of an arithmetic sequence from our formula booklet:\n\n$$u_n = a + (n-1)d$$\n\nwhere $a$ is the initial term and $d$ is the common difference.\n\nIn the 4th minute, the flow rate is $40\\text{ L/min}$. Substituting $n = 4$ and $u_4 = 40$ gives:\n\n$$a + (4-1)d = 40$$\n\n$$\\Rightarrow a + 3d = 40\\quad\\text{--- (Equation 1)}$$","highlights":[{"text":"In the 4th minute, the flow rate is 40 L/min","location":"specification"}]},{"type":"text","order":1,"title":"Set up second equation","content":"Next, we are given that the total volume of coolant delivered during the first $10$ minutes is $325\\text{ L}$. The sum of the first $n$ terms of an arithmetic series is given by:\n\n$$S_n = \\frac{n}{2}(2a + (n-1)d)$$\n\nSubstituting $n = 10$ and $S_{10} = 325$:\n\n$$S_{10} = \\frac{10}{2}(2a + (10-1)d) = 325$$\n\n$$5(2a + 9d) = 325$$\n\nDividing both sides by $5$ gives:\n\n$$2a + 9d = 65\\quad\\text{--- (Equation 2)}$$","highlights":[{"text":"The total volume of coolant delivered during the first 10 minutes is 325 L","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$$\\begin{array}{rcc}\\text{Multiply Equation 1 by 2:} & \\htmlData{anchor=eq1-scaled}{\\color{@sky}{2a + 6d = 80}} \\\\ \\text{Equation 2:} & \\htmlData{anchor=eq2}{\\color{@amber}{2a + 9d = 65}} \\end{array}$$"},{"latex":"$$$\\text{Subtracting the first equation from the second yields:}$$"},{"latex":"$$$\\htmlData{anchor=subtraction}{\\color{@purple}{3d = -15}}$$"},{"latex":"$$$\\htmlData{anchor=d-val}{\\color{@purple}{d = -5}}$$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"eq1-scaled"},{"color":"amber","style":"text-color","anchor":"eq2"},{"color":"purple","style":"box","anchor":"d-val"}],"connections":[{"to":"subtraction","from":"eq1-scaled","color":"sky","label":"subtraction","style":"solid"},{"to":"subtraction","from":"eq2","color":"amber","label":"subtraction","style":"solid"},{"to":"d-val","from":"subtraction","color":"purple","label":"solve","style":"solid"}]},"order":2,"title":"Solve the system","description":"We can solve this system of linear equations by elimination. We multiply Equation 1 by $2$ to match the $2a$ term in Equation 2."},{"type":"text","order":3,"title":"Find the initial rate","content":"Now substitute $d = -5$ back into Equation 1 to find the initial flow rate, $a$:\n\n$$a + 3(-5) = 40$$\n\n$$a - 15 = 40$$\n\n$$a = 55$$\n\nThus, the initial flow rate is $a = 55\\text{ L/min}$ and the constant change in flow rate per minute is $d = -5\\text{ L/min}^2$.","highlights":[{"text":"Determine the initial flow rate, a, and the constant change in flow rate per minute, d.","location":"specification"}]}]},{"id":"afd4057f-0012-4d53-bb22-1a0415147731:sequence_properties_mean","label":"(a) Properties of Average Term","steps":[{"type":"text","order":0,"title":"Analyze the given information","content":"Let the sequence of flow rates be an arithmetic progression with first term $a$ and common difference $d$.\n\nWe are given:\n- The flow rate in the 4th minute: $u_4 = 40\\text{ L/min}$\n- The total volume delivered in the first 10 minutes: $S_{10} = 325\\text{ L}$\n\nRecall that the general formula for the $n$-th term of an arithmetic progression is:\n$$u_n = a + (n-1)d$$\n\nSo for the 4th minute, we have:\n$$a + 3d = 40 \\quad \\text{--- (Equation 1)}$$","highlights":[{"text":"4th minute, the flow rate is 40 L/min","location":"specification"},{"text":"total volume of coolant delivered during the first 10 minutes is 325 L","location":"specification"}]},{"type":"text","order":1,"title":"Find the average term","content":"The sum of the first 10 terms, $S_{10}$, is $325\\text{ L}$. \n\nThe average value of these first 10 terms is simply the total sum divided by the number of terms:\n$$\\text{Average} = \\frac{S_{10}}{10} = \\frac{325}{10} = 32.5\\text{ L/min}$$\n\nIn any arithmetic progression with an even number of terms $N$, the average of the first $N$ terms is equal to the average of the two middle terms.\n\nFor $N = 10$, the two middle terms are the 5th term ($u_5$) and the 6th term ($u_6$). Therefore:\n$$\\text{Average} = \\frac{u_5 + u_6}{2} = 32.5$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\text{Since } u_5 = a + 4d \\text{ and } u_6 = a + 5d, \\text{ their average is:}$"},{"latex":"$$\\frac{(a + 4d) + (a + 5d)}{2} = \\htmlData{anchor=avg-form}{a + 4.5d}$"},{"latex":"$$\\htmlData{anchor=avg-val}{a + 4.5d = 32.5} \\quad \\text{--- (Equation 2)}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"avg-form"},{"color":"sky","style":"text-color","anchor":"avg-val"}],"connections":[{"to":"avg-val","from":"avg-form","color":"sky","label":"substitute average","style":"solid"}]},"order":2,"title":"Link to the middle point","description":"The average of the 5th and 6th terms lies exactly at the midpoint, which corresponds to $a + 4.5d$."},{"type":"text","order":3,"title":"Determine the common difference","content":"Now we can compare our two equations to find the common difference $d$:\n1. $a + 3d = 40$\n2. $a + 4.5d = 32.5$\n\nSubtract Equation 1 from Equation 2:\n$$(a + 4.5d) - (a + 3d) = 32.5 - 40$$\n$$1.5d = -7.5$$\n\nSolving for $d$:\n$$d = \\frac{-7.5}{1.5} = -5$$"},{"type":"text","order":4,"title":"Find the initial flow rate","content":"With $d = -5$, we can work backward to find the initial flow rate, $a$, by substituting $d$ back into Equation 1:\n$$a + 3(-5) = 40$$\n$$a - 15 = 40$$\n$$a = 55$$\n\nThus, the initial flow rate is $a = 55\\text{ L/min}$ and the constant change in flow rate per minute is $d = -5\\text{ L/min}^2$."}]},{"id":"afd4057f-0012-4d53-bb22-1a0415147731:gdc_system_solver","label":"(a) Using Graphing Calculator","steps":[{"type":"text","order":0,"title":"Set up the term equation","content":"An arithmetic progression has an $n$-th term formula given by:\n\n$$u_n = a + (n-1)d$$\n\nwhere $a$ is the initial term (initial flow rate) and $d$ is the common difference (constant change per minute).\n\nWe are given that in the 4th minute ($n = 4$), the flow rate is $40\\text{ L/min}$. Substituting these values into the formula gives:\n\n$$a + (4-1)d = 40 \\implies a + 3d = 40$$","highlights":[{"text":"4th minute, the flow rate is 40 L/min","location":"specification"}]},{"type":"text","order":1,"title":"Set up the sum equation","content":"The sum of the first $n$ terms of an arithmetic progression is given by:\n\n$$S_n = \\frac{n}{2}\\big(2a + (n-1)d\\big)$$\n\nWe are given that the total volume delivered over the first 10 minutes ($n = 10$) is $325\\text{ L}$. Substituting $n = 10$ and $S_{10} = 325$:\n\n$$\\frac{10}{2}(2a + 9d) = 325$$\n\n$$5(2a + 9d) = 325$$\n\nDividing both sides by 5 simplifies this to:\n\n$$2a + 9d = 65$$","highlights":[{"text":"total volume of coolant delivered during the first 10 minutes is 325 L","location":"specification"}]},{"type":"text","order":2,"title":"Solve using GDC","content":"We now have a system of two linear simultaneous equations:\n\n$$\\begin{cases} a + 3d = 40 \\\\ 2a + 9d = 65 \\end{cases}$$\n\nWe can use a graphing calculator (GDC) to solve this system directly instead of doing manual algebraic manipulation.","calculator":[{"gdc":{"expression":"x = 55, y = -5","instructions":"1. Go to the **Equation/Solver** menu (often under **MENU -> Equation** or **EQUA**).\n2. Choose **F1: Simultaneous**.\n3. Choose **F1: 2** unknowns.\n4. Input the coefficient matrix:\n - Row 1: `1`, `3`, `40`\n - Row 2: `2`, `9`, `65`\n5. Press **F6: SOLVE** to find the values."},"ti84":{"expression":"x1 = 55, x2 = -5","instructions":"1. Press **[APPS]** and select **PlySmlt2** (Polynomial Root Finder and Simultaneous Equation Solver).\n2. Select **2: Simul Eqn Solver**.\n3. Set **Number of Equations = 2** and **Number of Unknowns = 2**, then press **[NEXT]** (graph key).\n4. Enter the coefficients into the system matrix:\n - Row 1: `1`, `3`, `40`\n - Row 2: `2`, `9`, `65`\n5. Press **[SOLVE]** (graph key)."},"desmos":{"expression":"(a, d) = (55, -5)","instructions":"In the graphing calculator or matrix calculator, define the system of equations as equations:\n1. Type `a + 3d = 40`\n2. Type `2a + 9d = 65`\nDesmos will find the intersection of the two lines."},"description":"Use the built-in system solver or matrix solver to find the values of $a$ (as $X$) and $d$ (as $Y$)."}]},{"type":"text","order":3,"title":"State final answers","content":"From the calculator solver output, we get:\n\n$$a = 55$$\n$$d = -5$$\n\nThus, the initial flow rate is $55\\text{ L/min}$ and the constant change in flow rate is $-5\\text{ L/min}^2$ (decreasing by $5\\text{ L/min}$ every minute)."}]},{"id":"48384b40-dfd0-47ea-bf44-76ae83e6bfdd:algebraic","label":"(b) Algebraic Approach","steps":[{"type":"text","order":0,"title":"Define primary system sum","content":"To find the total volume delivered by the primary system over $n$ minutes, $S_n$, we use the sum formula for an arithmetic series from the formula booklet:\n\n$$S_n = \\frac{n}{2}[2a + (n-1)d]$$\n\nFor the primary system, the initial flow rate is $a = 55\\text{ L/min}$ and it decreases by $5\\text{ L/min}$ each minute, so $d = -5$.\n\nSubstituting these values into the formula gives:\n\n$$S_n = \\frac{n}{2}[2(55) + (n-1)(-5)]$$\n\n$$S_n = \\frac{n}{2}[110 - 5n + 5]$$\n\n$$S_n = \\frac{n}{2}(115 - 5n)$$\n\nThis earns the first mark **(M1)**.","highlights":[{"text":"total volume delivered by the primary system, S_n","location":"specification"}]},{"type":"text","order":1,"title":"Define auxiliary system sum","content":"For the secondary auxiliary system, we are given that the initial flow rate is $a = 6\\text{ L/min}$ and the common difference is $d = 2\\text{ L/min}$.\n\nUsing the same arithmetic series sum formula, we find the total volume delivered, $T_n$:\n\n$$T_n = \\frac{n}{2}[2(6) + (n-1)(2)]$$\n\n$$T_n = \\frac{n}{2}[12 + 2n - 2]$$\n\n$$T_n = \\frac{n}{2}(10 + 2n)$$\n\nThis earns the second mark **(M1)**.","highlights":[{"text":"initial flow rate of 6 L/min","location":"specification"},{"text":"increases by 2 L/min each minute","location":"specification"},{"text":"total volume delivered by this auxiliary system over n minutes is T_n","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\htmlData{anchor=s-expr}{\\frac{n}{2}(115 - 5n)} = \\htmlData{anchor=t-expr}{\\frac{n}{2}(10 + 2n)}$"},{"text":"Since we are given that $n > 0$, we can divide both sides of the equation by $\\frac{n}{2}$ to simplify it directly:"},{"latex":"$$\\htmlData{anchor=simp-expr}{115 - 5n = 10 + 2n}$"}],"highlights":[{"color":"sky","style":"box","anchor":"s-expr"},{"color":"amber","style":"box","anchor":"t-expr"}],"connections":[{"to":"simp-expr","from":"s-expr","color":"sky","label":"divide by n/2","style":"solid"}]},"order":2,"title":"Equate and simplify","description":"We set the two total volume expressions equal to each other, $S_n = T_n$, and simplify the resulting algebraic equation by eliminating the common factor of $n$."},{"type":"text","order":3,"title":"Solve for n","content":"Now, we solve the simplified linear equation for $n$ by isolating the variable terms:\n\n$$115 - 5n = 10 + 2n$$\n\n$$115 - 10 = 2n + 5n$$\n\n$$105 = 7n$$\n\n$$n = 15$$\n\nSolving this equation yields $n = 15$ minutes, earning the final marks **(M1 A1)**.","highlights":[{"text":"total volume delivered by the primary system, S_n, is equal to T_n, where n > 0","location":"specification"}]}]},{"id":"48384b40-dfd0-47ea-bf44-76ae83e6bfdd:graphical","label":"(b) Graphical/Calculator Approach","steps":[{"type":"text","order":0,"title":"Define primary system sum","content":"To find when the total volumes are equal, we first need the formula for the sum of an arithmetic series from our formula booklet:\n\n$$S_n = \\frac{n}{2}[2a + (n-1)d]$$\n\nFor the primary system, the initial flow rate is $a = 55\\text{ L/min}$ and it decreases by $d = -5\\text{ L/min}$ each minute. Substituting these into the formula gives:\n\n$$S_n = \\frac{n}{2}[2(55) + (n-1)(-5)] = \\frac{n}{2}(115 - 5n)$$\n\nWe can define this function as $y_1 = \\frac{x}{2}(115 - 5x)$ on our calculator.","highlights":[{"text":"total volume delivered by the primary system, S_n","location":"specification"}]},{"type":"text","order":1,"title":"Define auxiliary system sum","content":"For the auxiliary system, the initial flow rate is $a = 6\\text{ L/min}$ and it increases by $d = 2\\text{ L/min}$ each minute.\n\nUsing the same arithmetic series sum formula, we get:\n\n$$T_n = \\frac{n}{2}[2(6) + (n-1)(2)] = \\frac{n}{2}(10 + 2n)$$\n\nWe can define this second function as $y_2 = \\frac{x}{2}(10 + 2x)$ on our calculator.","highlights":[{"text":"initial flow rate of 6 L/min, which increases by 2 L/min each minute. The total volume delivered by this auxiliary system over n minutes is T_n","location":"specification"}]},{"type":"text","order":2,"title":"Graph and find intersection","content":"We want to find the value of $n > 0$ where $S_n = T_n$.\n\nInstead of solving the quadratic equation algebraically, we can plot both $y_1 = \\frac{x}{2}(115 - 5x)$ and $y_2 = \\frac{x}{2}(10 + 2x)$ on a graphing calculator and find their intersection point in the first quadrant ($x > 0$).","highlights":[{"text":"S_n, is equal to T_n, where n > 0","location":"specification"}]},{"type":"text","order":3,"title":"State the final solution","content":"The calculator displays the intersection point of the two curves as:\n\n$$(15, 300)$$\n\nSince $x$ represents the time in minutes $n$, we have:\n\n$$n = 15$$\n\n*(Note: There is another intersection at $n = 0$, but the question specifies $n > 0$.)*"}]}],"generatedAt":"2026-06-09T05:45:54.809Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":null,"walkthroughVideo":null},{"id":"5b6aade5-dc87-4be0-a6b6-53406cdbc334","revisionId":"fa949f03-2c0a-4c7d-809f-af14202c4cb8","questionNumber":null,"questionType":"LA","specification":"The sum of the first five terms of an arithmetic sequence is 40, and the ninth term is 20.","difficultyBand":"medium","tier":null,"totalMarks":4,"sourceTitle":null,"calculatorUse":null,"isCalculation":true,"parts":[{"id":"f5be3b8e-828f-4a62-829f-a7d5d7be315a","label":null,"content":"Find the first term and the common difference of the sequence.","metadata":null,"markscheme":"- Uses $S_5=\\frac{5}{2}(2a+4d)=40$ to obtain $a+2d=8$ M1\n- Uses the ninth term to obtain $a+8d=20$ M1\n- Subtracts the equations to obtain $6d=12$, so $d=2$ A1\n- $a=4$ A1","marks":4,"order":0,"markingRubric":null,"tableImages":[],"explanation":null,"options":[]}],"options":[],"explanation":null,"firstAttemptFullMarks":null,"walkthroughVideo":null},{"id":"5ea5f4ec-6760-4ab6-b3b3-671a5bf4f498","revisionId":"e4cbc168-5ac1-49b3-99bd-546f8c737dff","questionNumber":null,"questionType":"LA","specification":"The concentration of a specific catalyst, \$C_n\$ mmol/L, in a chemical reactor is recorded every hour. The process is modeled by the recurrence relation\n\\[ C_{n+1} = 15 - C_n, \\quad n \\ge 1 \\]\nwhere the initial concentration is \$C_1 = 4.5\$.","difficultyBand":"medium","tier":null,"totalMarks":6,"sourceTitle":null,"calculatorUse":null,"isCalculation":true,"parts":[{"id":"2de3f0f4-b555-4b03-bb7e-64575b350c33","label":"a","content":"Find the value of\n(i) \$C_2\$\n(ii) \$C_{85}\$","metadata":null,"markscheme":"- Calculates $C_2 = 15 - 4.5 = 10.5$ B1\n- Identifies that the sequence is periodic with period 2 ($4.5, 10.5, 4.5, 10.5, \\dots$) M1\n- Deduces that odd-indexed terms are $4.5$, so $C_{85} = 4.5$ A1","marks":3,"order":0,"markingRubric":null,"tableImages":[],"explanation":{"methods":[{"id":"pattern_parity","label":"Pattern Recognition and Parity Analysis","steps":[{"type":"text","order":0,"title":"Calculate the second term","content":"To find $C_2$, we substitute $n = 1$ into our recurrence relation:\n\n$$C_{n+1} = 15 - C_n$$\n\nGiven the initial term is $C_1 = 4.5$:\n\n$$C_2 = 15 - C_1$$\n$$C_2 = 15 - 4.5 = 10.5$$"},{"type":"text","order":1,"title":"Find the next few terms","content":"Let's calculate $C_3$ and $C_4$ to look for a pattern in the sequence.\n\nFor $C_3$, we use $C_2 = 10.5$:\n$$C_3 = 15 - C_2 = 15 - 10.5 = 4.5$$\n\nFor $C_4$, we use $C_3 = 4.5$:\n$$C_4 = 15 - C_3 = 15 - 4.5 = 10.5$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$C_{\\htmlData{anchor=odd-1}{\\color{@sky}{1}}} = \\htmlData{anchor=val-odd-1}{\\color{@sky}{4.5}}$"},{"latex":"$$C_{\\htmlData{anchor=even-1}{\\color{@amber}{2}}} = \\htmlData{anchor=val-even-1}{\\color{@amber}{10.5}}$"},{"latex":"$$C_{\\htmlData{anchor=odd-2}{\\color{@sky}{3}}} = \\htmlData{anchor=val-odd-2}{\\color{@sky}{4.5}}$"},{"latex":"$$C_{\\htmlData{anchor=even-2}{\\color{@amber}{4}}} = \\htmlData{anchor=val-even-2}{\\color{@amber}{10.5}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"val-odd-1"},{"color":"amber","style":"text-color","anchor":"val-even-1"},{"color":"sky","style":"text-color","anchor":"val-odd-2"},{"color":"amber","style":"text-color","anchor":"val-even-2"}],"connections":[{"to":"odd-2","from":"odd-1","color":"sky","label":"odd indices","style":"solid"},{"to":"even-2","from":"even-1","color":"amber","label":"even indices","style":"solid"}]},"order":2,"title":"Analyze the periodic pattern","description":"We can see that the sequence is periodic with a period of 2. It oscillates between two values depending on the parity (evenness or oddness) of the index $n$."},{"type":"text","order":3,"title":"Determine C_85 using parity","content":"Since the sequence alternates between $4.5$ and $10.5$:\n\n* If $n$ is **odd**, $C_n = 4.5$\n* If $n$ is **even**, $C_n = 10.5$\n\nWe need to find $C_{85}$. Since $85$ is an **odd** integer, the value of the term will match all other odd-indexed terms:\n\n$$C_{85} = 4.5$$"},{"type":"text","order":4,"title":"Calculate the transformed terms","content":"For Part 2, we need to evaluate the sum of the transformed terms:\n\n$$\\sum_{n=1}^{100} (4C_n - 5)$$\n\nLet's define a new sequence $b_n = 4C_n - 5$ and find its first two terms:\n\n* For odd terms ($C_n = 4.5$):\n$$b_1 = 4(4.5) - 5 = 18 - 5 = 13$$\n\n* For even terms ($C_n = 10.5$):\n$$b_2 = 4(10.5) - 5 = 42 - 5 = 37$$"},{"type":"text","order":5,"title":"Sum the periodic pairs","content":"Since the sequence $b_n$ is also periodic with a period of 2, it consists of alternating values of $13$ and $37$:\n\n$$b_n = \\{13, 37, 13, 37, \\dots\\}$$\n\nFor a sum up to $N = 100$ terms, we can group the terms into $50$ identical pairs of $(b_1 + b_2)$:\n\n$$\\sum_{n=1}^{100} b_n = 50 \\times (b_1 + b_2)$$\n$$\\sum_{n=1}^{100} b_n = 50 \\times (13 + 37)$$\n$$\\sum_{n=1}^{100} b_n = 50 \\times 50 = 2500$$"}]},{"id":"general_formula","label":"Explicit General Formula","steps":[{"type":"text","order":0,"title":"Understand the recurrence relation","content":"We are given the recurrence relation for the concentration of a catalyst:\n\n$$C_{n+1} = 15 - C_n, \\quad n \\ge 1$$\n\nwith the initial condition $C_1 = 4.5$. \n\nOur goal is to find an explicit general formula of the form:\n\n$$C_n = a + b(-1)^n$$\n\nwhere $a$ represents the steady-state (or fixed point) value of the sequence, and the term $b(-1)^n$ models the oscillation around this steady state."},{"type":"hyper_latex","block":{"nodes":[{"text":"The steady state $L$ is the value where the sequence remains constant from one step to the next, meaning $C_{n+1} = C_n = L$."},{"latex":"$$$\\htmlData{anchor=eq1}{L} = 15 - \\htmlData{anchor=eq2}{L}$$"},{"latex":"$$$2L = 15 \\implies \\htmlData{anchor=sol}{L = 7.5}$$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"eq1"},{"color":"sky","style":"text-color","anchor":"sol"}],"connections":[{"to":"sol","from":"eq1","color":"sky","label":"solve","style":"solid"}]},"order":1,"title":"Find the steady state","description":"We find the fixed point $L$ of the system by setting $C_{n+1} = C_n = L$."},{"type":"text","order":2,"title":"Find the general formula","content":"Since the steady state is $a = 7.5$, our explicit formula has the form:\n\n$$C_n = 7.5 + b(-1)^n$$\n\nWe can determine the constant $b$ using the initial term $C_1 = 4.5$:\n\n$$\\begin{aligned} C_1 &= 7.5 + b(-1)^1 \\\\ 4.5 &= 7.5 - b \\\\ b &= 7.5 - 4.5 = 3 \\end{aligned}$$\n\nSubstituting $b = 3$ back into our equation gives the complete explicit formula:\n\n$$C_n = 7.5 + 3(-1)^n$$"},{"type":"text","order":3,"title":"Calculate C2","content":"Now we can substitute $n = 2$ directly into our explicit formula to find $C_2$:\n\n$$\\begin{aligned} C_2 &= 7.5 + 3(-1)^2 \\\\ &= 7.5 + 3(1) \\\\ &= 10.5 \\end{aligned}$$\n\nThis gives us our answer for part (i)."},{"type":"text","order":4,"title":"Calculate C85","content":"To find $C_{85}$ for part (ii), we substitute $n = 85$ into our formula. Since $85$ is an odd number, $(-1)^{85} = -1$:\n\n$$\\begin{aligned} C_{85} &= 7.5 + 3(-1)^{85} \\\\ &= 7.5 + 3(-1) \\\\ &= 4.5 \\end{aligned}$$\n\nThis confirms the periodic behavior of the sequence, where odd terms are always $4.5$ and even terms are always $10.5$."},{"type":"text","order":5,"title":"Find the sum of the series","content":"Using our general formula, we can rewrite the term inside the summation for Part 2:\n\n$$4C_n - 5 = 4(7.5 + 3(-1)^n) - 5$$\n$$4C_n - 5 = 30 + 12(-1)^n - 5 = 25 + 12(-1)^n$$\n\nNow, we sum this expression from $n = 1$ to $100$:\n\n$$\\sum_{n=1}^{100} (25 + 12(-1)^n) = \\sum_{n=1}^{100} 25 + 12 \\sum_{n=1}^{100} (-1)^n$$\n\nSince $(-1)^n$ alternates between $-1$ and $1$ over $100$ terms, the sum of $(-1)^n$ cancels out completely to $0$. Thus:\n\n$$\\sum_{n=1}^{100} (4C_n - 5) = (100 \\times 25) + 12(0) = 2500$$"}]},{"id":"calculator_table","label":"Using Graphic Calculator Sequence Tool","steps":[{"type":"text","order":0,"title":"Set up the calculator","content":"To solve this problem using a graphing calculator, we can use the **Sequence** mode. This allows us to define the recurrence relation and generate a table of values automatically.\n\nThe recurrence relation is defined as:\n\n$$C_{n+1} = 15 - C_n \\quad \\text{with} \\quad C_1 = 4.5$$"},{"type":"text","order":1,"title":"Find the value of C_2","content":"Now we generate the sequence table to find the second term, $C_2$.\n\nUsing the formula manually or looking at the table at $n=2$:\n\n$$C_2 = 15 - C_1 = 15 - 4.5 = 10.5$$\n\nThis gives us our first mark.\n\n* **Markscheme check:** Calculating $C_2 = 10.5$ earns **1 mark (B1)**."},{"type":"hyper_latex","block":{"nodes":[{"text":"Scroll down the table to observe the pattern of the terms in the sequence:"},{"latex":"$$$\\begin{aligned} C_1 &= \\htmlData{anchor=c1}{\\color{@sky}{4.5}} \\\\ C_2 &= \\htmlData{anchor=c2}{\\color{@amber}{10.5}} \\\\ C_3 &= 15 - 10.5 = \\htmlData{anchor=c3}{\\color{@sky}{4.5}} \\\\ C_4 &= 15 - 4.5 = \\htmlData{anchor=c4}{\\color{@amber}{10.5}} \\end{aligned}$$"},{"text":"This confirms the sequence is periodic with a period of 2, alternating between 4.5 and 10.5."}],"highlights":[{"color":"sky","style":"text-color","anchor":"c1"},{"color":"amber","style":"text-color","anchor":"c2"}],"connections":[{"to":"c3","from":"c1","color":"sky","label":"odd terms","style":"dashed"},{"to":"c4","from":"c2","color":"amber","label":"even terms","style":"dashed"}]},"order":2,"title":"Analyze the periodic sequence","description":"By observing the repeating sequence of terms on the table screen, we can deduce a clear rule for any term $C_n$."},{"type":"text","order":3,"title":"Deduce the value of C_85","content":"Since the sequence is periodic with period 2:\n- All odd-indexed terms ($n = 1, 3, 5, \\dots$) have a value of $4.5$.\n- All even-indexed terms ($n = 2, 4, 6, \\dots$) have a value of $10.5$.\n\nSince $85$ is an odd number:\n\n$$C_{85} = 4.5$$\n\n* **Markscheme check:** Identifying that the sequence is periodic with period 2 earns **1 mark (M1)**, and deducing that $C_{85} = 4.5$ because 85 is odd earns the final **1 mark (A1)**."},{"type":"text","order":4,"title":"Use calculator to find the sum","content":"To find the sum $\\sum_{n=1}^{100} (4C_n - 5)$ on a graphing calculator, we can use the list summation function.\n\nFirst, compute the first two terms of the new sequence $b_n = 4C_n - 5$:\n* $b_1 = 4(4.5) - 5 = 13$\n* $b_2 = 4(10.5) - 5 = 37$\n\nSince we are summing 100 terms of this periodic sequence, we can enter the sum command directly on the home screen:\n\n$$\\text{Sum} = 50 \\times (13 + 37) = 2500$$","calculator":[{"gdc":{"expression":"\\sum_{x=1}^{100}(4a_x-5)","instructions":"1. In the Main menu, enter the Equation or Run-Matrix option.\n2. Navigate to the summation tool: **MATH** -> **$\\Sigma$**.\n3. Enter the variables to compute $\\sum_{x=1}^{100} (4a_x - 5)$ where $a_x$ is your recursive sequence.\n4. Press **EXE** to obtain `2500`."},"ti84":{"expression":"sum(seq(4*u(I)-5,I,1,100))","instructions":"1. Go to the home screen.\n2. Press **2nd** then **STAT** (LIST), arrow over to **MATH**, and select **5:sum(**.\n3. Press **2nd** then **STAT** (LIST), arrow over to **OPS**, and select **5:seq(**.\n4. Enter the expression: `sum(seq(4*u(N)-5, N, 1, 100))` where `u` is the defined sequence.\n5. Press **ENTER** to get `2500`."},"desmos":{"expression":"sum(50*(13+37))","instructions":"Define a sequence using list comprehension: `C = [4.5, 10.5, 4.5, 10.5, ...]` or simply evaluate: `sum(50 * (13 + 37))` to get `2500`."},"description":"Use the sum and sequence commands on your calculator to compute the series sum."}]}]}],"generatedAt":"2026-06-09T05:52:50.581Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]},{"id":"b2190b6c-cbf2-4121-8594-c1e0f51e2dd3","label":"b","content":"Determine the value of \$\\sum_{n=1}^{100} (4C_n - 5)\$.","metadata":null,"markscheme":"- Calculates the first two terms of the new sequence $b_n = 4C_n - 5$: $b_1 = 4(4.5) - 5 = 13$ and $b_2 = 4(10.5) - 5 = 37$ M1\n- Recognizes that the sum contains 50 pairs of $(b_1 + b_2)$ M1\n- Evaluates $50 \\times (13 + 37) = 50 \\times 50 = 2500$ A1","marks":3,"order":1,"markingRubric":null,"tableImages":[],"explanation":{"methods":[{"id":"pairing_terms","label":"Pairing Alternating Terms","steps":[{"type":"text","order":0,"title":"Define the terms","content":"Let's define the terms of the sum as a new sequence:\n\n$$b_n = 4C_n - 5$$\n\nTo find the overall sum $\\sum_{n=1}^{100} b_n$, we need to calculate the first few terms of $C_n$ and then find the corresponding values of $b_n$."},{"type":"hyper_latex","block":{"nodes":[{"text":"First, we find the values of the original sequence $C_n$:"},{"latex":"$$C_1 = \\htmlData{anchor=c1}{\\color{@sky}{4.5}}$"},{"latex":"$$C_2 = 15 - C_1 = 15 - \\htmlData{anchor=c1-sub}{\\color{@sky}{4.5}} = \\htmlData{anchor=c2}{\\color{@amber}{10.5}}$"},{"text":"Now, we substitute these to find the first two terms of our new sequence $b_n$:"},{"latex":"$$b_1 = 4(\\htmlData{anchor=c1-sub2}{\\color{@sky}{4.5}}) - 5 = \\htmlData{anchor=b1}{\\color{@purple}{13}}$"},{"latex":"$$b_2 = 4(\\htmlData{anchor=c2-sub}{\\color{@amber}{10.5}}) - 5 = \\htmlData{anchor=b2}{\\color{@green}{37}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"c1"},{"color":"amber","style":"text-color","anchor":"c2"},{"color":"purple","style":"text-color","anchor":"b1"},{"color":"green","style":"text-color","anchor":"b2"}],"connections":[{"to":"c1-sub","from":"c1","color":"sky","label":null,"style":"solid"},{"to":"c1-sub2","from":"c1","color":"sky","label":null,"style":"solid"},{"to":"c2-sub","from":"c2","color":"amber","label":null,"style":"solid"}]},"order":1,"title":"Calculate first two terms","description":"We calculate the first two values of $C_n$ using the recurrence relation, then plug them into $b_n = 4C_n - 5$."},{"type":"text","order":2,"title":"Observe the alternating pattern","content":"Using the recurrence relation $C_{n+1} = 15 - C_n$, we can see what happens for subsequent terms:\n\n$$C_3 = 15 - C_2 = 15 - 10.5 = 4.5$$\n\n$$C_4 = 15 - C_3 = 15 - 4.5 = 10.5$$\n\nBecause the values of $C_n$ alternate between $4.5$ and $10.5$, the terms of our sequence $b_n$ will also alternate:\n\n$$b_1 = 13, \\quad b_2 = 37, \\quad b_3 = 13, \\quad b_4 = 37, \\quad \\dots$$"},{"type":"text","order":3,"title":"Pair the adjacent terms","content":"Since the sequence alternates between $13$ and $37$, we can group adjacent terms into pairs:\n\n$$\\begin{aligned} \\sum_{n=1}^{100} b_n &= (b_1 + b_2) + (b_3 + b_4) + \\dots + (b_{99} + b_{100}) \\\\ &= (13 + 37) + (13 + 37) + \\dots + (13 + 37) \\end{aligned}$$\n\nSince we are summing $100$ terms in total, we will have exactly $50$ pairs of adjacent terms."},{"type":"text","order":4,"title":"Calculate the final sum","content":"Now we compute the value of a single pair and multiply by the total number of pairs:\n\n$$\\text{Value of one pair} = 13 + 37 = 50$$\n\n$$\\text{Total Sum} = 50 \\times 50 = 2500$$\n\nTherefore,\n\n$$\\sum_{n=1}^{100} (4C_n - 5) = 2500$$"}]},{"id":"separating_sum","label":"Separating the Sum","steps":[{"type":"text","order":0,"title":"Analyze the sequence","content":"First, let's understand how the concentration $C_n$ behaves using the recurrence relation given:\n\n$$C_{n+1} = 15 - C_n, \\quad C_1 = 4.5$$\n\nLet's calculate the first few terms:\n- For $n = 1$: $C_1 = 4.5$\n- For $n = 2$: $C_2 = 15 - 4.5 = 10.5$\n- For $n = 3$: $C_3 = 15 - 10.5 = 4.5$\n- For $n = 4$: $C_4 = 15 - 4.5 = 10.5$\n\nThis shows that the sequence alternates between $4.5$ (for odd $n$) and $10.5$ (for even $n$)."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$$\\sum_{n=1}^{100} (\\htmlData{anchor=four-c}{\\color{@sky}{4C_n}} - \\htmlData{anchor=five}{\\color{@amber}{5}})$$"},{"latex":"$$$= 4\\htmlData{anchor=sum-c}{\\color{@sky}{\\sum_{n=1}^{100} C_n}} - \\htmlData{anchor=sum-five}{\\color{@amber}{\\sum_{n=1}^{100} 5}}$$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"four-c"},{"color":"sky","style":"text-color","anchor":"sum-c"},{"color":"amber","style":"text-color","anchor":"five"},{"color":"amber","style":"text-color","anchor":"sum-five"}],"connections":[{"to":"sum-c","from":"four-c","color":"sky","label":"distribute constant","style":"solid"},{"to":"sum-five","from":"five","color":"amber","label":"distribute sum","style":"solid"}]},"order":1,"title":"Separate the sum","description":"We can distribute the summation operator across the terms, splitting the expression into two separate sums."},{"type":"text","order":2,"title":"Calculate the constant sum","content":"The second part of our separated expression is a sum of the constant $5$, added $100$ times:\n\n$$\\sum_{n=1}^{100} 5 = 100 \\times 5 = 500$$"},{"type":"text","order":3,"title":"Calculate the sequence sum","content":"Now we calculate the sum of the $C_n$ terms, $\\sum_{n=1}^{100} C_n$.\n\nSince the terms alternate between $C_1 = 4.5$ and $C_2 = 10.5$ for $100$ terms, we have exactly $50$ pairs of these two alternating values:\n\n$$\\begin{aligned} \\sum_{n=1}^{100} C_n &= 50 \\times (C_1 + C_2) \\\\ &= 50 \\times (4.5 + 10.5) \\\\ &= 50 \\times 15 \\\\ &= 750 \\end{aligned}$$"},{"type":"text","order":4,"title":"Evaluate the final value","content":"Now we substitute both components back into our separated summation formula:\n\n$$\\begin{aligned} \\sum_{n=1}^{100} (4C_n - 5) &= 4\\sum_{n=1}^{100} C_n - \\sum_{n=1}^{100} 5 \\\\ &= 4(750) - 500 \\\\ &= 3000 - 500 \\\\ &= 2500 \\end{aligned}$$\n\nThis gives us our final answer of **$2500$**."}]}],"generatedAt":"2026-06-09T05:54:10.058Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"2de3f0f4-b555-4b03-bb7e-64575b350c33:pattern_parity","label":"(a) Pattern Recognition and Parity Analysis","steps":[{"type":"text","order":0,"title":"Calculate the second term","content":"To find $C_2$, we substitute $n = 1$ into our recurrence relation:\n\n$$C_{n+1} = 15 - C_n$$\n\nGiven the initial term is $C_1 = 4.5$:\n\n$$C_2 = 15 - C_1$$\n$$C_2 = 15 - 4.5 = 10.5$$"},{"type":"text","order":1,"title":"Find the next few terms","content":"Let's calculate $C_3$ and $C_4$ to look for a pattern in the sequence.\n\nFor $C_3$, we use $C_2 = 10.5$:\n$$C_3 = 15 - C_2 = 15 - 10.5 = 4.5$$\n\nFor $C_4$, we use $C_3 = 4.5$:\n$$C_4 = 15 - C_3 = 15 - 4.5 = 10.5$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$C_{\\htmlData{anchor=odd-1}{\\color{@sky}{1}}} = \\htmlData{anchor=val-odd-1}{\\color{@sky}{4.5}}$"},{"latex":"$$C_{\\htmlData{anchor=even-1}{\\color{@amber}{2}}} = \\htmlData{anchor=val-even-1}{\\color{@amber}{10.5}}$"},{"latex":"$$C_{\\htmlData{anchor=odd-2}{\\color{@sky}{3}}} = \\htmlData{anchor=val-odd-2}{\\color{@sky}{4.5}}$"},{"latex":"$$C_{\\htmlData{anchor=even-2}{\\color{@amber}{4}}} = \\htmlData{anchor=val-even-2}{\\color{@amber}{10.5}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"val-odd-1"},{"color":"amber","style":"text-color","anchor":"val-even-1"},{"color":"sky","style":"text-color","anchor":"val-odd-2"},{"color":"amber","style":"text-color","anchor":"val-even-2"}],"connections":[{"to":"odd-2","from":"odd-1","color":"sky","label":"odd indices","style":"solid"},{"to":"even-2","from":"even-1","color":"amber","label":"even indices","style":"solid"}]},"order":2,"title":"Analyze the periodic pattern","description":"We can see that the sequence is periodic with a period of 2. It oscillates between two values depending on the parity (evenness or oddness) of the index $n$."},{"type":"text","order":3,"title":"Determine C_85 using parity","content":"Since the sequence alternates between $4.5$ and $10.5$:\n\n* If $n$ is **odd**, $C_n = 4.5$\n* If $n$ is **even**, $C_n = 10.5$\n\nWe need to find $C_{85}$. Since $85$ is an **odd** integer, the value of the term will match all other odd-indexed terms:\n\n$$C_{85} = 4.5$$"},{"type":"text","order":4,"title":"Calculate the transformed terms","content":"For Part 2, we need to evaluate the sum of the transformed terms:\n\n$$\\sum_{n=1}^{100} (4C_n - 5)$$\n\nLet's define a new sequence $b_n = 4C_n - 5$ and find its first two terms:\n\n* For odd terms ($C_n = 4.5$):\n$$b_1 = 4(4.5) - 5 = 18 - 5 = 13$$\n\n* For even terms ($C_n = 10.5$):\n$$b_2 = 4(10.5) - 5 = 42 - 5 = 37$$"},{"type":"text","order":5,"title":"Sum the periodic pairs","content":"Since the sequence $b_n$ is also periodic with a period of 2, it consists of alternating values of $13$ and $37$:\n\n$$b_n = \\{13, 37, 13, 37, \\dots\\}$$\n\nFor a sum up to $N = 100$ terms, we can group the terms into $50$ identical pairs of $(b_1 + b_2)$:\n\n$$\\sum_{n=1}^{100} b_n = 50 \\times (b_1 + b_2)$$\n$$\\sum_{n=1}^{100} b_n = 50 \\times (13 + 37)$$\n$$\\sum_{n=1}^{100} b_n = 50 \\times 50 = 2500$$"}]},{"id":"2de3f0f4-b555-4b03-bb7e-64575b350c33:general_formula","label":"(a) Explicit General Formula","steps":[{"type":"text","order":0,"title":"Understand the recurrence relation","content":"We are given the recurrence relation for the concentration of a catalyst:\n\n$$C_{n+1} = 15 - C_n, \\quad n \\ge 1$$\n\nwith the initial condition $C_1 = 4.5$. \n\nOur goal is to find an explicit general formula of the form:\n\n$$C_n = a + b(-1)^n$$\n\nwhere $a$ represents the steady-state (or fixed point) value of the sequence, and the term $b(-1)^n$ models the oscillation around this steady state."},{"type":"hyper_latex","block":{"nodes":[{"text":"The steady state $L$ is the value where the sequence remains constant from one step to the next, meaning $C_{n+1} = C_n = L$."},{"latex":"$$$\\htmlData{anchor=eq1}{L} = 15 - \\htmlData{anchor=eq2}{L}$$"},{"latex":"$$$2L = 15 \\implies \\htmlData{anchor=sol}{L = 7.5}$$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"eq1"},{"color":"sky","style":"text-color","anchor":"sol"}],"connections":[{"to":"sol","from":"eq1","color":"sky","label":"solve","style":"solid"}]},"order":1,"title":"Find the steady state","description":"We find the fixed point $L$ of the system by setting $C_{n+1} = C_n = L$."},{"type":"text","order":2,"title":"Find the general formula","content":"Since the steady state is $a = 7.5$, our explicit formula has the form:\n\n$$C_n = 7.5 + b(-1)^n$$\n\nWe can determine the constant $b$ using the initial term $C_1 = 4.5$:\n\n$$\\begin{aligned} C_1 &= 7.5 + b(-1)^1 \\\\ 4.5 &= 7.5 - b \\\\ b &= 7.5 - 4.5 = 3 \\end{aligned}$$\n\nSubstituting $b = 3$ back into our equation gives the complete explicit formula:\n\n$$C_n = 7.5 + 3(-1)^n$$"},{"type":"text","order":3,"title":"Calculate C2","content":"Now we can substitute $n = 2$ directly into our explicit formula to find $C_2$:\n\n$$\\begin{aligned} C_2 &= 7.5 + 3(-1)^2 \\\\ &= 7.5 + 3(1) \\\\ &= 10.5 \\end{aligned}$$\n\nThis gives us our answer for part (i)."},{"type":"text","order":4,"title":"Calculate C85","content":"To find $C_{85}$ for part (ii), we substitute $n = 85$ into our formula. Since $85$ is an odd number, $(-1)^{85} = -1$:\n\n$$\\begin{aligned} C_{85} &= 7.5 + 3(-1)^{85} \\\\ &= 7.5 + 3(-1) \\\\ &= 4.5 \\end{aligned}$$\n\nThis confirms the periodic behavior of the sequence, where odd terms are always $4.5$ and even terms are always $10.5$."},{"type":"text","order":5,"title":"Find the sum of the series","content":"Using our general formula, we can rewrite the term inside the summation for Part 2:\n\n$$4C_n - 5 = 4(7.5 + 3(-1)^n) - 5$$\n$$4C_n - 5 = 30 + 12(-1)^n - 5 = 25 + 12(-1)^n$$\n\nNow, we sum this expression from $n = 1$ to $100$:\n\n$$\\sum_{n=1}^{100} (25 + 12(-1)^n) = \\sum_{n=1}^{100} 25 + 12 \\sum_{n=1}^{100} (-1)^n$$\n\nSince $(-1)^n$ alternates between $-1$ and $1$ over $100$ terms, the sum of $(-1)^n$ cancels out completely to $0$. Thus:\n\n$$\\sum_{n=1}^{100} (4C_n - 5) = (100 \\times 25) + 12(0) = 2500$$"}]},{"id":"2de3f0f4-b555-4b03-bb7e-64575b350c33:calculator_table","label":"(a) Using Graphic Calculator Sequence Tool","steps":[{"type":"text","order":0,"title":"Set up the calculator","content":"To solve this problem using a graphing calculator, we can use the **Sequence** mode. This allows us to define the recurrence relation and generate a table of values automatically.\n\nThe recurrence relation is defined as:\n\n$$C_{n+1} = 15 - C_n \\quad \\text{with} \\quad C_1 = 4.5$$"},{"type":"text","order":1,"title":"Find the value of C_2","content":"Now we generate the sequence table to find the second term, $C_2$.\n\nUsing the formula manually or looking at the table at $n=2$:\n\n$$C_2 = 15 - C_1 = 15 - 4.5 = 10.5$$\n\nThis gives us our first mark.\n\n* **Markscheme check:** Calculating $C_2 = 10.5$ earns **1 mark (B1)**."},{"type":"hyper_latex","block":{"nodes":[{"text":"Scroll down the table to observe the pattern of the terms in the sequence:"},{"latex":"$$$\\begin{aligned} C_1 &= \\htmlData{anchor=c1}{\\color{@sky}{4.5}} \\\\ C_2 &= \\htmlData{anchor=c2}{\\color{@amber}{10.5}} \\\\ C_3 &= 15 - 10.5 = \\htmlData{anchor=c3}{\\color{@sky}{4.5}} \\\\ C_4 &= 15 - 4.5 = \\htmlData{anchor=c4}{\\color{@amber}{10.5}} \\end{aligned}$$"},{"text":"This confirms the sequence is periodic with a period of 2, alternating between 4.5 and 10.5."}],"highlights":[{"color":"sky","style":"text-color","anchor":"c1"},{"color":"amber","style":"text-color","anchor":"c2"}],"connections":[{"to":"c3","from":"c1","color":"sky","label":"odd terms","style":"dashed"},{"to":"c4","from":"c2","color":"amber","label":"even terms","style":"dashed"}]},"order":2,"title":"Analyze the periodic sequence","description":"By observing the repeating sequence of terms on the table screen, we can deduce a clear rule for any term $C_n$."},{"type":"text","order":3,"title":"Deduce the value of C_85","content":"Since the sequence is periodic with period 2:\n- All odd-indexed terms ($n = 1, 3, 5, \\dots$) have a value of $4.5$.\n- All even-indexed terms ($n = 2, 4, 6, \\dots$) have a value of $10.5$.\n\nSince $85$ is an odd number:\n\n$$C_{85} = 4.5$$\n\n* **Markscheme check:** Identifying that the sequence is periodic with period 2 earns **1 mark (M1)**, and deducing that $C_{85} = 4.5$ because 85 is odd earns the final **1 mark (A1)**."},{"type":"text","order":4,"title":"Use calculator to find the sum","content":"To find the sum $\\sum_{n=1}^{100} (4C_n - 5)$ on a graphing calculator, we can use the list summation function.\n\nFirst, compute the first two terms of the new sequence $b_n = 4C_n - 5$:\n* $b_1 = 4(4.5) - 5 = 13$\n* $b_2 = 4(10.5) - 5 = 37$\n\nSince we are summing 100 terms of this periodic sequence, we can enter the sum command directly on the home screen:\n\n$$\\text{Sum} = 50 \\times (13 + 37) = 2500$$","calculator":[{"gdc":{"expression":"\\sum_{x=1}^{100}(4a_x-5)","instructions":"1. In the Main menu, enter the Equation or Run-Matrix option.\n2. Navigate to the summation tool: **MATH** -> **$\\Sigma$**.\n3. Enter the variables to compute $\\sum_{x=1}^{100} (4a_x - 5)$ where $a_x$ is your recursive sequence.\n4. Press **EXE** to obtain `2500`."},"ti84":{"expression":"sum(seq(4*u(I)-5,I,1,100))","instructions":"1. Go to the home screen.\n2. Press **2nd** then **STAT** (LIST), arrow over to **MATH**, and select **5:sum(**.\n3. Press **2nd** then **STAT** (LIST), arrow over to **OPS**, and select **5:seq(**.\n4. Enter the expression: `sum(seq(4*u(N)-5, N, 1, 100))` where `u` is the defined sequence.\n5. Press **ENTER** to get `2500`."},"desmos":{"expression":"sum(50*(13+37))","instructions":"Define a sequence using list comprehension: `C = [4.5, 10.5, 4.5, 10.5, ...]` or simply evaluate: `sum(50 * (13 + 37))` to get `2500`."},"description":"Use the sum and sequence commands on your calculator to compute the series sum."}]}]},{"id":"b2190b6c-cbf2-4121-8594-c1e0f51e2dd3:pairing_terms","label":"(b) Pairing Alternating Terms","steps":[{"type":"text","order":0,"title":"Define the terms","content":"Let's define the terms of the sum as a new sequence:\n\n$$b_n = 4C_n - 5$$\n\nTo find the overall sum $\\sum_{n=1}^{100} b_n$, we need to calculate the first few terms of $C_n$ and then find the corresponding values of $b_n$."},{"type":"hyper_latex","block":{"nodes":[{"text":"First, we find the values of the original sequence $C_n$:"},{"latex":"$$C_1 = \\htmlData{anchor=c1}{\\color{@sky}{4.5}}$"},{"latex":"$$C_2 = 15 - C_1 = 15 - \\htmlData{anchor=c1-sub}{\\color{@sky}{4.5}} = \\htmlData{anchor=c2}{\\color{@amber}{10.5}}$"},{"text":"Now, we substitute these to find the first two terms of our new sequence $b_n$:"},{"latex":"$$b_1 = 4(\\htmlData{anchor=c1-sub2}{\\color{@sky}{4.5}}) - 5 = \\htmlData{anchor=b1}{\\color{@purple}{13}}$"},{"latex":"$$b_2 = 4(\\htmlData{anchor=c2-sub}{\\color{@amber}{10.5}}) - 5 = \\htmlData{anchor=b2}{\\color{@green}{37}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"c1"},{"color":"amber","style":"text-color","anchor":"c2"},{"color":"purple","style":"text-color","anchor":"b1"},{"color":"green","style":"text-color","anchor":"b2"}],"connections":[{"to":"c1-sub","from":"c1","color":"sky","label":null,"style":"solid"},{"to":"c1-sub2","from":"c1","color":"sky","label":null,"style":"solid"},{"to":"c2-sub","from":"c2","color":"amber","label":null,"style":"solid"}]},"order":1,"title":"Calculate first two terms","description":"We calculate the first two values of $C_n$ using the recurrence relation, then plug them into $b_n = 4C_n - 5$."},{"type":"text","order":2,"title":"Observe the alternating pattern","content":"Using the recurrence relation $C_{n+1} = 15 - C_n$, we can see what happens for subsequent terms:\n\n$$C_3 = 15 - C_2 = 15 - 10.5 = 4.5$$\n\n$$C_4 = 15 - C_3 = 15 - 4.5 = 10.5$$\n\nBecause the values of $C_n$ alternate between $4.5$ and $10.5$, the terms of our sequence $b_n$ will also alternate:\n\n$$b_1 = 13, \\quad b_2 = 37, \\quad b_3 = 13, \\quad b_4 = 37, \\quad \\dots$$"},{"type":"text","order":3,"title":"Pair the adjacent terms","content":"Since the sequence alternates between $13$ and $37$, we can group adjacent terms into pairs:\n\n$$\\begin{aligned} \\sum_{n=1}^{100} b_n &= (b_1 + b_2) + (b_3 + b_4) + \\dots + (b_{99} + b_{100}) \\\\ &= (13 + 37) + (13 + 37) + \\dots + (13 + 37) \\end{aligned}$$\n\nSince we are summing $100$ terms in total, we will have exactly $50$ pairs of adjacent terms."},{"type":"text","order":4,"title":"Calculate the final sum","content":"Now we compute the value of a single pair and multiply by the total number of pairs:\n\n$$\\text{Value of one pair} = 13 + 37 = 50$$\n\n$$\\text{Total Sum} = 50 \\times 50 = 2500$$\n\nTherefore,\n\n$$\\sum_{n=1}^{100} (4C_n - 5) = 2500$$"}]},{"id":"b2190b6c-cbf2-4121-8594-c1e0f51e2dd3:separating_sum","label":"(b) Separating the Sum","steps":[{"type":"text","order":0,"title":"Analyze the sequence","content":"First, let's understand how the concentration $C_n$ behaves using the recurrence relation given:\n\n$$C_{n+1} = 15 - C_n, \\quad C_1 = 4.5$$\n\nLet's calculate the first few terms:\n- For $n = 1$: $C_1 = 4.5$\n- For $n = 2$: $C_2 = 15 - 4.5 = 10.5$\n- For $n = 3$: $C_3 = 15 - 10.5 = 4.5$\n- For $n = 4$: $C_4 = 15 - 4.5 = 10.5$\n\nThis shows that the sequence alternates between $4.5$ (for odd $n$) and $10.5$ (for even $n$)."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$$\\sum_{n=1}^{100} (\\htmlData{anchor=four-c}{\\color{@sky}{4C_n}} - \\htmlData{anchor=five}{\\color{@amber}{5}})$$"},{"latex":"$$$= 4\\htmlData{anchor=sum-c}{\\color{@sky}{\\sum_{n=1}^{100} C_n}} - \\htmlData{anchor=sum-five}{\\color{@amber}{\\sum_{n=1}^{100} 5}}$$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"four-c"},{"color":"sky","style":"text-color","anchor":"sum-c"},{"color":"amber","style":"text-color","anchor":"five"},{"color":"amber","style":"text-color","anchor":"sum-five"}],"connections":[{"to":"sum-c","from":"four-c","color":"sky","label":"distribute constant","style":"solid"},{"to":"sum-five","from":"five","color":"amber","label":"distribute sum","style":"solid"}]},"order":1,"title":"Separate the sum","description":"We can distribute the summation operator across the terms, splitting the expression into two separate sums."},{"type":"text","order":2,"title":"Calculate the constant sum","content":"The second part of our separated expression is a sum of the constant $5$, added $100$ times:\n\n$$\\sum_{n=1}^{100} 5 = 100 \\times 5 = 500$$"},{"type":"text","order":3,"title":"Calculate the sequence sum","content":"Now we calculate the sum of the $C_n$ terms, $\\sum_{n=1}^{100} C_n$.\n\nSince the terms alternate between $C_1 = 4.5$ and $C_2 = 10.5$ for $100$ terms, we have exactly $50$ pairs of these two alternating values:\n\n$$\\begin{aligned} \\sum_{n=1}^{100} C_n &= 50 \\times (C_1 + C_2) \\\\ &= 50 \\times (4.5 + 10.5) \\\\ &= 50 \\times 15 \\\\ &= 750 \\end{aligned}$$"},{"type":"text","order":4,"title":"Evaluate the final value","content":"Now we substitute both components back into our separated summation formula:\n\n$$\\begin{aligned} \\sum_{n=1}^{100} (4C_n - 5) &= 4\\sum_{n=1}^{100} C_n - \\sum_{n=1}^{100} 5 \\\\ &= 4(750) - 500 \\\\ &= 3000 - 500 \\\\ &= 2500 \\end{aligned}$$\n\nThis gives us our final answer of 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3.4 Sequences and Series (A-level only)

Question 89

3.4 Sequences and Series (A-level only) Questions

3.4 Sequences and Series (A-level only)

Question 89

3.4 Sequences and Series (A-level only) Questions