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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\$.","difficulty":4,"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}\$","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)\$.","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 **$2500$**."}]}],"generatedAt":"2026-06-09T05:54:21.100Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":null,"walkthroughVideo":null},{"id":"60d0d265-689d-4a15-9b12-449b4211b338","revisionId":"47908fd1-c96d-4c83-8db2-6b06cb646add","questionNumber":null,"questionType":"LA","specification":"The fifth term of an arithmetic sequence is 5 and the eighth term of the sequence is $-16$.","difficulty":null,"difficultyBand":"medium","tier":null,"totalMarks":4,"sourceTitle":null,"calculatorUse":null,"isCalculation":true,"parts":[{"id":"db05c669-cc29-4df7-bd0c-0f60264cd425","label":"a","content":"Find the common difference of the sequence.","markscheme":"- Forms $a + 4d = 5$ and $a + 7d = -16$ M1\n- $3d = -21$, so $d = -7$ A1","marks":2,"order":0,"markingRubric":null,"tableImages":[],"explanation":null,"options":[]},{"id":"66bb1dbb-59e6-4c0d-940a-3e8800042cb8","label":"b","content":"Find the first term of the sequence.","markscheme":"- Substitutes $d = -7$ into either equation M1\n- $a = 33$ A1","marks":2,"order":1,"markingRubric":null,"tableImages":[],"explanation":null,"options":[]}],"options":[],"explanation":null,"firstAttemptFullMarks":null,"walkthroughVideo":null},{"id":"6174aaa8-8897-4e9e-883d-a49fb3bd7c3b","revisionId":"a9e62285-ccbb-44be-baa3-6fed3595ad55","questionNumber":null,"questionType":"LA","specification":"","difficulty":3,"difficultyBand":"medium","tier":null,"totalMarks":6,"sourceTitle":null,"calculatorUse":null,"isCalculation":true,"parts":[{"id":"c1bd8462-5859-45e6-8a3e-3b71a21ef61f","label":"a","content":"A deep-sea mining drone operates for 400 consecutive shifts. The net yield of minerals on the first shift was $-1.5$ kg, representing a net energy loss, while the net yield on the 400th shift was $13.5$ kg. Given that the yields form an arithmetic sequence, calculate the total net yield over the 400 shifts.","markscheme":"- Uses $S_n = \\frac{n}{2}(a + l)$ with $n = 400$, $a = -1.5$, and $l = 13.5$ M1\n- Calculates total net yield as $200(12) = 2400$ kg A1","marks":2,"order":0,"markingRubric":null,"tableImages":[],"explanation":{"methods":[{"id":"direct_sum_formula","label":"Direct Sum Formula","steps":[{"type":"text","order":0,"title":"Identify key information","content":"To find the total net yield, we first need to extract the parameters of our arithmetic sequence from the question:\n\n* The total number of shifts is $n = 400$.\n* The first term (yield on the first shift) is $a = -1.5$ kg.\n* The last term (yield on the 400th shift) is $l = 13.5$ kg.","highlights":[{"text":"400 consecutive shifts","location":"specification"},{"text":"-1.5","location":"specification"},{"text":"13.5","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\{ \\htmlData{anchor=num-shifts}{\\color{@sky}{n = 400}}, \\; \\htmlData{anchor=first-val}{\\color{@amber}{a = -1.5}}, \\; \\htmlData{anchor=last-val}{\\color{@purple}{l = 13.5}} \\}$"},{"latex":"$$S_{\\htmlData{anchor=n-sub1}{\\color{@sky}{n}}} = \\frac{\\htmlData{anchor=n-sub2}{\\color{@sky}{n}}}{2}(\\htmlData{anchor=a-sub}{\\color{@amber}{a}} + \\htmlData{anchor=l-sub}{\\color{@purple}{l}})$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"num-shifts"},{"color":"amber","style":"text-color","anchor":"first-val"},{"color":"purple","style":"text-color","anchor":"last-val"}],"connections":[{"to":"n-sub1","from":"num-shifts","color":"sky","label":"number of terms","style":"solid"},{"to":"a-sub","from":"first-val","color":"amber","label":"first term","style":"solid"},{"to":"l-sub","from":"last-val","color":"purple","label":"last term","style":"solid"}]},"order":1,"title":"Select the sum formula","description":"Since we are given the first term $a$, the last term $l$, and the total number of terms $n$, we can directly use the arithmetic series sum formula: \n\n$$S_n = \\frac{n}{2}(a + l)$$"},{"type":"text","order":2,"title":"Calculate the total sum","content":"Now, substitute the values into the formula to calculate the total net yield:\n\n$$S_{400} = \\frac{400}{2}(-1.5 + 13.5)$$\n\n$$S_{400} = 200(12)$$\n\n$$S_{400} = 2400 \\text{ kg}$$\n\nThis calculation awards **1 mark** (M1) for setting up the substitution and **1 mark** (A1) for the correct final answer of $2400$ kg.","calculator":[{"gdc":{"expression":"400/2 * (-1.5 + 13.5)","instructions":"Enter the complete formula expression on the home screen:\n`400 / 2 * (-1.5 + 13.5)` and press `EXE`."},"ti84":{"expression":"(400/2)*(-1.5 + 13.5)","instructions":"Type the expression directly on the main screen:\n1. Press `400 / 2` or `(400 / 2)`\n2. Multiply by opening parenthesis `(`\n3. Enter `-1.5 + 13.5` (use the negative key `(-)` for `-1.5`)\n4. Close parenthesis `)` and press `ENTER`"},"desmos":{"expression":"\\frac{400}{2}\\left(-1.5+13.5\\right)","instructions":"Type the expression into an active line to see the output instantly."},"description":"Evaluate the arithmetic series sum formula with the given parameters."}]}]},{"id":"find_d_then_sum","label":"Finding Common Difference First","steps":[{"type":"text","order":0,"title":"Identify given values","content":"From the question, we are given the following values for our arithmetic sequence:\n- Number of shifts, $n = 400$\n- First term, $a = -1.5$\n- 400th term, $u_{400} = 13.5$\n\nWe need to find the total net yield ($S_{400}$) by first finding the common difference, $d$, and then using the sum formula.","highlights":[{"text":"400 consecutive shifts","location":"specification"},{"text":"-1.5","location":"specification"},{"text":"13.5","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$u_{n} = a + (n-1)d$"},{"latex":"$$\\htmlData{anchor=term-val}{\\color{@sky}{13.5}} = \\htmlData{anchor=first-val}{\\color{@red}{-1.5}} + (\\htmlData{anchor=n-val}{\\color{@purple}{400}} - 1)d$"},{"latex":"$$13.5 = -1.5 + 399d$"},{"latex":"$$15 = 399d$"},{"latex":"$$d = \\frac{15}{399} = \\htmlData{anchor=diff-val}{\\color{@teal}{\\frac{5}{133}}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"term-val"},{"color":"red","style":"text-color","anchor":"first-val"},{"color":"purple","style":"text-color","anchor":"n-val"},{"color":"teal","style":"text-color","anchor":"diff-val"}],"connections":[{"to":"diff-val","from":"term-val","color":"sky","label":null,"style":"solid"}]},"order":1,"title":"Find common difference","description":"We use the $n$-th term formula of an arithmetic sequence to find the common difference $d$."},{"type":"text","order":2,"title":"Calculate total sum","content":"Now we substitute $a = -1.5$, $n = 400$, and the common difference $d = \\frac{5}{133}$ into the arithmetic series sum formula:\n\n$$S_n = \\frac{n}{2}[2a + (n-1)d]$$\n\n$$S_{400} = \\frac{400}{2} \\left[ 2(-1.5) + (400 - 1)\\left(\\frac{5}{133}\\right) \\right]$$\n\n$$S_{400} = 200 \\left[ -3 + 399\\left(\\frac{5}{133}\\right) \\right]$$\n\n$$S_{400} = 200 \\left[ -3 + 3(5) \\right]$$\n\n$$S_{400} = 200 \\left[ -3 + 15 \\right]$$\n\n$$S_{400} = 200(12)$$\n\n$$S_{400} = 2400 \\text{ kg}$$"},{"type":"text","order":3,"title":"State final answer","content":"The total net yield over the 400 shifts is **$2400$ kg**."}]}],"generatedAt":"2026-06-09T05:16:10.418Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]},{"id":"c041a5b7-a0d5-44e3-a180-19a120e8556d","label":"b","content":"A corporate incentive fund distributes a total of $\\pounds 13,500$ among 15 employees. The individual rewards form an arithmetic sequence. The reward for the highest-performing employee is exactly eight times the reward for the lowest-performing employee. Determine the value of the reward for the highest-performing employee.","markscheme":"- Sets up equation for the sum: $13500 = \\frac{15}{2}(a + l)$ or $a + l = 1800$ M1\n- Identifies the relationship between the first and last terms as $l = 8a$ B1\n- Substitutes $a = \\frac{1}{8}l$ into the sum equation to get $\\frac{1}{8}l + l = 1800$ or $\\frac{9}{8}l = 1800$ M1\n- Solves to find the value of the highest-performing reward $l = 1600$ A1","marks":4,"order":1,"markingRubric":null,"tableImages":[],"explanation":{"methods":[{"id":"first_last_formula","label":"First and Last Term Formula","steps":[{"type":"text","order":0,"title":"Identify the key values","content":"First, let's extract the key values from the question:\n- The total sum of rewards is $S_n = 13500$\n- The total number of employees is $n = 15$\n- Let the reward for the lowest-performing employee (the first term) be $a$\n- Let the reward for the highest-performing employee (the last term) be $l$","highlights":[{"text":"total of $\\pounds 13,500$","location":"specification"},{"text":"15 employees","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$$S_n = \\frac{n}{2}(a + l)$$"},{"latex":"$$\\htmlData{anchor=sum-val}{\\color{@sky}{13500}} = \\frac{\\htmlData{anchor=n-val}{\\color{@amber}{15}}}{2}(a + l)$"},{"latex":"$$$1800 = a + l$$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"sum-val"},{"color":"amber","style":"text-color","anchor":"n-val"}],"connections":[{"to":"n-val","from":"sum-val","color":"sky","label":null,"style":"solid"}]},"order":1,"title":"Set up sum equation","description":"We can use the arithmetic series sum formula containing the first term $a$ and last term $l$:\n\n$$S_n = \\frac{n}{2}(a + l)$$\n\nSubstituting $S_n = 13500$ and $n = 15$:\n\n$$13500 = \\frac{15}{2}(a + l)$$\n\nMultiplying both sides by $2$ and dividing by $15$ gives:\n\n$$a + l = 1800$$\n\n*(This earns the first method mark **M1**)*"},{"type":"text","order":2,"title":"Relate first and last terms","content":"The question states that the highest reward is exactly eight times the lowest reward.\n\nThis gives us the relationship:\n$$l = 8a$$\n\n*(This earns the accuracy mark **B1**)*","highlights":[{"text":"exactly eight times the reward for the lowest-performing employee","location":"specification"}]},{"type":"text","order":3,"title":"Express a in terms of l","content":"Since we want to find the value of the highest reward $l$, we can express $a$ in terms of $l$ to solve for $l$ directly:\n\n$$a = \\frac{1}{8}l$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\htmlData{anchor=a-expr}{\\color{@purple}{a}} + l = 1800$"},{"latex":"$$\\htmlData{anchor=sub-val}{\\color{@purple}{\\frac{1}{8}l}} + l = 1800$"},{"latex":"$$$\\frac{9}{8}l = 1800$$"},{"latex":"$$$l = 1600$$"}],"highlights":[{"color":"purple","style":"text-color","anchor":"a-expr"},{"color":"purple","style":"text-color","anchor":"sub-val"}],"connections":[{"to":"sub-val","from":"a-expr","color":"purple","label":"Substitute","style":"solid"}]},"order":4,"title":"Substitute and solve for l","description":"Now substitute $a = \\frac{1}{8}l$ into our sum equation:\n\n$$\\frac{1}{8}l + l = 1800$$\n\n*(This earns the method mark **M1**)*\n\nCombine the terms:\n\n$$\\frac{9}{8}l = 1800$$\n\nNow, solve for $l$:\n\n$$l = 1800 \\times \\frac{8}{9}$$\n$$l = 1600$$\n\n*(This earns the final accuracy mark **A1**)*"},{"type":"text","order":5,"title":"State the final answer","content":"The reward for the highest-performing employee is **£1,600**."}]},{"id":"common_difference_formula","label":"Using Common Difference","steps":[{"type":"text","order":0,"title":"Define sequence parameters","content":"Let the individual rewards form an arithmetic sequence with:\n- First term (lowest reward) = $a$\n- Common difference = $d$\n- Number of terms ($n$) = $15$\n\nThe last term $l$ (the highest reward) is the $15^{\\text{th}}$ term. Using the $n^{\\text{th}}$ term formula, $$u_n = a + (n-1)d$$, we can express the last term as:\n\n$$l = a + 14d$$","highlights":[{"text":"15 employees","location":"specification"}]},{"type":"text","order":1,"title":"Find relation between $a$ and $d$","content":"We are given that the highest reward is exactly eight times the lowest reward:\n\n$$l = 8a$$\n\nSubstituting $l = a + 14d$ into this relation gives:\n\n$$a + 14d = 8a$$\n\nSubtracting $a$ from both sides:\n\n$$14d = 7a \\implies a = 2d$$","highlights":[{"text":"exactly eight times the reward for the lowest-performing employee","location":"specification"}]},{"type":"text","order":2,"title":"Set up sum equation","content":"The sum of the arithmetic series is given by the formula:\n\n$$S_n = \\frac{n}{2}[2a + (n-1)d]$$\n\nGiven that the total sum distributed is $S_{15} = 13500$ and $n = 15$, we substitute these along with $a = 2d$:\n\n$$\\begin{aligned} 13500 &= \\frac{15}{2}[2(2d) + 14d] \\\\ 13500 &= \\frac{15}{2}[4d + 14d] \\\\ 13500 &= \\frac{15}{2}(18d) \\\\ 13500 &= 135d \\end{aligned}$$","highlights":[{"text":"total of £ 13,500","location":"specification"}]},{"type":"text","order":3,"title":"Solve for parameters","content":"Now we solve for the common difference $d$:\n\n$$d = \\frac{13500}{135} = 100$$\n\nUsing our relationship $a = 2d$, we find the lowest reward:\n\n$$a = 2(100) = 200$$"},{"type":"text","order":4,"title":"Calculate highest reward","content":"Finally, we calculate the highest reward $l$ using the relationship $l = 8a$:\n\n$$l = 8 \\times 200 = 1600$$\n\nThus, the reward for the highest-performing employee is **$\\pounds 1,600$**.","highlights":[{"text":"Determine the value of the reward for the highest-performing employee.","location":"specification"}]}]},{"id":"average_value","label":"Average Value Approach","steps":[{"type":"text","order":0,"title":"Calculate average reward","content":"In an arithmetic sequence, the average (mean) value of all terms is equal to the average of the first term $a$ and the last term $l$.\n\nFirst, let's find the average reward distributed to the 15 employees by dividing the total fund by the number of employees:\n\n$$\\text{Average} = \\frac{13500}{15} = 900$$","highlights":[{"text":"total of £ 13,500 among 15 employees","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$$a = \\text{first term (lowest reward)}$$"},{"latex":"$$$l = \\text{last term (highest reward)}$$"},{"latex":"$$$\\text{Average} = \\frac{\\htmlData{anchor=a-term}{a} + \\htmlData{anchor=l-term}{l}}{2} = \\htmlData{anchor=mean-val}{900}$$"}],"highlights":[{"color":"sky","style":"box","anchor":"mean-val"}],"connections":[{"to":"a-term","from":"mean-val","color":"sky","label":null,"style":"dashed"}]},"order":1,"title":"Equate averages","description":"Since the individual rewards form an arithmetic sequence, the average of all 15 terms is exactly equal to the average of the first and last terms: $\\frac{a + l}{2}$."},{"type":"text","order":2,"title":"Set up relationship","content":"We are given that the reward for the highest-performing employee ($l$) is exactly eight times the reward for the lowest-performing employee ($a$):\n\n$$l = 8a$$\n\nTo find $l$, we can express $a$ in terms of $l$:\n\n$$a = \\frac{1}{8}l$$","highlights":[{"text":"The reward for the highest-performing employee is exactly eight times the reward for the lowest-performing employee.","location":"specification"}]},{"type":"text","order":3,"title":"Substitute and solve","content":"Now substitute $a = \\frac{1}{8}l$ and our average value of $900$ into the average equation:\n\n$$\\frac{\\frac{1}{8}l + l}{2} = 900$$\n\nMultiply both sides by 2:\n\n$$\\frac{1}{8}l + l = 1800$$\n\n$$\\frac{9}{8}l = 1800$$\n\nSolve for $l$:\n\n$$l = 1800 \\times \\frac{8}{9}$$\n\n$$l = 200 \\times 8 = 1600$$\n\nThus, the reward for the highest-performing employee is $\\pounds 1,600$."}]}],"generatedAt":"2026-06-09T05:16:58.478Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"c1bd8462-5859-45e6-8a3e-3b71a21ef61f:direct_sum_formula","label":"(a) Direct Sum Formula","steps":[{"type":"text","order":0,"title":"Identify key information","content":"To find the total net yield, we first need to extract the parameters of our arithmetic sequence from the question:\n\n* The total number of shifts is $n = 400$.\n* The first term (yield on the first shift) is $a = -1.5$ kg.\n* The last term (yield on the 400th shift) is $l = 13.5$ kg.","highlights":[{"text":"400 consecutive shifts","location":"specification"},{"text":"-1.5","location":"specification"},{"text":"13.5","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\{ \\htmlData{anchor=num-shifts}{\\color{@sky}{n = 400}}, \\; \\htmlData{anchor=first-val}{\\color{@amber}{a = -1.5}}, \\; \\htmlData{anchor=last-val}{\\color{@purple}{l = 13.5}} \\}$"},{"latex":"$$S_{\\htmlData{anchor=n-sub1}{\\color{@sky}{n}}} = \\frac{\\htmlData{anchor=n-sub2}{\\color{@sky}{n}}}{2}(\\htmlData{anchor=a-sub}{\\color{@amber}{a}} + \\htmlData{anchor=l-sub}{\\color{@purple}{l}})$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"num-shifts"},{"color":"amber","style":"text-color","anchor":"first-val"},{"color":"purple","style":"text-color","anchor":"last-val"}],"connections":[{"to":"n-sub1","from":"num-shifts","color":"sky","label":"number of terms","style":"solid"},{"to":"a-sub","from":"first-val","color":"amber","label":"first term","style":"solid"},{"to":"l-sub","from":"last-val","color":"purple","label":"last term","style":"solid"}]},"order":1,"title":"Select the sum formula","description":"Since we are given the first term $a$, the last term $l$, and the total number of terms $n$, we can directly use the arithmetic series sum formula: \n\n$$S_n = \\frac{n}{2}(a + l)$$"},{"type":"text","order":2,"title":"Calculate the total sum","content":"Now, substitute the values into the formula to calculate the total net yield:\n\n$$S_{400} = \\frac{400}{2}(-1.5 + 13.5)$$\n\n$$S_{400} = 200(12)$$\n\n$$S_{400} = 2400 \\text{ kg}$$\n\nThis calculation awards **1 mark** (M1) for setting up the substitution and **1 mark** (A1) for the correct final answer of $2400$ kg.","calculator":[{"gdc":{"expression":"400/2 * (-1.5 + 13.5)","instructions":"Enter the complete formula expression on the home screen:\n`400 / 2 * (-1.5 + 13.5)` and press `EXE`."},"ti84":{"expression":"(400/2)*(-1.5 + 13.5)","instructions":"Type the expression directly on the main screen:\n1. Press `400 / 2` or `(400 / 2)`\n2. Multiply by opening parenthesis `(`\n3. Enter `-1.5 + 13.5` (use the negative key `(-)` for `-1.5`)\n4. Close parenthesis `)` and press `ENTER`"},"desmos":{"expression":"\\frac{400}{2}\\left(-1.5+13.5\\right)","instructions":"Type the expression into an active line to see the output instantly."},"description":"Evaluate the arithmetic series sum formula with the given parameters."}]}]},{"id":"c1bd8462-5859-45e6-8a3e-3b71a21ef61f:find_d_then_sum","label":"(a) Finding Common Difference First","steps":[{"type":"text","order":0,"title":"Identify given values","content":"From the question, we are given the following values for our arithmetic sequence:\n- Number of shifts, $n = 400$\n- First term, $a = -1.5$\n- 400th term, $u_{400} = 13.5$\n\nWe need to find the total net yield ($S_{400}$) by first finding the common difference, $d$, and then using the sum formula.","highlights":[{"text":"400 consecutive shifts","location":"specification"},{"text":"-1.5","location":"specification"},{"text":"13.5","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$u_{n} = a + (n-1)d$"},{"latex":"$$\\htmlData{anchor=term-val}{\\color{@sky}{13.5}} = \\htmlData{anchor=first-val}{\\color{@red}{-1.5}} + (\\htmlData{anchor=n-val}{\\color{@purple}{400}} - 1)d$"},{"latex":"$$13.5 = -1.5 + 399d$"},{"latex":"$$15 = 399d$"},{"latex":"$$d = \\frac{15}{399} = \\htmlData{anchor=diff-val}{\\color{@teal}{\\frac{5}{133}}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"term-val"},{"color":"red","style":"text-color","anchor":"first-val"},{"color":"purple","style":"text-color","anchor":"n-val"},{"color":"teal","style":"text-color","anchor":"diff-val"}],"connections":[{"to":"diff-val","from":"term-val","color":"sky","label":null,"style":"solid"}]},"order":1,"title":"Find common difference","description":"We use the $n$-th term formula of an arithmetic sequence to find the common difference $d$."},{"type":"text","order":2,"title":"Calculate total sum","content":"Now we substitute $a = -1.5$, $n = 400$, and the common difference $d = \\frac{5}{133}$ into the arithmetic series sum formula:\n\n$$S_n = \\frac{n}{2}[2a + (n-1)d]$$\n\n$$S_{400} = \\frac{400}{2} \\left[ 2(-1.5) + (400 - 1)\\left(\\frac{5}{133}\\right) \\right]$$\n\n$$S_{400} = 200 \\left[ -3 + 399\\left(\\frac{5}{133}\\right) \\right]$$\n\n$$S_{400} = 200 \\left[ -3 + 3(5) \\right]$$\n\n$$S_{400} = 200 \\left[ -3 + 15 \\right]$$\n\n$$S_{400} = 200(12)$$\n\n$$S_{400} = 2400 \\text{ kg}$$"},{"type":"text","order":3,"title":"State final answer","content":"The total net yield over the 400 shifts is **$2400$ kg**."}]},{"id":"c041a5b7-a0d5-44e3-a180-19a120e8556d:first_last_formula","label":"(b) First and Last Term Formula","steps":[{"type":"text","order":0,"title":"Identify the key values","content":"First, let's extract the key values from the question:\n- The total sum of rewards is $S_n = 13500$\n- The total number of employees is $n = 15$\n- Let the reward for the lowest-performing employee (the first term) be $a$\n- Let the reward for the highest-performing employee (the last term) be $l$","highlights":[{"text":"total of $\\pounds 13,500$","location":"specification"},{"text":"15 employees","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$$S_n = \\frac{n}{2}(a + l)$$"},{"latex":"$$\\htmlData{anchor=sum-val}{\\color{@sky}{13500}} = \\frac{\\htmlData{anchor=n-val}{\\color{@amber}{15}}}{2}(a + l)$"},{"latex":"$$$1800 = a + l$$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"sum-val"},{"color":"amber","style":"text-color","anchor":"n-val"}],"connections":[{"to":"n-val","from":"sum-val","color":"sky","label":null,"style":"solid"}]},"order":1,"title":"Set up sum equation","description":"We can use the arithmetic series sum formula containing the first term $a$ and last term $l$:\n\n$$S_n = \\frac{n}{2}(a + l)$$\n\nSubstituting $S_n = 13500$ and $n = 15$:\n\n$$13500 = \\frac{15}{2}(a + l)$$\n\nMultiplying both sides by $2$ and dividing by $15$ gives:\n\n$$a + l = 1800$$\n\n*(This earns the first method mark **M1**)*"},{"type":"text","order":2,"title":"Relate first and last terms","content":"The question states that the highest reward is exactly eight times the lowest reward.\n\nThis gives us the relationship:\n$$l = 8a$$\n\n*(This earns the accuracy mark **B1**)*","highlights":[{"text":"exactly eight times the reward for the lowest-performing employee","location":"specification"}]},{"type":"text","order":3,"title":"Express a in terms of l","content":"Since we want to find the value of the highest reward $l$, we can express $a$ in terms of $l$ to solve for $l$ directly:\n\n$$a = \\frac{1}{8}l$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\htmlData{anchor=a-expr}{\\color{@purple}{a}} + l = 1800$"},{"latex":"$$\\htmlData{anchor=sub-val}{\\color{@purple}{\\frac{1}{8}l}} + l = 1800$"},{"latex":"$$$\\frac{9}{8}l = 1800$$"},{"latex":"$$$l = 1600$$"}],"highlights":[{"color":"purple","style":"text-color","anchor":"a-expr"},{"color":"purple","style":"text-color","anchor":"sub-val"}],"connections":[{"to":"sub-val","from":"a-expr","color":"purple","label":"Substitute","style":"solid"}]},"order":4,"title":"Substitute and solve for l","description":"Now substitute $a = \\frac{1}{8}l$ into our sum equation:\n\n$$\\frac{1}{8}l + l = 1800$$\n\n*(This earns the method mark **M1**)*\n\nCombine the terms:\n\n$$\\frac{9}{8}l = 1800$$\n\nNow, solve for $l$:\n\n$$l = 1800 \\times \\frac{8}{9}$$\n$$l = 1600$$\n\n*(This earns the final accuracy mark **A1**)*"},{"type":"text","order":5,"title":"State the final answer","content":"The reward for the highest-performing employee is **£1,600**."}]},{"id":"c041a5b7-a0d5-44e3-a180-19a120e8556d:common_difference_formula","label":"(b) Using Common Difference","steps":[{"type":"text","order":0,"title":"Define sequence parameters","content":"Let the individual rewards form an arithmetic sequence with:\n- First term (lowest reward) = $a$\n- Common difference = $d$\n- Number of terms ($n$) = $15$\n\nThe last term $l$ (the highest reward) is the $15^{\\text{th}}$ term. Using the $n^{\\text{th}}$ term formula, $$u_n = a + (n-1)d$$, we can express the last term as:\n\n$$l = a + 14d$$","highlights":[{"text":"15 employees","location":"specification"}]},{"type":"text","order":1,"title":"Find relation between $a$ and $d$","content":"We are given that the highest reward is exactly eight times the lowest reward:\n\n$$l = 8a$$\n\nSubstituting $l = a + 14d$ into this relation gives:\n\n$$a + 14d = 8a$$\n\nSubtracting $a$ from both sides:\n\n$$14d = 7a \\implies a = 2d$$","highlights":[{"text":"exactly eight times the reward for the lowest-performing employee","location":"specification"}]},{"type":"text","order":2,"title":"Set up sum equation","content":"The sum of the arithmetic series is given by the formula:\n\n$$S_n = \\frac{n}{2}[2a + (n-1)d]$$\n\nGiven that the total sum distributed is $S_{15} = 13500$ and $n = 15$, we substitute these along with $a = 2d$:\n\n$$\\begin{aligned} 13500 &= \\frac{15}{2}[2(2d) + 14d] \\\\ 13500 &= \\frac{15}{2}[4d + 14d] \\\\ 13500 &= \\frac{15}{2}(18d) \\\\ 13500 &= 135d \\end{aligned}$$","highlights":[{"text":"total of £ 13,500","location":"specification"}]},{"type":"text","order":3,"title":"Solve for parameters","content":"Now we solve for the common difference $d$:\n\n$$d = \\frac{13500}{135} = 100$$\n\nUsing our relationship $a = 2d$, we find the lowest reward:\n\n$$a = 2(100) = 200$$"},{"type":"text","order":4,"title":"Calculate highest reward","content":"Finally, we calculate the highest reward $l$ using the relationship $l = 8a$:\n\n$$l = 8 \\times 200 = 1600$$\n\nThus, the reward for the highest-performing employee is **$\\pounds 1,600$**.","highlights":[{"text":"Determine the value of the reward for the highest-performing employee.","location":"specification"}]}]},{"id":"c041a5b7-a0d5-44e3-a180-19a120e8556d:average_value","label":"(b) Average Value Approach","steps":[{"type":"text","order":0,"title":"Calculate average reward","content":"In an arithmetic sequence, the average (mean) value of all terms is equal to the average of the first term $a$ and the last term $l$.\n\nFirst, let's find the average reward distributed to the 15 employees by dividing the total fund by the number of employees:\n\n$$\\text{Average} = \\frac{13500}{15} = 900$$","highlights":[{"text":"total of £ 13,500 among 15 employees","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$$a = \\text{first term (lowest reward)}$$"},{"latex":"$$$l = \\text{last term (highest reward)}$$"},{"latex":"$$$\\text{Average} = \\frac{\\htmlData{anchor=a-term}{a} + \\htmlData{anchor=l-term}{l}}{2} = \\htmlData{anchor=mean-val}{900}$$"}],"highlights":[{"color":"sky","style":"box","anchor":"mean-val"}],"connections":[{"to":"a-term","from":"mean-val","color":"sky","label":null,"style":"dashed"}]},"order":1,"title":"Equate averages","description":"Since the individual rewards form an arithmetic sequence, the average of all 15 terms is exactly equal to the average of the first and last terms: $\\frac{a + l}{2}$."},{"type":"text","order":2,"title":"Set up relationship","content":"We are given that the reward for the highest-performing employee ($l$) is exactly eight times the reward for the lowest-performing employee ($a$):\n\n$$l = 8a$$\n\nTo find $l$, we can express $a$ in terms of $l$:\n\n$$a = \\frac{1}{8}l$$","highlights":[{"text":"The reward for the highest-performing employee is exactly eight times the reward for the lowest-performing employee.","location":"specification"}]},{"type":"text","order":3,"title":"Substitute and solve","content":"Now substitute $a = \\frac{1}{8}l$ and our average value of $900$ into the average equation:\n\n$$\\frac{\\frac{1}{8}l + l}{2} = 900$$\n\nMultiply both sides by 2:\n\n$$\\frac{1}{8}l + l = 1800$$\n\n$$\\frac{9}{8}l = 1800$$\n\nSolve for $l$:\n\n$$l = 1800 \\times \\frac{8}{9}$$\n\n$$l = 200 \\times 8 = 1600$$\n\nThus, the reward for the highest-performing employee is $\\pounds 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Sequences and series (A-level only)","topicId":"053492e1-c402-4158-af07-bb52b29684d7","topicSidebar":"$L3c","topicSlug":"sequences-and-series-4","topicTitle":"3.3 Sequences and series (A-level only)"}]}],"$L3d","$L3e"]}]

3.3 Sequences and series (A-level only)

Question 88

3.3 Sequences and series (A-level only) Questions

3.3 Sequences and series (A-level only)

Question 88

3.3 Sequences and series (A-level only) Questions