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3.3.6 Sequences and series (A-level only)

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Question 5

The first three terms of a geometric series are (3k+3)(3k+3)(3k+3), (k+3)(k+3)(k+3), and k k\,k respectively, where k k\,k is a positive constant.

a.

Show that 2k2−3k−9=02k^2 - 3k - 9 = 02k2−3k−9=0.

[4]
b.

Hence show that k=3k = 3k=3.

[2]
c.

Find the common ratio.

[2]
d.

Find the sum to infinity of the series.

[2]

3.3.6 Sequences and series (A-level only) Questions

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