Solution of Past Paper of CAIE Math 9709/13 Oct/Nov 2019 Question # 9
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Course
CAIE AS & A Level Mathematics P1 (9709/1)
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CAIE AS & A Level Mathematics P1 (9709/1)
This document presents most carefully worked out and thoroughly explained complete Solution of Past Paper of Cambridge AS/A Level Mathematics (9709), Pure Mathematics 1.
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solution of past paper 9709 13 october november 19
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CIE/GCE/AS/Math/P1/19/Nov/13/Q#9
Question
The first, second and third terms of a geometric progression are 3𝑘, 5𝑘 − 6 and 6𝑘 − 4
respectively.
(i) Show that 𝑘 satisfies the equation 7k2 − 48k + 36 = 0.
(i) Find, showing all necessary working, the exact values of the common
ratio corresponding to each of the possible values of k.
(ii) One of these ratios gives a progression which is convergent. Find the
sum to infinity.
Solution
i.
From the given information, we can collect following information about this Geometric
Progression (G.P).
𝑎1 = 3𝑘
𝑎2 = 5𝑘 − 6
𝑎3 = 6𝑘 − 4
Expression for Common Ratio (𝑟) in a Geometric Progression (G.P) is;
𝑎𝑛
𝑟=
𝑎𝑛−1
Hence;
𝑎𝑛
𝑟=
𝑎𝑛−1
We are required to solve the following equation obtained in (i);
7𝑘2 − 48𝑘 + 36 = 0
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