# GP

CIE 2020 winter paper 22 no 7

A geometric progression has a first term of 3 and a second term of 2.4. For this progression, find

(a) the sum of the first 8 terms,[3]

(b) the sum to infinity, [1]

(c) the least number of terms for which the sum is greater than 95% of the sum to infinity. [4]

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*********math solution*************

$u_1=3,u_2=u_1r=2.4$ Thus $r=2.4/3=0.8.$

(a) $S_8=\dfrac{3(1-0.8^8)}{1-0.8}=12.48$

Ans: 12.5

(b) Sum to infinity=$\dfrac{3}{1-0.8}=15.$

(c) $\begin{array}[t]{rcll}S_n=\dfrac{3(1-0.8^n)}{1-0.8}&>& 0.95(15)\\ \dfrac{3(1-0.8^n)}{0.2}&>&0.95\times 15\\1-0.8^n&>&\dfrac{0.2}{3}\times 0.95\times 15\\0.8^n&<&0.05\\n\log 0.8&<&\log 0.05\\ n&>& \dfrac{\log 0.05}{\log 0.8}=13.42&(\because \log 0.8<0)\end{array}$

Ans: 14

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