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AP and GP

(CIE 0606/2020/w/11/Q10)
(a) An arithmetic progression has a second term of 8 and a fourth term of 18. Find the least number of terms for which the sum of this progression is greater than 1560

(b) A geometric progression has a sum to infinity of 72. The sum of the first 3 terms of this progression is $\dfrac{333}{8}.$

(i) Find the value of the common ratio. [5] 

(ii) Hence find the value of the first term. [1]


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

(a) $ \begin{array}[t]{rlll}  a+d&=8 &\cdots (1)\\ a+3 d&=18 &\cdots (2)\\ (2)-(1): &2d=10&\rightarrow d=5,a=3\\ \end{array}$ 

Since $S_n>1560,\begin{array}[t]{rlll}\frac{n}{2}(6+(n-1) 5)&>&1560 \\ 5 n^{2}+n-3120&>&0 \end{array}$ 

 Critical values $=\dfrac{-1\pm\sqrt{62401}}{10}.$

Positive critical value $= 24.9$. Thus required term=25.  

(b) (i)Since $S=72$ and $S_3=\dfrac{333}{8},$

$\begin{array}[t]{lll}\dfrac{a}{1-r}&=72 &\cdots (1) \\ \dfrac{a\left(1-r^{3}\right)}{1-r}& =\dfrac{333}{8}&\cdots (2)\\(2)\div (1)&1-r^3&=\dfrac{333}{8\times 72}=\dfrac{333}{576}\\&r^3&=1-\dfrac{333}{576}=\dfrac{243}{576}\\&r=\dfrac 34.\end{array} $

(ii) First term $=a=72(1-\dfrac 34)=18.$

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