(CIE 0606/2021/m/12/Q6)
(a) A geometric progression has first term 10 and sum to infinity 6.
(i) Find the common ratio of this progression.[2]
(ii) Hence find the sum of the first 7 terms, giving your answer correct to 2 decimal places. [2] ©
(b) The first three terms of an arithmetic progression are $\log_x3 , \log_x (3^2),\log_x(3^3).$
(i) Find the common difference of this progression.[1]
(ii) Find, in terms of $n$ and $\log_x3,$ the sum to $n$ terms of this progression. Simplify your answer. [2]
(iii) Given that the sum to $n$ terms is $3081 log_x 3,$ find the value of $n.$ [2]
(iv) Hence, given that the sum to $n$ terms is also equal to 1027, find the value of $x.$ [2]
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*********math solution*************
$ \begin{aligned}&\text { (i) } GP, \quad a=10, \quad S_{\infty}=6=\frac{a}{1-r}=\frac{10}{1-r} \\&\qquad \begin{aligned}1-r=\frac{10}{6} \Rightarrow r=-\frac{2}{3} \\\text { (ii) } S_{7}=\frac{10\left(1-\left(-\frac{2}{3}\right)^{7}\right.}{1-\left(-\frac{2}{3}\right)}=6.35 \end{aligned}\end{aligned}$ $\begin{array}{ll}\text { b) } A P, a=\log _{x} 3 \\\text { (i) } d=\log _{x}\left(3^{2}\right)-\log _{x}(3)&=\log _{x} \dfrac{3^{2}}{3}=\log _{x} 3 \\\text { (ii) } S_{n} &=\dfrac{n}{2}\left[2 \log _{x} 3+(n-1) \log _{x} 3\right] \\&=\dfrac{n}{2}\left[\log _{x} 3(2+(n-1))\right] \\&=\dfrac{n}{2}(n+1) \log _{x}3\end{array}$ (iii) $\begin{array}[t]{rll}\frac{n}{2}(n+1) \log _{x} 3 &=3081 \log _{x} 3 \\\frac{n}{2}(n+1) &=3081 \\n^{2}+n-6162=0 \\(n-78)(n+79)=0 \\\therefore n &=78 \quad \text { or } \quad n=-79(\text { reject }) \end{array}$ (iv) $\begin{array}[t]{rl} 1027 &=\frac{78}{2}\left(2 \log _{x} 3+(78-1) \log _{x} 3\right) \\&=\frac{78}{2} \times 79 \log _{x} 3 \\\log _{x} 3 &=\frac{1027 \times 2}{78 \times 79}=\frac{1}{3} \\3 &=x^{\frac{1}{3}} \\\therefore x &=3^{3}=27\end{array}$
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