Let $a, a r, a r^2$, ............ be an infinite G.P. If $\sum_\limits{n=0}^{\infty} a r^n=57$ and $\sum_\limits{n=0}^{\infty} a^3 r^{3 n}=9747$, then $a+18 r$ is equal to
Solution
<p>$$\begin{array}{ll}
\sum_{n=0}^{\infty} a r^n=57 & \Rightarrow \frac{a}{1-r}=57 \quad \text{.... (i)}\\
\sum_{n=0}^{\infty} a^3 r^{3 n}=9747 & \Rightarrow \frac{a^3}{1-r^3}=9747 \quad \text{.... (ii)}
\end{array}$$</p>
<p>$$\begin{aligned}
& \frac{\left(1-r^3\right)}{(1-r)^3}=\frac{(57)^3}{9747}=19 \\
& \Rightarrow \quad \frac{(1-r)\left(1+r+r^2\right)}{(1-r)^3}=19 \\
& \Rightarrow \quad 18 r^2-39 r+18=0 \\
& \Rightarrow \quad r=\frac{2}{3}, \frac{3}{2} \text { (rejected) } \\
& \therefore \quad a=19 \\
& \quad a+18 r \\
& \quad=19+12=31
\end{aligned}$$</p>
About this question
Subject: Mathematics · Chapter: Sequences and Series · Topic: Arithmetic Progression
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