JEE MAIN - Mathematics (2024 - 9th April Evening Shift - No. 18)
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
27
38
31
46
Explanation
$$\begin{array}{ll} \sum_{n=0}^{\infty} a r^n=57 & \Rightarrow \frac{a}{1-r}=57 \quad \text{.... (i)}\\ \sum_{n=0}^{\infty} a^3 r^{3 n}=9747 & \Rightarrow \frac{a^3}{1-r^3}=9747 \quad \text{.... (ii)} \end{array}$$
$$\begin{aligned} & \frac{\left(1-r^3\right)}{(1-r)^3}=\frac{(57)^3}{9747}=19 \\ & \Rightarrow \quad \frac{(1-r)\left(1+r+r^2\right)}{(1-r)^3}=19 \\ & \Rightarrow \quad 18 r^2-39 r+18=0 \\ & \Rightarrow \quad r=\frac{2}{3}, \frac{3}{2} \text { (rejected) } \\ & \therefore \quad a=19 \\ & \quad a+18 r \\ & \quad=19+12=31 \end{aligned}$$
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