Q84 / 180 · KCET 2023 · Mathematics
$n$th term of the series $1+\frac{3}{7}+\frac{5}{7^2}+\frac{1}{7^2}+\ldots$ is
- A$\frac{2 n+1}{7^n}$
- B$\frac{2 n-1}{7^n}$
- C$\frac{2 n+1}{7^n-1}$
- Correct answerD$\frac{2 n-1}{7^n-1}$
Solution
We have,
$\begin{aligned} & 1+\frac{3}{7}+\frac{5}{72}+\frac{7}{73}+\ldots \\ & \frac{1}{70}+\frac{3}{7}+\frac{5}{72}+\frac{7}{73}+\ldots\end{aligned}$
Numerator are in AP
So, $T_n$ of $\mathrm{AP}=1+(n-1)^2=2 n-1$
Denominator are in GP
So, $T_n$ of $\mathrm{GP}=7^{n-1}$
$T_n$ of given series $=\frac{2 n-1}{7^{n-1}}$
Hence, the correct answer is option: 'D'.