Mathematics - Sequence & Series Question with Solution | TestHub

According to Question $$\Rightarrow \frac{S_5}{S_5^{\prime }}=49$$ (here, $S_5=$ Sum of first 5 terms and $S_5=$ Sum of their reciprocals) $$\begin{array}{rl}& \Rightarrow \frac{\frac{a\left(r^5-1\right)}{(r-1)}}{\frac{a^{-1}\left(r^{-5}-1\right)}{\left(r^{-1}-1\right)}}=49\\ & \Rightarrow \frac{a\left(r^5-1\right)\times \left(r^{-1}-1\right)}{a^{-1}\left(r^{-5}-1\right)\times (r-1)}=49\\ & \text{ or }\frac{a^2\left(1-r^5\right)\times (1-r)\times r^5}{\left(1-r^5\right)\times (1-\mathrm{r})\times r}=49\\ & \Rightarrow a^2{r}^4=49\Rightarrow a^2{r}^4=7^2\\ & \Rightarrow a{r}^2=7\end{array}$$ Also, given, $S_1+S_3=35$ $a+a{r}^2=35$ Now substituting the value of eq. (1) in eq. (2) $a+7=35$ $a=28$