In a geometric progression, if the ratio of the sum of first 5 terms to the sum of their reciprocals is 49 , and the sum of the first and the third…

According to Question $$\Rightarrow \frac{S_5}{S_5^{\prime}}=49$$⇒S5′​S5​​=49 (here, $S_5=$S5​= Sum of first 5 terms and $S_5=$S5​= Sum of their reciprocals) $$\begin{aligned} &\Rightarrow \frac{\frac{a\left(r^5-1\right)}{(r-1)}}{\frac{a^{-1}\left(r^{-5}-1\right)}{\left(r^{-1}-1\right)}}=49 \\ &\Rightarrow \frac{a\left(r^5-1\right) \times\left(r^{-1}-1\right)}{a^{-1}\left(r^{-5}-1\right) \times(r-1)}=49 \\ &\text { or } \frac{a^2\left(1-r^5\right) \times(1-r) \times r^5}{\left(1-r^5\right) \times(1-\mathrm{r}) \times r}=49 \\ &\Rightarrow a^2 r^4=49 \Rightarrow a^2 r^4=7^2 \\ &\Rightarrow a r^2=7 \end{aligned}$$​⇒(r−1−1)a−1(r−5−1)​(r−1)a(r5−1)​​=49⇒a−1(r−5−1)×(r−1)a(r5−1)×(r−1−1)​=49 or (1−r5)×(1−r)×ra2(1−r5)×(1−r)×r5​=49⇒a2r4=49⇒a2r4=72⇒ar2=7​ Also, given, $S_1+S_3=35$S1​+S3​=35 $a+a r^2=35$a+ar2=35 Now substituting the value of eq. (1) in eq. (2) $a+7=35$a+7=35 $a=28$a=28