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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 term is 35. Then the first term of this geometric progression is:
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Verified Answer
The correct answer is:
28
28
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)
$$
\begin{aligned}
&\Rightarrow \frac{\frac{a\left(r^5-1\right)}{(r-1)}}{\frac{a^{-1}\left(r^{-5}-1\right)}{\left(r^{-1}-1\right)}}=49 \\
&\Rightarrow \frac{a\left(r^5-1\right) \times\left(r^{-1}-1\right)}{a^{-1}\left(r^{-5}-1\right) \times(r-1)}=49 \\
&\text { or } \frac{a^2\left(1-r^5\right) \times(1-r) \times r^5}{\left(1-r^5\right) \times(1-\mathrm{r}) \times r}=49 \\
&\Rightarrow a^2 r^4=49 \Rightarrow a^2 r^4=7^2 \\
&\Rightarrow a r^2=7
\end{aligned}
$$
Also, given, $S_1+S_3=35$ $a+a r^2=35$
Now substituting the value of eq. (1) in eq. (2)
$a+7=35$
$a=28$
$$
\Rightarrow \frac{S_5}{S_5^{\prime}}=49
$$
(here, $S_5=$ Sum of first 5 terms and $S_5=$ Sum of their reciprocals)
$$
\begin{aligned}
&\Rightarrow \frac{\frac{a\left(r^5-1\right)}{(r-1)}}{\frac{a^{-1}\left(r^{-5}-1\right)}{\left(r^{-1}-1\right)}}=49 \\
&\Rightarrow \frac{a\left(r^5-1\right) \times\left(r^{-1}-1\right)}{a^{-1}\left(r^{-5}-1\right) \times(r-1)}=49 \\
&\text { or } \frac{a^2\left(1-r^5\right) \times(1-r) \times r^5}{\left(1-r^5\right) \times(1-\mathrm{r}) \times r}=49 \\
&\Rightarrow a^2 r^4=49 \Rightarrow a^2 r^4=7^2 \\
&\Rightarrow a r^2=7
\end{aligned}
$$
Also, given, $S_1+S_3=35$ $a+a r^2=35$
Now substituting the value of eq. (1) in eq. (2)
$a+7=35$
$a=28$
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