In a geometric progression, if the ratio of t | vedclass

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Accorrding to Question

$ \Rightarrow \frac{{{S_5}}}{{S{'_5}}} = 49$

(here, ${S_5} = $ Sum of $5$ terms and ${S_5} = $ Sum of their

Vedclass · Accepted Answer

$28$

Vedclass · Answer

Accorrding to Question

$ \Rightarrow \frac{{{S_5}}}{{S{'_5}}} = 49$

(here, ${S_5} = $ Sum of $5$ terms and ${S_5} = $ Sum of their reciprocals)

$ \Rightarrow \frac{{\frac{{a\left( {{r^5} - 1} ight)}}{{\left( {r - 1} ight)}}}}{{\frac{{{a^{ - 1}}\left( {{r^{ - 5}} - 1} ight)}}{{\left( {{r^{ - 1}} - 1} ight)}}}} = 49$

$ \Rightarrow \frac{{a\left( {{r^5} - 1} ight) 	imes \left( {{r^{ - 1}} - 1} ight)}}{{{a^{ - 1}}\left( {{r^{ - 5}} - 1} ight) 	imes \left( {r - 1} ight)}} = 49$

$\frac{{{a_2}\left( {1 - {r^5}} ight) 	imes \left( {1 - r} ight) 	imes {r^5}}}{{\left( {1 - {r^5}} ight) 	imes \left( {1 - r} ight) 	imes r}} = 49$

$ \Rightarrow {a^2}{r^4} = 49 \Rightarrow {a^2}{r^4} = {7^2}$

$ \Rightarrow \boxed{a{r^2} = 7}\,\,\,\,\,\,\,\,\,\,...\left( 1 ight)$

Also, given ${S_1} + {S_3} = 35$

$a + a{r^2} = 35\,\,\,\,\,\,.....\left( 2 ight)$

Now substituting the value of eq. $(1)$ in eq. $(2)$

$a + 7 = 35$

$\boxed{a = 28}$

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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