The difference between the fourth term and th | vedclass

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Question

Let $a,ar,a{r^2},a{r^3},a{r^4},a{r^5}$ be six terms of a $G.P.$ where 'a' is first term andr is  common ratio.

According to given

Vedclass · Accepted Answer

$728$

Vedclass · Answer

Let $a,ar,a{r^2},a{r^3},a{r^4},a{r^5}$ be six terms of a $G.P.$ where 'a' is first term andr is  common ratio.

According to given condition, we have 

$a{r^3} - a = 5 \Rightarrow a\left( {{r^3} - 1} ight) = 52\,\,\,\,......\left( 1 ight)$

and $a + ar + a{r^2} = 26$

$ \Rightarrow a\left( {1 + r + {r^2}} ight) = 26\,\,\,\,\,\,......\left( 2 ight)$

To find : $a\left( {1 + r + {r^2} + {r^3} + {r^4} + {r^5}} ight)$

Consider

$a\left[ {1 + r + {r^2} + {r^3} + {r^4} + {r^5}} ight]$

$ = a\left[ {1 + r + {r^2} + {r^3}\left( {1 + r + {r^2}} ight)} ight]$

$ = a\left[ {1 + r + {r^2}} ight]\left[ {1 + {r^3}} ight]\,\,\,\,\,\,......\left( 3 ight)$

Divide $(1)$ by $(2)$, we get

$\frac{{{r^3} - 1}}{{1 + r + {r^2}}} = 2$

we know ${r^3} - 1 = \left( {1 + {r^3}} ight)\left( {1 + r + {r^2}} ight)$

$	herefore r - 1 = 2 \Rightarrow r = 3\,\,\,\,$  and $a = 2$

$	herefore a\left( {1 + r + {r^2} + {r^3} + {r^4} + {r^5}} ight)$

$ = a\left( {1 + r + {r^2}} ight)\left( {1 + {r^3}} ight)$

$ = 2\left( {1 + 3 + 9} ight)\left( {1 + 27} ight)$

$ = 26 	imes 28 = 728$

The difference between the fourth term and the first term of a Geometrical Progresssion is $52.$ If the sum of its first three terms is $26,$ then the sum of the first six terms of the progression is

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