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(C) Let the terms of the $G.P.$ be $a, ar, ar^2, ar^3, ar^4, ar^5$,where $a$ is the first term and $r$ is the common ratio.According to the given cond

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$728$

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(C) Let the terms of the $G.P.$ be $a, ar, ar^2, ar^3, ar^4, ar^5$,where $a$ is the first term and $r$ is the common ratio.According to the given conditions:$ar^3 - a = 52 \Rightarrow a(r^3 - 1) = 52 \quad ......(1)$$a + ar + ar^2 = 26 \Rightarrow a(1 + r + r^2) = 26 \quad ......(2)$We know that $r^3 - 1 = (r - 1)(r^2 + r + 1)$.Dividing equation $(1)$ by equation $(2)$:$\frac{a(r - 1)(r^2 + r + 1)}{a(1 + r + r^2)} = \frac{52}{26}$$r - 1 = 2 \Rightarrow r = 3$.Substituting $r = 3$ in equation $(2)$:$a(1 + 3 + 9) = 26$ $\Rightarrow 13a = 26$ $\Rightarrow a = 2$.The sum of the first six terms is $S_6 = a(1 + r + r^2 + r^3 + r^4 + r^5) = a(1 + r + r^2)(1 + r^3)$.$S_6 = 26 	imes (1 + 3^3) = 26 	imes (1 + 27) = 26 	imes 28 = 728$.

The difference between the fourth term and the first term of a Geometric Progression 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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