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

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$728$

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(C) Let the terms of the Geometric Progression $(G.P.)$ 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:$1$) The difference between the fourth term and the first term is $52$: $ar^3 - a = 52 \Rightarrow a(r^3 - 1) = 52$ ... $(1)$$2$) The sum of the first three terms is $26$: $a + ar + ar^2 = 26 \Rightarrow a(1 + r + r^2) = 26$ ... $(2)$We know that $r^3 - 1 = (r - 1)(r^2 + r + 1)$.Dividing equation $(1)$ by equation $(2)$:$\frac{a(r^3 - 1)}{a(1 + r + r^2)} = \frac{52}{26}$$\frac{(r - 1)(r^2 + r + 1)}{(r^2 + r + 1)} = 2$$r - 1 = 2 \Rightarrow r = 3$.Substituting $r = 3$ into equation $(2)$:$a(1 + 3 + 3^2) = 26 \Rightarrow a(1 + 3 + 9) = 26 \Rightarrow 13a = 26 \Rightarrow a = 2$.The sum of the first six terms is $S_6 = a(1 + r + r^2 + r^3 + r^4 + r^5) = a(1 + r + r^2)(1 + r^3)$.$S_6 = 2(1 + 3 + 9)(1 + 3^3) = 2(13)(1 + 27) = 26 	imes 28 = 728$.

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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