Let a_1, a_2, a_3, ldots be a G.P. of increas | vedclass

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(C) Let the first term of the $G.P.$ be $a$ and the common ratio be $r$ ($r > 1$ since the terms are increasing).$a_3 a_5 = (ar^2)(ar^4) = a^2 r^6 = 7

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$129$

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(C) Let the first term of the $G.P.$ be $a$ and the common ratio be $r$ ($r > 1$ since the terms are increasing).$a_3 a_5 = (ar^2)(ar^4) = a^2 r^6 = 729 \Rightarrow ar^3 = 27 \dots (i)$$a_2 + a_4 = ar + ar^3 = \frac{111}{4} \dots (ii)$Substitute $(i)$ into $(ii)$:$ar + 27 = \frac{111}{4} \Rightarrow ar = \frac{111}{4} - 27 = \frac{111 - 108}{4} = \frac{3}{4} \dots (iii)$Divide $(i)$ by $(iii)$:$\frac{ar^3}{ar} = \frac{27}{3/4}$ $\Rightarrow r^2 = 27 	imes \frac{4}{3} = 36$ $\Rightarrow r = 6$ (since terms are positive).From $(iii)$,$a(6) = \frac{3}{4} \Rightarrow a = \frac{3}{24} = \frac{1}{8}$.Now,$24(a_1 + a_2 + a_3) = 24(a + ar + ar^2) = 24a(1 + r + r^2)$.$= 24 	imes \frac{1}{8} (1 + 6 + 36) = 3(43) = 129$.

Let $a_1, a_2, a_3, \ldots$ be a $G.P.$ of increasing positive numbers. If $a_3 a_5 = 729$ and $a_2 + a_4 = \frac{111}{4}$,then $24(a_1 + a_2 + a_3)$ is equal to

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