Let a_{1}, a_{2}, a_{3}, dots be a G.P. of in | vedclass

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(B) Let the $G$.$P$. be $a, ar, ar^2, \dots$. Given $a_2 \cdot a_3 \cdot a_4 = 64$, we have $(ar)(ar^2)(ar^3) = 64$, which implies $a^3 r^6 = 64$, so

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$3252$

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(B) Let the $G$.$P$. be $a, ar, ar^2, \dots$. Given $a_2 \cdot a_3 \cdot a_4 = 64$, we have $(ar)(ar^2)(ar^3) = 64$, which implies $a^3 r^6 = 64$, so $ar^2 = 4$. Given $a_1 + a_3 + a_5 = \frac{813}{7}$, we have $a + ar^2 + ar^4 = \frac{813}{7}$. Substituting $a = \frac{4}{r^2}$, we get $\frac{4}{r^2} + 4 + 4r^2 = \frac{813}{7}$. Let $x = r^2$. Then $\frac{4}{x} + 4 + 4x = \frac{813}{7}$ $\Rightarrow 4 + 4x + 4x^2 = \frac{813}{7}x$ $\Rightarrow 28 + 28x + 28x^2 = 813x$. $28x^2 - 785x + 28 = 0$. Solving this quadratic, $x = \frac{785 \pm \sqrt{785^2 - 4(28)(28)}}{2(28)} = \frac{785 \pm 783}{56}$. Since the terms are increasing, $r > 1$, so $x = r^2 = \frac{1568}{56} = 28$. We need $a_3 + a_5 + a_7 = ar^2 + ar^4 + ar^6 = ar^2(1 + r^2 + r^4) = 4(1 + 28 + 28^2) = 4(1 + 28 + 784) = 4(813) = 3252$.

Let $a_{1}, a_{2}, a_{3}, \dots$ be a $G$.$P$. of increasing positive terms such that $a_{2} \cdot a_{3} \cdot a_{4} = 64$ and $a_{1} + a_{3} + a_{5} = \frac{813}{7}$. Then $a_{3} + a_{5} + a_{7}$ is equal to:

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