The sum of the first three terms of a G.P. is | vedclass

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Question

Let the first three terms of the $G.P.$ be $\frac{a}{r}, a, ar$.Given the product is $-1$:$\left(\frac{a}{r}\right)(a)(ar) = -1$$a^3 = -1 \implies a =

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Let the first three terms of the $G.P.$ be $\frac{a}{r}, a, ar$.
Given the product is $-1$:
$\left(\frac{a}{r}\right)(a)(ar) = -1$
$a^3 = -1 \implies a = -1$.
Given the sum is $\frac{13}{12}$:
$\frac{a}{r} + a + ar = \frac{13}{12}$
Substituting $a = -1$:
$-\frac{1}{r} - 1 - r = \frac{13}{12}$
$-\frac{1+r+r^2}{r} = \frac{13}{12}$
$-12 - 12r - 12r^2 = 13r$
$12r^2 + 25r + 12 = 0$
Solving the quadratic equation:
$12r^2 + 16r + 9r + 12 = 0$
$4r(3r + 4) + 3(3r + 4) = 0$
$(4r + 3)(3r + 4) = 0$
$r = -\frac{3}{4}$ or $r = -\frac{4}{3}$.
For $r = -\frac{3}{4}$,the terms are $\frac{4}{3}, -1, \frac{3}{4}$.
For $r = -\frac{4}{3}$,the terms are $\frac{3}{4}, -1, \frac{4}{3}$.

The sum of the first three terms of a $G.P.$ is $\frac{13}{12}$ and their product is $-1$. Find the common ratio and the terms.

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