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

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Question

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

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$\frac{5}{2}, 1, \frac{2}{5}$

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(A) Let the first three terms of the $G.P.$ be $\frac{a}{r}, a, ar$.Given,the product of the terms is $1$:$(\frac{a}{r}) 	imes a 	imes (ar) = 1$$a^3 = 1 \Rightarrow a = 1$.Given,the sum of the terms is $\frac{39}{10}$:$\frac{a}{r} + a + ar = \frac{39}{10}$Substitute $a = 1$:$\frac{1}{r} + 1 + r = \frac{39}{10}$$1 + r + r^2 = \frac{39}{10}r$$10r^2 + 10r + 10 = 39r$$10r^2 - 29r + 10 = 0$$10r^2 - 25r - 4r + 10 = 0$$5r(2r - 5) - 2(2r - 5) = 0$$(5r - 2)(2r - 5) = 0$$r = \frac{2}{5}$ or $r = \frac{5}{2}$.If $r = \frac{5}{2}$,the terms are $\frac{1}{5/2}, 1, \frac{5}{2}$,which are $\frac{2}{5}, 1, \frac{5}{2}$.If $r = \frac{2}{5}$,the terms are $\frac{1}{2/5}, 1, \frac{2}{5}$,which are $\frac{5}{2}, 1, \frac{2}{5}$.

The sum of the first three terms of a $G.P.$ is $\frac{39}{10}$ and their product is $1$. Find the common ratio and the terms.

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