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

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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 \Ri

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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 $1$:$(\frac{a}{r}) 	imes a 	imes (ar) = 1 \Rightarrow a^3 = 1 \Rightarrow a = 1$.Given, the sum of the terms is $\frac{39}{10}$:$\frac{a}{r} + a + ar = \frac{39}{10} \Rightarrow \frac{1}{r} + 1 + r = \frac{39}{10}$.Multiplying by $10r$: $10 + 10r + 10r^2 = 39r \Rightarrow 10r^2 - 29r + 10 = 0$.Factoring the quadratic: $10r^2 - 25r - 4r + 10 = 0 \Rightarrow 5r(2r - 5) - 2(2r - 5) = 0$.$(5r - 2)(2r - 5) = 0 \Rightarrow r = \frac{2}{5}$ or $r = \frac{5}{2}$.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 terms.

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