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12. The sum of 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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Given : The sum of first three terms of a G.P. is $\frac{39}{10}$ and their product is 1.

Let three terms be $\frac{a}{r},a,ar$ .

$S_n=\frac{a(1-r^n)}{1-r}$

$S_3=\frac{a(1-r^3)}{1-r}=\frac{39}{10}$

$\frac{a}{r}+a+ar=\frac{39}{10}.........1$

Product of 3 terms is 1.

$\frac{a}{r}\times a\times ar=1$

$\Rightarrow a^3=1$

$\Rightarrow a=1$

Put value of a in equation 1,

$\frac{1}{r}+1+r=\frac{39}{10}$

$10(1+r+r^2)=39(r)$

$\Rightarrow 10r^2-29r+10=0$

$\Rightarrow 10r^2-25r-4r+10=0$

$\Rightarrow 5r(2r-5)-2(2r-5)=0$

$\Rightarrow (2r-5)(5r-2)=0$

$\Rightarrow r=\frac{5}{2},r=\frac{2}{5}$

The three terms of AP are $\frac{5}{2},1,\frac{2}{5}$ .

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