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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}.

Posted by

seema garhwal

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