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16.  Find a G.P. for which sum of the first two terms is – 4 and the fifth term is 4 times the third term

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Given : sum of the first two terms is – 4 and the fifth term is 4 times the third term

Let first term be a and common ratio be r

a_5=4.a_3

\Rightarrow a.r^{5-1}=4.a.r^{3-1}

\Rightarrow a.r^{4}=4.a.r^{2}

\Rightarrow r^{2}=4

\Rightarrow r=\pm 2

If r=2, then

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

\Rightarrow \frac{a(1-2^2)}{1-2}=-4

\Rightarrow \frac{a(1-4)}{-1}=-4

\Rightarrow a(-3)=4

\Rightarrow a=\frac{-4}{3}

 

If r= - 2, then

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

\Rightarrow \frac{a(1-(-2)^2)}{1-(-2)}=-4

\Rightarrow \frac{a(1-4)}{3}=-4

\Rightarrow a(-3)=-12

\Rightarrow a=\frac{-12}{-3}=4

Thus, required GP is \frac{-4}{3},\frac{-8}{3},\frac{-16}{3},.........      or    4,-8,-16,-32,..........

Posted by

seema garhwal

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