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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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