The sum of first three terms of a G.P. is 14 and the sum of next three terms is 112 . Then $100^{ e

Question

The sum of first three terms of a G.P. is 14 and the sum of next three terms is 112 . Then $100^{	ext {th }}$ term of the G.P. is\

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(c) : Let $a, a r, a r^{2}, a r^{3}, a r^{4}, a r^{5}$ be the first six terms of the G.P., then \ a+a r+a r$^{2}$=14 ...(i)\ and $a r^{3}+a r^{4}+a r^{5}=112$ \Divide equation (i) by equation (ii), we get $\frac{1}{r^{3}}=\frac{1}{8}$\ $\Rightarrow \quad r^{3}=8 \quad \Rightarrow r=2$\ Substitute $r=2$ in equation (i), we get\ $a+2 a+4 a=14 \Rightarrow a=2$\ Hence, $100^{	ext {th }}$ term of the G.P. $=2.2^{(100-1)}=2.2^{99}=2^{100}$\