The 4^{th} term of a G.P. is the square of it | vedclass

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(A) Let $a$ be the first term and $r$ be the common ratio of the $G.P.$Given,$a = -3$.The $n^{th}$ term of a $G.P.$ is given by $a_n = a r^{n-1}$.Ther

Vedclass · Accepted Answer

$-2187$

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(A) Let $a$ be the first term and $r$ be the common ratio of the $G.P.$Given,$a = -3$.The $n^{th}$ term of a $G.P.$ is given by $a_n = a r^{n-1}$.Therefore,$a_4 = a r^3 = -3 r^3$ and $a_2 = a r = -3 r$.According to the given condition,$a_4 = (a_2)^2$.Substituting the values: $-3 r^3 = (-3 r)^2$.$-3 r^3 = 9 r^2$.Dividing both sides by $-3 r^2$ (assuming $r 
eq 0$),we get $r = -3$.Now,the $7^{th}$ term is $a_7 = a r^6$.$a_7 = (-3) 	imes (-3)^6 = (-3)^1 	imes (-3)^6 = (-3)^7$.$a_7 = -2187$.Thus,the $7^{th}$ term of the $G.P.$ is $-2187$.

The $4^{th}$ term of a $G.P.$ is the square of its second term,and the first term is $-3$. Determine its $7^{th}$ term.

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