Question:
The first term of a GP is -3 and the square of the second term is equal to its $4^{\text {th }}$ term. Find its $7^{\text {th }}$ term.
Solution:
It is given that the first term of GP is -3.
So, a = -3
It is also given that the square of the second term is equal to its 4th term.
$\therefore\left(a_{2}\right)^{2}=a_{4}$
$n^{\text {th }}$ term of GP, $a_{n}=a r^{n-1}$
So, $a_{2}=a r ; a_{4}=a r^{3}$
$(a r)^{2}=a r^{3} \rightarrow a=r=-3$
Now, the $7^{\text {th }}$ term in the GP, a7 $=a r^{6}$
$a_{7}=(-3)^{7}=-2187$
Hence, the $7^{\text {th }}$ term of GP is $-2187$.
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