Question

The 4th term of a G.P is square of its second term and the first term is -3, determine bits 7th term.​

Answer 1

Hi mate

Given a= -3 , T4=( T2)^2 this implies ar^3= (ar)^2 -> r= a .now T7 = ar^n-1 = a×a^6 = -2187.

Answer 2

$\huge\boxed{ \underline{ \underline{ \bf{Answer}}}}$

GIVEN :

4th term = (2nd term)^2,

a = -3

TO FIND :

7th term of the GP

PROCESS :

$\implies \: a_{4} \: = \: {(a_ {2})}^{2}$

$\implies \: a {r}^{3} \: = \: {(ar)}^{2}$

$\implies \: a {r}^{3} \: = \: {a}^{2} {r}^{2}$

$\implies \: ( - 3) {r}^{3} \: = \: {( - 3)}^{2} {r}^{2}$

$\implies \: - 3r \: = \: 9$

$\implies \: r \: = \: - 3$

Now,

$\implies \: a _{7} \: = \: a {r}^{6}$

$\implies \:a_{7} \: = \: ( - 3) {( - 3)}^{6}$

$\implies \: a _{7} \: = \: {( - 3)}^{7}$

$\boxed{ \sf{Hence, \: the \: 7th \: term \: is \: {( - 3)}^{7} }}$