Let’s consider a to be the first term and r to be the common ratio of the G.P.

Given, a = –3

And we know that,

a_{n} = ar^{n–1}

So, a_{4 }= ar^{3} = (–3) r^{3}

a_{2} = a r^{1} = (–3) r

Then from the question, we have

(–3) r^{3} = [(–3) r]^{2}

⇒ –3r^{3} = 9 r^{2}

⇒ r = –3

a_{7} = a r ^{7–1 }= a r^{6} = (–3) (–3)^{6} = – (3)^{7} = –2187

Therefore, the seventh term of the G.P. is –2187.

Answered by Pragya Singh | 1 year agoConstruct a quadratic in x such that A.M. of its roots is A and G.M. is G.

Find the geometric means of the following pairs of numbers:

**(i) **2 and 8

**(ii) **a^{3}b and ab^{3}

**(iii) **–8 and –2

Insert 5 geometric means between \( \dfrac{32}{9}\) and \( \dfrac{81}{2}\).