a = 729 and a_{7} = 64

Let r be the common ratio of the G.P.

Then we know that, a_{n} = a r^{n–1}

a_{7} = ar^{7–1} = (729)r^{6}

⇒ 64 = 729 r^{6}

r^{6} = \( \dfrac{64}{729}\)

r^{6} = (\( \dfrac{2}{3}\))^{6}

r = \( \dfrac{2}{3}\)

And, we know that

\( s_n=\dfrac{a(1-r^n)}{1-r}\)

\( s_7=\dfrac{729[1-(\dfrac{2}{3})^7]}{1-\dfrac{2}{3}}\)

\(3\times 729=[1-(\dfrac{2}{3})^7]\)

\( (3)^7=\dfrac{(3)^7-(2)^7}{(3)^7}\)

= (3)^{7} - (2)^{7}

= 2187 - 128

= 2059

Answered by Abhisek | 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 \( \dfrac{81}{2}\).