Given G.P. is \( \dfrac{5}{2}\), \( \dfrac{5}{4}\), \( \dfrac{5}{8}\), ………

Here, a = First term = \( \dfrac{1}{2} \)

r = Common ratio = (5/4)/(5/2) = \( \dfrac{1}{2} \)

Thus, the 20^{th} term and n^{th} term

\( a_{20}=ar^{20-1} =\dfrac{5}{2}(\dfrac{1}{2})^{19} \)

= \( \dfrac{5}{(2)^{20}}\)

\( a_n=ar^{n-1} =\dfrac{5}{2}(\dfrac{1}{2})^{n-1} \)

= \( \dfrac{5}{(2)^n}\)

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 \( \dfrac{81}{2}\).