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 20th term and nth 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) a3b and ab3
(iii) –8 and –2
Insert 5 geometric means between \( \dfrac{32}{9}\) and \( \dfrac{81}{2}\).