nth term of the G.P. \( \sqrt{3}\), \( \dfrac{1}{\sqrt{3}}\), \( \dfrac{1}{3\sqrt{3}}\), …
We know that,
t1 = a = \( \sqrt{3}\), r = \( \dfrac{\dfrac{1}{\sqrt{3}}}{\sqrt{3}}\)
= \( \dfrac{1}{\sqrt{3}\times \sqrt{3}}\) = \( \dfrac{1}{3}\)
By using the formula,
Tn = arn-1
Tn = \( \sqrt{3}(\dfrac{1}{3})^{n-1}\)
Answered by Sakshi | 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 \( \dfrac{81}{2}\).