\( a_n=\dfrac{a_{n-1}}{n}\) and a1 = -1
Then,
a2 = \( \dfrac{a_1}{2}\) = \(- \dfrac{1}{2}\)
a3 = \( \dfrac{a_2}{3}\) = \( - \dfrac{1}{6}\)
a4 = \( \dfrac{a_3}{4}\) = \( - \dfrac{1}{24}\)
a5 = \( \dfrac{a_4}{5}\) = \( - \dfrac{1}{120}\)
Thus, the first 5 terms of the sequence are -1, \( - \dfrac{1}{2}\), \( - \dfrac{1}{6}\), \( - \dfrac{1}{24}\) and \( - \dfrac{1}{120}\).
Hence, the corresponding series is
-1 + (\( - \dfrac{1}{2}\)) + (\( - \dfrac{1}{6}\)) + (\( - \dfrac{1}{24}\)) + (\( - \dfrac{1}{120}\)) + …….
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 \( \dfrac{81}{2}\).