0.5 + 0.55 + 0.555 + …. to n terms
Let us take 5 as a common term so we get,
5(0.1 + 0.11 + 0.111 + …n terms)
Now multiply and divide by 9 we get,
\( \dfrac{5}{9}\) [0.9 + 0.99 + 0.999 + …+ to n terms]
\( \dfrac{5}{9}[\dfrac{9}{10} + \dfrac{9}{100} + \dfrac{9}{1000} + … + n terms]\)
This can be written as
\( \dfrac{5}{9}(\dfrac{n – 1}{9} )(1 – \dfrac{1}{10^n})\)
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 \( \dfrac{81}{2}\).