0.6 + 0.66 + 0.666 + …. to n terms.
Let us take 6 as a common term so we get,
6(0.1 + 0.11 + 0.111 + …n terms)
Now multiply and divide by 9 we get,
\( \dfrac{6}{9}\) [0.9 + 0.99 + 0.999 + …+ n terms]
\( \dfrac{6}{9}\) [\( \dfrac{9}{10}\) + \( \dfrac{9}{100}\) + \( \dfrac{9}{1000}\) + …+ n terms]
This can be written as
\( \dfrac{6}{9}(1 – \dfrac{1}{10}) + (1 – \dfrac{1}{100}) + (1 -\dfrac{1}{1000}) + … + n terms]\)
\( \dfrac{6}{9}( \dfrac{n – 1}{9}) (1 – \dfrac{1}{10n})\)
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 \( \dfrac{81}{2}\).