9 + 99 + 999 + … to n terms.
The given terms can be written as
(10 – 1) + (100 – 1) + (1000 – 1) + … + n terms
(10 + 102 + 103 + … n terms) – n
By using the formula,
Sum of GP for n terms =
Where, a = 10, r = 10, n = n
=\( \dfrac{10 (10^n – 1)}{(10-1) }-n\)
= \( \dfrac{10}{9}\)(10n – 1) – n
=\( \dfrac{1}{9}\) [10n+1 – 10 – 9n]
= \( \dfrac{1}{9}\) [10n+1 – 9n – 10]
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 \( \dfrac{81}{2}\).