9 + 99 + 999 + … to n terms.

The given terms can be written as

(10 – 1) + (100 – 1) + (1000 – 1) + … + n terms

(10 + 10^{2} + 10^{3} + … n terms) – n

By using the formula,

Sum of GP for n terms =

Where, a = 10, r = 10, n = n

=\( \dfrac{10 (10^n – 1)}{(10-1) }-n\)

= \( \dfrac{10}{9}\)(10^{n} – 1) – n

=\( \dfrac{1}{9}\) [10^{n+1} – 10 – 9n]

= \( \dfrac{1}{9}\) [10^{n+1} – 9n – 10]

Construct 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) **a^{3}b and ab^{3}

**(iii) **–8 and –2

Insert 5 geometric means between \( \dfrac{32}{9}\) and \( \dfrac{81}{2}\).