Question

Find the sum of GP 0.9 + 0.99 + 0.999 + ...... To n terms

Answer 1

We have to determine the sum of the GP

0.9+0.99+0.999+.....

= (1-0.1) + (1-0.01) +(1-0.001)+......... n

= (1+1+1+1+....... n) +( -0.1-0.01-0.001- ......... )

= $(1 \times n) - (0.1+0.01+0.01+......)$

= $(n) - (0.1+0.01+0.01+......)$

Sum of geometric progression upto 'n' terms with first term as 'a', common ratio as 'r' is given by the formula $\frac{a(1-r^n)}{(1-r)}$

= $n - \frac{0.1(1-(0.1)^n)}{(1-0.1)}$

= $n - \frac{0.1(1-(0.1)^n)}{0.9}$

= $n - \frac{1}{9}(1-(0.1)^n)}$ is the sum of the given geometric progression.

Answer 2

here's your solution dear...