If $${\text{A}} = \frac{1}{{0.4}} + \frac{1}{{0.04}} + \frac{1}{{0.004}} + ....{\text{ upto 8 terms,}}$$         then what is the value of A?

Solution (By Examveda Team)

$$\eqalign{ & \frac{1}{{0.4}} + \frac{1}{{0.04}} + \frac{1}{{0.004}} + ....{\text{ upto 8 terms}} \cr & = \frac{{10}}{4} + \frac{{100}}{4} + \frac{{1000}}{4} + ....{\text{ upto 8 terms}} \cr & = \frac{1}{4}\left[ {10 + 100 + 1000 + ....{\text{ upto 8 terms}}} \right] \cr & = \frac{1}{4}\left[ {\frac{{10\left( {{{10}^8} - 1} \right)}}{{10 - 1}}} \right]\,\,\,\,\,\,\,\,\,\,\left( {\because {S_n} = \frac{{a\left( {{r^n} - 1} \right)}}{{r - 1}}} \right) \cr & = \frac{1}{{4 \times 9}} \times 10\left( {{{10}^4} + 1} \right)\left( {{{10}^4} - 1} \right) \cr & = \frac{{10}}{{4 \times 9}} \times \left( {{{10}^4} + 1} \right)\left( {{{10}^2} + 1} \right)\left( {10 + 1} \right)\left( {10 - 1} \right) \cr & = \frac{{10}}{{4 \times 9}} \times \left( {{{10}^4} + 1} \right)\left( {{{10}^2} + 1} \right)\left( {11} \right)\left( 9 \right) \cr & = \frac{{10 \times 10001 \times 101 \times 11}}{4} \cr & = 27777777.5 \cr} $$