MCQ
The sum of infinite terms of the following series $1 + \frac{4}{5} + \frac{7}{{{5^2}}} + \frac{{10}}{{{5^3}}} + .........$ will be
- A$\frac{3}{{16}}$
- B$\frac{{35}}{8}$
- C$\frac{{35}}{4}$
- ✓$\frac{{35}}{{16}}$
$ \Rightarrow $ $\frac{1}{5}S = {\rm{ }}\frac{1}{5} + 4.\frac{1}{{{5^2}}} + 7.\frac{1}{{{5^3}}} + .........$
Subtracting $\left( {1 - \frac{1}{5}} \right)S = 1 + 3.\frac{1}{5} + 3.\frac{1}{{{5^2}}} + 3.\frac{1}{{{5^3}}} + ........$
$ = 1 + 3\left( {\frac{1}{5} + \frac{1}{{{5^2}}} + ......} \right)$
$ \Rightarrow $$\frac{4}{5}.S = 1 + 3.\frac{1}{5}\left( {\frac{1}{{1 - \frac{1}{5}}}} \right) = 1 + \frac{3}{4} = \frac{7}{4} \Rightarrow S = \frac{{35}}{{16}}$.
Aliter : Use direct formula ${S_\infty } = \frac{{ab}}{{1 - r}} + \frac{{dbr}}{{{{(1 - r)}^2}}}$
Here $a = 1,\;b = 1,\;d = 3,\;r = \frac{1}{5}$, therefore
${S_\infty } = \frac{1}{{1 - \frac{1}{5}}} + \frac{{3 \times 1 \times \frac{1}{5}}}{{{{\left( {1 - \frac{1}{5}} \right)}^2}}} = \frac{5}{4} + \frac{{\frac{3}{5}}}{{\frac{{16}}{{25}}}} = \frac{5}{4} + \frac{{15}}{{16}} = \frac{{35}}{{16}}$.
Aliter : Use $S = \left[ {1 + \frac{r}{{1 - r}} \times {\rm{diff}}{\rm{.}}\;{\rm{of}}\;{\rm{A}}{\rm{.P}}{\rm{.}}} \right]\frac{1}{{1 - r}}$
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