A) \[\frac{3}{16}\]
B) \[\frac{35}{8}\]
C) \[\frac{35}{4}\]
D) \[\frac{35}{16}\]
Correct Answer: D
Solution :
Let the sum to infinity of the arithmetico-geometric series be \[S=1+4.\frac{1}{5}+7.\frac{1}{{{5}^{2}}}+10.\frac{1}{{{5}^{3}}}+........\] \[\Rightarrow \] \[\frac{1}{5}S=\text{ }\frac{1}{5}+4.\frac{1}{{{5}^{2}}}+7.\frac{1}{{{5}^{3}}}+.........\] 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 \[S=\left[ 1+\frac{r}{1-r}\times \text{diff}\text{.}\ \text{of}\ \text{A}\text{.P}\text{.} \right]\frac{1}{1-r}\]
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