Sequence And Series Question 390
Question: The sum of infinite terms of the following series $ 1+\frac{4}{5}+\frac{7}{5^{2}}+\frac{10}{5^{3}}+……… $ will be
[MP PET 1981; RPET 1997; Roorkee 1992; DCE 1996, 2000]
Options:
A) $ \frac{3}{16} $
B) $ \frac{35}{8} $
C) $ \frac{35}{4} $
D) $ \frac{35}{16} $
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Answer:
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=\frac{1}{5}+4.\frac{1}{5^{2}}+7.\frac{1}{5^{3}}+……… $ Subtracting $ ( 1-\frac{1}{5} )S=1+3.\frac{1}{5}+3.\frac{1}{5^{2}}+3.\frac{1}{5^{3}}+…….. $ $ =1+3( \frac{1}{5}+\frac{1}{5^{2}}+…… ) $
$ \Rightarrow $ $ \frac{4}{5}.S=1+3.\frac{1}{5}( \frac{1}{1-\frac{1}{5}} )=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}}{{{( 1-\frac{1}{5} )}^{2}}}=\frac{5}{4}+\frac{\frac{3}{5}}{\frac{16}{25}}=\frac{5}{4}+\frac{15}{16}=\frac{35}{16} $ . Aliter : Use $ S=[ 1+\frac{r}{1-r}\times diff\text{.}\ of\ A\text{.P}\text{.} ]\frac{1}{1-r} $
BETA
