A) \[\frac{4}{5}\]
B) \[\frac{5}{4}\]
C) \[\frac{6}{5}\]
D) \[\frac{5}{6}\]
Correct Answer: B
Solution :
Given series is \[1+\frac{4}{5}+\frac{7}{{{5}^{2}}}+\frac{10}{{{5}^{3}}}+....\] Here, \[a=1,\,r=\frac{1}{5},\,d=3\] We know that, sum of n terms of an arithmetic-geometric series is \[{{S}_{n}}=\frac{a}{1-r}+\frac{dr(1-{{r}^{n-1}})}{{{(1-r)}^{2}}}-\frac{\{a+(n-1)d\}{{r}^{n}}}{1-r}\] \[\therefore \] \[{{S}_{n}}=\frac{1}{1-\frac{1}{5}}+\frac{3.\frac{1}{5}\left[ 1-{{\left( \frac{1}{5} \right)}^{n-1}} \right]}{{{\left( 1-\frac{1}{5} \right)}^{2}}}\] \[-\frac{\{1+(n-1)3\}}{1-\frac{1}{5}}{{\left( \frac{1}{5} \right)}^{n}}\] \[=\frac{5}{4}+\frac{15}{16}\left( 1-\frac{1}{{{5}^{n-1}}} \right)-\frac{1}{4}.\frac{(3n-2)}{{{5}^{n-1}}}\] But it is given that \[{{S}_{n}}=l+\frac{15}{16}\left( 1-\frac{1}{{{5}^{n-1}}} \right)-\frac{(3n-2)}{{{4.5}^{n-1}}}\] \[\therefore \] \[l=\frac{5}{4}\]
You need to login to perform this action.
You will be redirected in
3 sec
