A) \[\frac{1}{5}\]
B) \[\frac{2}{5}\]
C) \[\frac{4}{5}\]
D) \[\frac{3}{5}\]
Correct Answer: D
Solution :
Given, \[\frac{\text{sum of Ist three terms}}{\text{sum of 1st six terms}}=\frac{125}{152}\] \[\Rightarrow \] \[\frac{a+ar+a{{r}^{2}}}{a+ar+a{{r}^{2}}+a{{r}^{3}}+a{{r}^{4}}+a{{r}^{5}}}=\frac{125}{152}\] \[\Rightarrow \] \[\frac{1+r+{{r}^{2}}}{1+r+{{r}^{2}}+{{r}^{3}}+{{r}^{4}}+{{r}^{5}}}=\frac{125}{152}\] \[\Rightarrow \] \[\frac{1+r+{{r}^{2}}}{(1+r+{{r}^{2}})(1+{{r}^{3}})}=\frac{125}{152}\] \[\Rightarrow \] \[\frac{1}{1+{{r}^{3}}}=\frac{125}{152}\] \[\Rightarrow \] \[1+{{r}^{3}}=\frac{152}{125}\] \[\Rightarrow \] \[{{r}^{3}}=\frac{152}{125}-1=\frac{27}{125}={{\left( \frac{3}{5} \right)}^{3}}\] \[\Rightarrow \] \[r=\frac{3}{5}\]
You need to login to perform this action.
You will be redirected in
3 sec
