A) \[A{{(1-A)}^{z}}\]
B) \[{{\left( \frac{A-1}{A} \right)}^{1/z}}\]
C) \[{{\left( \frac{1}{A}-1 \right)}^{1/z}}\]
D) \[A{{(1-A)}^{1/z}}\]
Correct Answer: B
Solution :
\[A=1+{{r}^{z}}+{{r}^{2z}}+{{r}^{3z}}+........\infty \] \[A=1+[{{r}^{z}}+{{r}^{2z}}+{{r}^{3z}}+........\infty ]\] We know that sum of infinite G.P. is \[{{S}_{\infty }}=\frac{a}{1-r}(-1<r<1)\] Therefore, \[A=1+\left[ \frac{{{r}^{z}}}{1-{{r}^{z}}} \right]\Rightarrow A=\frac{1-{{r}^{z}}+{{r}^{z}}}{1-{{r}^{z}}}\] \[\therefore \] \[A=\frac{1}{1-{{r}^{z}}}\Rightarrow 1-{{r}^{z}}=\frac{1}{A}\Rightarrow {{r}^{z}}=\frac{A-1}{A}\] Hence\[r={{\left[ \frac{A-1}{A} \right]}^{1/z}}\].
You need to login to perform this action.
You will be redirected in
3 sec
