Sequence And Series Question 490
Question: If $ A=1+r^{z}+r^{2z}+r^{3z}+…….\infty $ , then the value of r will be
Options:
A) $ A{{(1-A)}^{z}} $
B) $ {{( \frac{A-1}{A} )}^{1/z}} $
C) $ {{( \frac{1}{A}-1 )}^{1/z}} $
D) $ A{{(1-A)}^{1/z}} $
Show Answer
Answer:
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+[ \frac{r^{z}}{1-r^{z}} ]\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={{[ \frac{A-1}{A} ]}^{1/z}} $ .
BETA
