Sequence And Series Question 612
Question: The sum of an infinite geometric series is 3. A series, which is formed by squares of its terms, have the sum also 3. First series will be
[UPSEAT 1999]
Options:
A) $ \frac{3}{2},\frac{3}{4},\frac{3}{8},\frac{3}{16},….. $
B) $ \frac{1}{2},\frac{1}{4},\frac{1}{8},\frac{1}{16},….. $
C) $ \frac{1}{3},\frac{1}{9},\frac{1}{27},\frac{1}{81},….. $
D) $ 1,-\frac{1}{3},,\frac{1}{3^{2}},-\frac{1}{3^{3}},….. $
Show Answer
Answer:
Correct Answer: A
Solution:
$ {{(S_1)}{\infty }}=\frac{a}{1-r}=3 $ or $ a=3,(,1-r) $ ?..(i) $ {{(S_2)}{\infty }}=\frac{a^{2}}{1-r^{2}}=3 $ or $ a^{2}=3,(1-r^{2}) $ or $ 9,{{(1-r)}^{2}}=3,(1-r^{2}) $ [by (i)] or $ 3,(1-2r+r^{2})=1-r^{2} $ or $ 2r^{2}-3r+1=0 $ or $ (r-1),(2r-1)=0 $ ,
$ \therefore $ $ r=1,\frac{1}{2} $ If $ r=1, $ then $ a=3(1-1)=0 $ which is impossible. If $ r=\frac{1}{2}, $ then $ a=3,( 1-\frac{1}{2} )=3/2 $ So first series is 3/2, 3/4, 3/8, 3/16,…..
BETA
