logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  Sequence and Series
0 votes

$a_{1}=50$ and $a_{1}+a_{2}+....+a_{n}=n^2a_{n}$ $\forall\;n \geq 1$, value of $a_{50}$ equal to

$(a)\;\large\frac{1}{51}\qquad(b)\;\large\frac{2}{51}\qquad(c)\;\large\frac{1}{101}\qquad(d)\;\large\frac{2}{101}$

Can you answer this question?
 
 

1 Answer

0 votes
Answer : (b) $\;\large\frac{2}{51}$
Explanation : $\;a_{1}+a_{2}+......+a_{n}=n^2a_{n}$
$\;a_{1}+a_{2}+......+a_{n-1}=(n^2-1)a_{n}$
$(n-1)^2 a_{n-1}=(n^2-1)\;a_{n}$
$a_{n}=\large\frac{n-1}{n+1}\;a_{n-1}$
$a_{n}=\large\frac{(n-1)}{(n+1)}\;.\large\frac{(n-2)}{(n)}\;a_{n-2}$
$=\large\frac{(n-1)}{(n+1)}\;.\large\frac{(n-2)}{(n)}\;.\large\frac{(n-3)}{(n-1)}\;a_{n-3}$
$=\large\frac{(n-1)(n-2)(n-3)....1}{(n+1) n (n-1).....3}\;a_{1}$
$a_{n}=\large\frac{2.1}{(n+1) . n}\;a_{1}$
$a_{n}=\large\frac{2 (50)}{(n+1) . n}$
$a_{n}=\large\frac{2 (50)}{(51)(50)}$
$a_{50}=\large\frac{2}{51}\;.$
answered Jan 23, 2014 by yamini.v
 

Related questions

Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...