A sequence is defined by $a_{1} + a_{2} + . . . . + a_{n} = n^{2} a_{n}$ and $a_{1} = 50$ ,then $a_{200}$ is

\frac{1}{401}

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\frac{1}{402}

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\frac{1}{400}

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None of these

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Solution

The correct option is A $\frac{1}{402}$
We know that,
$a_{1} + a_{2} + . . . . + a_{n} = n^{2} a_{n}$
Bringing $a_{n}$ from LHS to RHS.
$a_{1} + a_{2} + . . . + a_{n - 1} = (n^{2} - 1) a_{n}$ $. . . (1)$
Also, $a_{1} + a_{2} + . . . + a_{n - 1} = (n - 1)^{2} a_{n - 1}$ $. . . (2)$
Comparing equation $(1)$ and $(2)$ ; we get
$(n^{2} - 1) a_{n} = (n - 1)^{2} a_{n - 1}$
$\Rightarrow \frac{a_{n}}{a_{n - 1}} = \frac{(n - 1)^{2}}{(n^{2} - 1)}$
$\Rightarrow \frac{a_{n}}{a_{n - 1}} = \frac{(n - 1)}{(n + 1)}$
$\Rightarrow \frac{a_{n - 1}}{a_{n - 2}} = \frac{(n - 2)}{n}$ , $\frac{a_{n - 2}}{a_{n - 3}} = \frac{(n - 3)}{(n - 1)}$ $, . . .,$ $\frac{a_{4}}{a_{3}} = \frac{3}{5}$ , $\frac{a_{3}}{a_{2}} = \frac{2}{4}$ , and $\frac{a_{2}}{a_{1}} = \frac{1}{3}$
Now, multiplying all these; we get
$\frac{a_{n}}{a_{1}} = \frac{2}{n (n + 1)}$
$\Rightarrow a_{n} = \frac{2 a_{1}}{n (n + 1)}$
Now, substitute the given value $a_{1} = 50$ and $n = 200$ ; we get
$a_{200} = \frac{2 \times 50}{200 \times (200 + 1)}$
$\Rightarrow a_{200} = \frac{100}{200 \times 201}$
$\Rightarrow a_{200} = \frac{1}{402}$

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A sequence is defined by a1+a2+....+an=n2an and a1=50,then a200 is

A sequence is defined by $a_{1} + a_{2} + . . . . + a_{n} = n^{2} a_{n}$ and $a_{1} = 50$ ,then $a_{200}$ is