If the sum of the first $2n$ terms of the A.P. $2,\ 5,\ 8\ ..$ is equal to the sum of the first $n$ terms of the A.P. $57,\ 59,\ 61,\ ...$, then find $n$

Given: the sum of the first $2n$ terms of the A.P. $2,\ 5,\ 8\ ..$ is equal to the sum of the first $n$ terms of the A.P. $57,\ 59,\ 61,\ ...$.

To do: To find $n$.

Solution:

In $2,\ 5,\ 8,...$ The sum of first $2n$ terms

$=\frac{2n}{2}[2( 2)+( 2n−1)( 3)]$

$=n( 6n+1)$

In $57,\ 59,\ 61,\ ...$ is The sum of n terms

$=\frac{n}{2}[2(57)+(n−1)(2)]$

$=n(56+n)$

Also given, $n(6n+1)=n(56+n)$

$\Rightarrow 6n+1=56+n$

$\Rightarrow 5n=55$

$\Rightarrow n=11$

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