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# If the sum of n terms of the AP 55, 57, 59------ is equal to the sum of first 2n terms of the AP 1, 4, 710--------then n equals

$\begin{array}{1 1}100 \\ 12 \\ 11 \\ 13\end{array}$

Explanation : Sum of 2n terms of AP 1,4,7----
$=\large\frac{2n}{2}\;[2*1+(2n-1)3]$
$=n(6n-1)$
Sum of n terms of AP 55,57,59--------
$=\large\frac{n}{2}\;[55*2+(n-1)2]$
$=n(54+n)$
$=n(6n-1)=n(54+n)$
$6n-1=54+n$
$5n=55\quad\;n=11.$
edited Jan 24, 2014