Question

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 n equals

• A. 10
• B. 12
• C. 11
• D. 13

Answer 1

The correct option is C 11

$\frac{2n}{2}[4+(2n-1\left)3\right]=\frac{n}{2}[2\times 57+(n-1{)}^2]$
$2(6n+1)=(144+2n-2)$
$2(6n+1)=(2n+112)$
$12n+2=2n+112$
${10}_n=110$
$n=11$