Question

If the sum to $2n$ terms of the $ A.P. \text{ }2,\text{ }5,\text{ }8,11,... $ is equal to the sum to $n$ terms of the $\text{ }57,\text{ }59,\text{ }61,\text{ }63,\text{ }...\text{ }, $ then $n$ =

• A. $ 10 $
• B. $ 11 $
• C. $ 12 $
• D. $ 13 $

Answer 1

Answer: (B)

Let sum of 2n terms of the $ AP\,\,\,57,59,61,63 $
is $ {{S}_{n}} $ .
$ \therefore $ $ {{S}_{n}}=\frac{n}{2}[2\times 57+(n-1)2] $
$ \frac{n}{2}(2n+112) $
According to question $ {{S}_{2n}}={{S}_{n}} $
$ \Rightarrow $ $ n(6n+1)=\frac{n}{2}(2n+112) $
$ \Rightarrow $ $ 12n+2=2n+112 $
$ \Rightarrow $ $ 10n=110 $
$ \Rightarrow $ $ n=11 $