Question

Find the sum of first $51$ terms of an AP, whose second and third terms are $14$ and $18$ respectively.

• A. $5610$
  
• B. $5800$
• C. $6120$
• D. $6680$

Answer 1

The correct option is A $5610$

As we know $n^{th}$ term, $a_n=a+(n-1)d$

Also the sum of first $n$ terms, $S_n=\frac{n}{2}(2a+(n-1\left)d\right)$, where $a$ & $d$ are the first term amd common difference of an AP respectively.

Since, Second term $=a_2=a+d=14$................(1)

Third term $=a_3=a+2d=18$ ......................(2)
On subtracting equation $\left(1\right)$ from $\left(2\right)$, we get
$d=4$
Putting $d=4$ in equation $\left(1\right)$, we get
$a=14-4=10$
Now, Required sum=$\frac{n}{2}[2a+(n-1\left)d\right]$
=$\frac{51}{2}[2\times 10+(51-1)\times 4]$
=$51\times 110=5610$