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

Given:

The second term of an A.P. is 14 and the third term is 18. 

To do:

We have to find the sum of the first 51 terms of the A.P.

Solution: 

Let the first term and the common difference of the A.P. be $a$ and $d$ respectively.

We know that,

$a_n=a+(n-1)d$

This implies,

$a_2=a+(2-1)d$

$14=a+d$

$a=14-d$.......(i)

$a_3=a+(3-1)d$

$18=a+2d$

$18=14-d+2d$       (From (i))

$d=18-14$

$d=4$

Therefore,

$a=14-d$

$=14-4$

$=10$

We know that,

$S_{n}=\frac{n}{2}[2 n+(n-1) d]$

$S_{51}=\frac{51}{2}[2 \times(10)+(51-1) \times 4]$

$=\frac{51}{2}[20+50 \times 4]$

$=\frac{51}{2}(20+200)$

$=\frac{51}{2} \times 220$

$=51 \times 110$

$=5610$

The sum of the first 51 terms of the A.P. is $5610$.