Find the sum of the first51 terms of the A.P.: whose second term is 2 and fourth term is 8.

Given:

The second term of an A.P. is 2 and the fourth term is 8.
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,

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

$2=a+d$

$a=2-d$.......(i)

$a_4=a+(4-1)d$

$8=a+3d$

$8=2-d+3d$       (From (i))

$2d=8-2$

$d=\frac{6}{2}$

$d=3$

\( \therefore a=2-d=2-3=-1 \)

We know that,

\( S_{n}=\frac{n}{2}[2 n+(n-1) d] \)
\( S_{51}=\frac{51}{2}[2 \times(-1)+(51-1) \times 3] \)
\( =\frac{51}{2}[-2+50 \times 3] \)
\( =\frac{51}{2}(-2+150) \)
\( =\frac{51}{2} \times 148 \)

\( =51 \times 74 \)

\( =3774 \)

The sum of the first 51 terms of the A.P. is $3774$.