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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 $3774$.
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