Question

Find the sum of the first
(i) 11 terms of the A.P.: 2,6,10,14,...
(ii) 13 terms of the A.P.: -6,0,6,12,...
(iii) 51 terms of the A.P.: whose second term is 2 and fourth term is 8.

Answer 1

1-
A = 2, d = 6 - 2 = 4,
n = 11
Sum of n terms = $S_n=\frac{n}{2}[2a+(n-1\left)d\right]$
Substituting = $\frac{11}{2}\left[2\right(2)+(11-1\left)4\right]$
= $11/2[4+(10\left)4\right]=\frac{11}{2}[4+40]$
= $11\times 22$
= $242$

Hence Sum of terms of AP 2,6,10,14..........,32. is 242

2-
A = -6, d = 0 - (-6) = 6
n = 13
Sum of n terms = $S_n=\frac{n}{2}[2a+(n-1\left)d\right]$
=$\frac{13}{2}[-12+(12\left)6\right]$
=$390$

3-
2 = a + d
8 = a + 3d

6 = 2d
d = 3
a = - 1
$S_{51}=\frac{51}{2}\left(2\right(-1)+(51-1\left)3\right)$

$S_{51}=\frac{51}{2}(-2+150)$
$S_{51}=51\times 74=3774$