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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.

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Solution

1-
A = 2, d = 6 - 2 = 4,
n = 11
Sum of n terms = Sn=n2[2a+(n1)d]
Substituting = 112[2(2)+(111)4]
= 11/2[4+(10)4]=112[4+40]
= 11×22
= 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 = Sn=n2[2a+(n1)d]
=132[12+(12)6]
=390

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

6 = 2d
d = 3
a = - 1
S51=512(2(1)+(511)3)

S51=512(2+150)
S51=51×74=3774


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