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Question
Find the sum of first 28 terms of an A.P. whose nth term is 4n – 3
Solution
n = 28
tn = 4n – 3
t1 = 4 × 1 – 3 = 1
t2 = 4 × 2 – 3 = 5
t28 = 4 × 28 – 3
= 112 – 3 = 109
∴ a = 1, d = t2 – t1 = 5 – 1 = 4
l = 109
Sn = `"n"/2` (2a + (n – 1)d)
S28 = `28/2` (2 × 1 + 27 × 4)
= 14(2 + 108)
= 14 × 110
= 1540
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