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Question
The sum of first n terms of a certain series is given as 2n2 – 3n. Show that the series is an A.P.
Solution
Given Sn = 2n2 – 3n
S1 = 2(1)2 – 3(1) = 2 – 3 = – 1
⇒ t1 = a = – 1
S2 = 2(22) – 3(2) = 8 – 6 = 2
t2 = S2 – S1 = 2 – (– 1) = 3
∴ d = t2 – t1 = 3 – (– 1) = 4
Consider a, a + d, a + 2d, ….….
– 1, – 1 + 4, – 1 + 2(4), …..…
– 1, 3, 7, ….
Clearly, this is an A.P. with a = – 1, and d = 4.
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