Sum

The sum of first n terms of a certain series is given as 2n^{2} – 3n. Show that the series is an A.P.

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#### Solution

Given S_{n} = 2n^{2} – 3n

S_{1} = 2(1)^{2} – 3(1) = 2 – 3 = – 1

⇒ t_{1} = a = – 1

S_{2} = 2(2^{2}) – 3(2) = 8 – 6 = 2

t_{2} = S_{2} – S_{1} = 2 – (– 1) = 3

∴ d = t_{2} – t_{1} = 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.

Concept: Series

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