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Question

If sn, the sum of first n terms of an AP is given by sn= (3n2-4n), then find its nth term.

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Solution

Let the sum of first n terms of the A.P.=Sn

Given: Sn = 3n2 – 4n ...(i)

Now,

Replacing n by (n –1) in (i), we get,

Sn – 1 = 3(n – 1)2 – 4(n – 1)

nth term of the A.P. an = Sn – Sn – 1

an = (3n2 – 4n) – [3(n – 1)2 – 4(n – 1)]

an = 3 [ n2 – (n – 1)2] – 4 [n – (n – 1)]

an = 3 (n2 n2 + 2n – 1) – 4 (n n + 1)

an = 3(2n –1) – 4

an = 6n – 3 – 4

an= 6n – 7

Thus, the nth term of the A.P = 6n – 7.


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