Q. 45.0( 1 Vote )

# If the sum of the first n terms of an A.P. is given by Sn = (3n2- n), find its(i) first term (ii) common difference(iii) nth term.

Sn = 3n2 – n

Taking n = 1, we get

S1 = 3(1)2 - (1)

S1 = 3 – 1

S1 = 2

a1 = 2

Taking n = 2, we get

S2 = 3(2)2 – 2

S2 = 12 – 2

S2 = 10

a2 = S2 – S1 = 10 – 2 = 8

Taking n = 3, we get

S3 = 3(3)2 – 3

S3 = 27 – 3

S3 = 24

a3 = 24 – 10 = 14

So, a = 1,

d = a2 – a1 = 8 - 2 = 6

Now, we have to find the 15th term

an = a + (n – 1)d

an = 2 + (n – 1)6

an = 2 + 6n – 6

an = - 4 + 6n

Hence, the nth term is 4n - 3.

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