Question

the sum of first n terms of an AP is given by Sn=7n²-3n. find the nth term of the AP.

Answer 1

Let a=first term

d=common difference.

a1=S1......(1)

a1+a2=S2.............(2)

Given Sn=7n²-3n

Putting n as 1 in the 1st term ,

S1=7*1²-3*1

=7-3

=4

So first term is 4.

S2=7*2²-3*2

=28-6

=22

S2-S1=a2[By subtracting (1) and (2)]

So 2nd term=22-4

=18.

So d=18-4

=14

We also know that:

nth term is a+(n-1)d

=4+(n-1)*14

=4+14n-14

=14n-10

The answer is 14n-10.

Hope it helps.

Answer 2

we have Sn=7n²-3n

putting n=4
S4=7×4²-3×4=100

putting n=2

S2=7×2²-3×2=22

we know Sn=n/2 {2a+(n-1)d}

S2=2/2 {2a+(2-1)d}
22=2a+d ----------(1)

again S4=4/2 {2a+(4-1)d}
100=2{2a+3d}
50=2a+3d-----------(2)

substracting( 1)from (2)


we have d=14
so a=4

so nth term of AP=4+(n-1)14=4+14n-14





Hence nth term is 4+14n-14