Question

If the sum of the first n terms of an AP is 4n-n2 what is the first term that is S1 ? What is the sum of first two terms? What is the second term similarly find the third and 10th and nth terms

Answer 1

Given sum of first n terms of the AP is

Sn = 4n - n²

Put n = 1, we get

S1 = 4*1 - 1²

= 4 – 1

= 3

So first term = 3

Now, sum of first two terms S2 = 4*2−2² (Put n=2)

= 8−4

= 4

So sum of first two terms = 4

Therefore Second term =S2 −S1

=4−3

=1

So second term = 1

Again S3 = 4×3 - 3² (Put n= 3)

= 12 – 9

= 3

Therefore Third term = S3 − S2

= 3 – 4

= – 1

So third term = -1

Again

S9 = 4×9−9² (Put n = 9)

= 36 – 81

= – 45

and

S10 = 4×10−10² (Put n = 10)

= 40 – 100

= – 60

Therefore Tenth term = S10 − S9

= – 60 – (– 45)

= – 60 + 45

= – 15

Now

Sn = 4n−n²

and Sn-1 = 4(n−1)−(n−1)2

= 4n − 4 − (n2 - 2n + 1)

= 4n − 4 − n2 + 2n - 1

= −n2 + 6n - 5

Therefore, nth term = Sn − Sn-1

= 4n−n2 −(−n2 +6n−5)

= 4n − n2 + n2 − 6n + 5

= 5 − 2n

So nth term of AP is 5 − 2n