Question

if the sum of first n terms of an AP is 1/2(3n^2+7n) then find its nth term. hence write its 20th term

Answer 1

Sn = 1/2(3n^2+7n)
Sn = 3n^2/2+7n/2 = 3n^2+7n/2
Put n = n-1
Sn-1 = {3n^2-6n+3/2} +{7n-7/2}
= 3n^2+n-4/2
Nth term = Sn-Sn-1
= {3n^2+7n/2} - {3n^2-n+4}
= 6n+4/2
=2(3n+2)/2
=3n+2
We have to find the twentieth term i.e. a20
Put n=20
a20= 3×20+2
=62
HOPE IT HELPS

Answer 2

Answer:

Step-by-step explanation:

Given :-

The sum of first n terms of an AP is 1/2(3n² + 7n).

To Find :-

Its nth term and 20th term.

Formula to be used :-

S(n) = 1/2[2a + (n - 1)d]

a(n) = a + (n - 1)d

Solution :-

S(n) = 1/2[3n² + 7n]

S(1) = 1/2[3 × (1)² + 7(1)]

= 1/2[3 + 7]

= 1/2 × 10

= 5

S(2) = 1/2[3 (2)² + 7 × 2]

= 1/2[12 + 14]

= 1/2 × 26

= 13

a(1) = S(1) = 5

a(2) = S(2) - S(1)

= 13 - 5

= 8

a(2) = S(2) - a(1)

= 8 - 5

= 3

Now, A.P. is 5, 8, 11....

nth term a(n) = a + (n - 1)d

= 5 + (n - 1)3

= 3n + 2

a(20) = 3 × 20 + 2

= 62

Hence, the 20th term is 62.