Question

if the nth term of an A.P is (3n+1), then the sum of its first 'n' terms is​

Answer 1

Answer:

let n=1

3*1+1=4=a

let n=2

3*2+1=7=a2

let n=3

3*3+1=10=a3

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Sn=n/2(2a+(n-1)d)

=n/2(2*4+(n-1)5)

=n/2(8+5n-5)

=n/2(3+5n)

Answer 2

Given:

A.P. with the nth term (3n+1)

To Find:

sum of its first 'n' terms

Solution:

nth term of A.P. is  (3n+1)

So, For 1st term put n = 1

a = 3(1) + 1 = 4

2nd term, n =2

a₂ = 3(2) + 1 = 7

3ed term, n = 3

a₃ = 3(3) + 1 = 10

Therefore,

common differance = d = a₂ - a = 7 - 4 = 3

First term = a = 4

The sum of n terms of A.P. is given by

= n/2 × ( 2a + (n-1) d )

=  n/2 × ( 2 (4)+ (n-1) 3 )

=  n/2 × ( 8+ 3n-3 )

=  n/2 × ( 5+ 3n)

= (3n² + 5n ) / 2

Hence,  sum of its first 'n' terms is​ (3n² + 5n ) / 2.