Question

find the sum of first n terms of an a.p ..whose n trem is 3n+ 1​

Answer 1

3n+1

put n=1

3×1+1

3+1

4 = a1

put n= 2

3×2+1

6+1

7=a2

put n=3

3×3+1

9+1

10=a3

a2-a1 = d

7-4=d

3=d

an= a+(n-1)d

= 4+(n-1)3

= 4+(n-1)

3= 4+(n-1)

3+1 = 4+n

4= 4n

1=n

so, the sum of first n terms is 1...

hope it helps you