Question

Find Sum of all terms of an A.P whose nth term is (6n-5)

Answer 1

Answer:

3(n)^2 - 2n where n = no. of terms of AP

Step-by-step explanation:

nth term is given by = [6n-5]

1st term = 6(1) - 5 = 1

2nd term = 6(2) - 5 = 7

3rd term = 6(3) - 5 = 13

and the AP goes on with a common difference of 6

Sum of an AP

= n/2×{2a+(n-1)d}

=n/2×{2(1)+(n-1)6}

=n/2×{2+6n-6}

=n/2×{6n-4}

=n{3n-2}

=3(n)^2 - 2n

Answer 2

Answer:

let n =1

=6(1)-5

=6-5

=1

let n=2

=6(2)-5

=12-5

=7

d=7-1=6

A P =1,7,13.......

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

nth term =6n-5

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

Sn=6n-5/2(2×1+(6n-5-1)6

Sn=6n-5(1+(6n-6)3)

Sn=6n-5(1+18n-18)

Sn=6n-5(18n-17)