Question

Find the sum of the first 40 terms of an arithmetic sequence with general formula 3n+6​

Answer 1

Answer:

term is given to be 3n + 7.

First term = 3*1 + 7 = 10;

Second term = 3*2 + 7 = 13;

Third term = 3*3 + 7 = 16;

Fourth term = 3*4 + 7 = 19;

The arithmetic progression is, therefore, 10, 13, 16, 19,…. with the first term a = 10, and common difference d = 3.

Sum of the first “n” terms of an arithmetic progression is given by

S = [n/2]*{2*a + (n-1)d}.

Sum of the first 40 terms of the arithmetic progression with “a” = 10 and “d” = 3 is

S = {40/2]{2*10 + (40–1)*3} = 20*{20 + 39*3} = 20*{20 + 117} = 20 * 137 = 2740

Sum of the first forty terms = 2740.

Alternately,

Sum of the first “n” terms of the series whose “n”th term is 3*n + 7 will be

3*(1+2+3+4+5+…..+n) + (7+7+7+7+….n times) = 3* n(n+1)/2 + 7n = n* [3(n+1)/2 + 7] = n(3n + 17)/2.

Hence, the sum of the first 40 terms will be 40 * (3*40 + 17)/2 = 40*137 / 2 = 2740.