Question

Find the sum of the first 40 positive integers divisible by 6.

Answer 1

The positive integers that are divisible by 6 are
\(6, 12, 18, 24 ….\)
It can be observed that these are making an A.P. whose first term is 6 and common difference is 6.
\(a = 6\) and \(d = 6\)
\(S_{40 }=?\)
\(S_n = \frac n2 [2a + (n-1)d]\)

\(S_{40} = \frac {40}{2} [2(6) + (40-1)6]\)
\(S_{40} = 20[12 + (39) (6)]\)
\(S_{40} = 20(12 + 234)\)
\(S_{40} = 20 × 246\)
\(S_{40 }= 4920 \)