One of the most important and exam oriented question from arithmetic progression chapter as it was already asked in previous year paper of 2012 in which we are to find the sum of first forty positive integers divisible by 6.
Book – RS Aggarwal, Class 10, chapter 5C, question no 14.
First forty positive integers which are divisible by 6 are
6,12,18,24,.... to 40 terms
Here, a=6,d=12−6=6, and n=40.
Sum of 40 terms:
S40=n/2[2a+(n−1)d]
=40/2[2×6+(40−1)×6]
=20[12+39+6]
=20[12+234]
=20×246
=4920.