find the sum of the first 40 positive integers divisible by 6
Answers
Answered by
0
sum of first 40 positive integer divisible by 6 are:-
6+12+18 + 24 +30+36+42+48+54+60+66+72+78+84+ 90+96 +102+108 and so on up to 40
Answered by
0
Answer:
Step-by-step explanation:
Sn=n/2[2a+{n-1}d]
=40/2[2*6+{40-1}6]
=20[12+39*6]
=20[12+234]
=20[246]
=4920
hope it helps you
Similar questions