Question
Find the sum of first 40 positive integers divisible by 6.
Solution
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
d = 6
S40 =?
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
d = 6
S40 =?
`S_n = n/2[2a+(n-1)d]`
`S_40 = 40/2[2(6)+(40-1)6]`
= 20[12 + (39) (6)]
= 20(12 + 234)
= 20 × 246
= 4920
Is there an error in this question or solution?
Solution Find the Sum of First 40 Positive Integers Divisible by 6. Concept: Sum of First n Terms of an AP.
