find the sum of first 4o positive integers divisible by 3
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Hi !
First 40 positive integers divisible by 3 :-
3,6,9,12,15......
a = 3
d = 3
n = 40
Sn = n/2(2a + (n - 1)d)
S₄₀ = 40/2(2*3 + (40-1)*3)
= 20*(6 + 117)
= 20*123
= 2460
Sum of first 40 positive integers divisible by 3 is 2460
First 40 positive integers divisible by 3 :-
3,6,9,12,15......
a = 3
d = 3
n = 40
Sn = n/2(2a + (n - 1)d)
S₄₀ = 40/2(2*3 + (40-1)*3)
= 20*(6 + 117)
= 20*123
= 2460
Sum of first 40 positive integers divisible by 3 is 2460
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