find the sum of first 18 terms divisible by 3
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The number series 3, 6, 9, 12, 15, 18, 21, . . . . , 120.
The first term a = 3
The common difference d = 3
Total number of terms n = 40
step 2 apply the input parameter values in the AP formula
Sum = n/2 x (a + Tn)
= 40/2 x (3 + 120)
= (40 x 123)/ 2
= 4920/2
3 + 6 + 9 + 12 + 15 + 18 + 21 + . . . . + 120 = 2460
Therefore, 2460 is the sum of first 40 positive integers which are divisible by 3
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