Question

give sum of first 40 integers divisible by 3


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Answer 1

Answer:

2460

Step-by-step explanation:

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

Hope it helps.

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Answer 2

Answer:

2460 is your answer

hope it will help you