Question

find the sum of first 40 positive integers divisible by 3​

Answer 1

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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Answer 2

Answer:

Sn = 2460

Step-by-step explanation:

Given

AP :- 3,6,9,.........,120

a = 3

an = 120

d = a2 - a1

= 6 - 3

= 3

n = 40

Sn = ?

By Sn Formula :-

Sn = n÷ 2 [a + an ]

Sn = 40 ÷ 2 [ 3 + 120 ]

Sn = 20 [ 123 ]

Sn = 2460

Thus,

Sn = 2460

Hope it helps!!!

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