Question

Find the sum of the first 40 positive integers divisible by (a) 3 (b) 5 (c) 6.

Answer 1

We know that the sum of terms for an A.P is given by

S_n = \(\frac{n}{2}\)[2a + (n − 1)d]

Where; a = first term for the given A.P. d = common difference of the given A.P. n = number of terms

(a) First 40 positive integers divisible by 3.

Hence, the first multiple is 3 and the 40^th multiple is 120.

And, these terms will form an A.P. with the common difference of 3.

Here, First term (a) = 3

Number of terms (n) = 40

Common difference (d) = 3

So, the sum of 40 terms

S₄₀  = \(\frac{40}{2}\)[2(3) + (40 − 1)3]

= 20[6 + (39)3]

= 20(6 + 117)

= 20(123) = 2460

Thus, the sum of first 40 multiples of 3 is 2460.

(b) First 40 positive integers divisible by 5

Hence, the first multiple is 5 and the 40^th multiple is 200.

And, these terms will form an A.P. with the common difference of 5.

Here, First term (a) = 5

Number of terms (n) = 40

Common difference (d) = 5

So, the sum of 40 terms

S₄₀  = \(\frac{40}{2}\)[2(5) + (40 − 1)5]

= 20[10 + (39)5]

= 20 (10 + 195)

= 20 (205) = 4100

Hence, the sum of first 40 multiples of 5 is 4100.

(c) First 40 positive integers divisible by 6

Hence, the first multiple is 6 and the 40^th multiple is 240.

And, these terms will form an A.P. with the common difference of 6.

Here, First term (a) = 6

Number of terms (n) = 40

Common difference (d) = 6

So, the sum of 40 terms

S₄₀  = \(\frac{40}{2}\)[2(6) + (40 − 1)6]

= 20[12 + (39)6]

=20(12 + 234)

= 20(246) = 4920

Hence, the sum of first 40 multiples of 6 is 4920.