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# Question 12. Find the sum of(ii) the first 40 positive integers divisible by (a) 3 (b) 5 (c) 6.

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This is the basic and conceptual question from Chapter name- Arithmetic Progression
Chapter number- 9
Exercise 9.6

In this question we have been asked to find the sum of the first 40 positive integers

CBSE DHANPAT RAI publications
Class:- 10th
Solutions of CBSE Mathematics
Question 11(ii)

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1. We know sum of n terms of an A.P. is given by, Sn = n[2a + (n − 1)d] / 2.

(a) First 40 positive integers divisible by 3 are 3, 6, 9, 12,…… ,120.

These integers form an A.P. with first term(a) = 3,

Common difference(d) = 6 − 3 = 3 and number of terms(n) = 40.

S40 = 40[2(3) + (40 − 1)3]/2

= 40(6 + 117)/2

= 20(123) = 2460

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

(b) First 40 positive integers divisible by 5 are 5, 10, 15, 20,…… ,200.

These integers form an A.P. with first term(a) = 5,

Common difference(d) = 10 − 5 = 5 and number of terms(n) = 40.

S40 = 40[2(5) + (40 − 1)5]/2

= 40(10 + 195)/2

= 20(205) = 4100

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

(c) First 40 positive integers divisible by 6 are 6, 12, 18, 24,…… ,240.

These integers form an A.P. with first term(a) = 6,

Common difference(d) = 12 − 6 = 6 and number of terms(n) = 40.

S40 = 40[2(6) + (40 − 1)6]/2

= 40(12 + 234)/2

= 20(246) = 4920

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

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