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Find the Sum Of The First 40 Positive Integers Divisible by 3 - Mathematics

Find the sum of the first 40 positive integers divisible by 3

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Solution

In the given problem, we need to find the sum of terms for different arithmetic progressions. So, here we use the following formula for the sum of n terms of an A.P.,

`S_n = n/2 [2a + (n - 1)d]`

Where; a = first term for the given A.P.

d = common difference of the given A.P.

First 40 positive integers divisible by 3

= number of terms

First 40 positive integers divisible by 3

So, we know that the first multiple of 3 is 3 and the last multiple of 3 is 120.

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

So here,

First term (a) = 3

Number of terms (n) = 40

Common difference (d) = 3

Now, using the formula for the sum of n terms, we get

`S_n = 40/2 [2(3) + (40 - 1)3]`

= 20[6 + (39)3]

= 20(6 + 117)

= 20(123)

= 2460

Therefore, the sum of first 40 multiples of 3 is 2460

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APPEARS IN

RD Sharma Class 10 Maths
Chapter 5 Arithmetic Progression
Exercise 5.6 | Q 12.2 | Page 51
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