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Question
Find the sum of the first 40 positive integers divisible by 5
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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.
n = number of terms
First 40 positive integers divisible by 5
So, we know that the first multiple of 5 is 5 and the last multiple of 5 is 200.
Also, all these terms will form an A.P. with the common difference of 5.
So here,
First term (a) = 5
Number of terms (n) = 40
Common difference (d) = 5
Now, using the formula for the sum of n terms, we get
`S_n = 40/2 [2(5) = (40 - 1)5]`
= 20[10 + (39)5]
= 20(10 + 195)
= 20(205)
= 4100
Therefore, the sum of first 40 multiples of 3 is 4100
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