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Question
Find the sum of all integers between 84 and 719, which are multiples of 5.
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Solution
In this problem, we need to find the sum of all the multiples of 5 lying between 84 and 719.
So, we know that the first multiple of 5 after 84 is 85 and the last multiple of 5 before 719 is 715.
Also, all these terms will form an A.P. with the common difference of 5.
So here
First term (a) = 85
Last term (l) = 715
Common difference (d) = 5
So, here the first step is to find the total number of terms. Let us take the number of terms as n.
Now as we know
`a_n = a + (n - 1)d`
So, for the last term,
715 = 85 + (n - 1)5
715 = 85 + 5n - 5
715 = 80 + 5n
Further simplifying
635 = 5n
`n = 635/5`
n = 127
Now, using the formula for the sum of n terms,
`S_n = n/2 [2a + (n - 1)d]`
We get
`S_n = 127/2 = [2(85) + (127 - 1)5]`
`= 127/2 [170 + (126)5]`
`= 127/2 (170 + 630)`
`= (127(800))/2`
On further simplification, we get,
`S_n = 127(400)`
= 50800
Therefore, the sum of all the multiples of 5 lying between 84 and 719 is `S_n = 50800`
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