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Question
Find the sum of all integers between 50 and 500, which are divisible by 7.
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Solution
In this problem, we need to find the sum of all the multiples of 7 lying between 50 and 500.
So, we know that the first multiple of 7 after 50 is 56 and the last multiple of 7 before 500 is 497.
Also, all these terms will form an A.P. with the common difference of 7.
So here,
First term (a) = 56
Last term (l) = 497
Common difference (d) = 7
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,
497 = 56 + (n -1)7
497 = 56+ 7n - 7
497 = 49 +7n
497 - 49 = 7n
Further simplifying
448 = 7n
`n = 448/7`
n = 64
Now, using the formula for the sum of n terms,
`S_n = n/2 [2a + (n -1)d]`
For n = 64 we get
`S_n = 64/2 [2(56) + (64 - 1) 7]`
= 32 [112 + (63)7]
= 32(1123 + 441)
= 32(112 + 441)
= 32(553)
= 17696
Therefore the sum of all the multiples of 7 lying betwenn 50 and 500 is `S_n = 17696`
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