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Question
Find the sum of all integers between 100 and 550, which are divisible by 9.
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Solution
In this problem, we need to find the sum of all the multiples of 9 lying between 100 and 550.
So, we know that the first multiple of 9 after 100 is 108 and the last multiple of 9 before 550 is 549.
Also, all these terms will form an A.P. with the common difference of 9.
So here,
First term (a) = 108
Last term (l) = 549
Common difference (d) = 9
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
`549 = 108 + (n - 1)9
549 = 108 + 9n - 9
549 = 99 + 9n
Further simplifying,
450 = 9n
`n= 450/9`
n = 50
Now, using the formula for the sum of n terms,
`S_n =n/2[2a + (n -1)d]`
We get
`S_n = 50/2 [2(108) + (50 - 1)d]`
= 25[216 + (49)9]
= 25(216 + 441)
= 25(657)
= 16425
Therefore, the sum of all the multiples of 9 lying between 100 and 550 is `S_n= 16425`
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