Find the sum of all integers between 100 and 550, which are divisible by 9.

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Solution

The integers between 100 and 550 that are divisible by 9 are:
108, 117...549
Here, we have:
$a = 108 d = 9 a_{n} = 549 \Rightarrow 108 + (n - 1) (9) = 549 \Rightarrow 9 n - 9 = 441 \Rightarrow 9 n = 450 \Rightarrow n = 50 S_{n} = \frac{n}{2} [2 a + (n - 1) d] \Rightarrow S_{50} = \frac{50}{2} [2 \times 108 + (50 - 1) \times 9] \Rightarrow S_{50} = 25 (657) = 16425$

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Q. Question 5 (i)
Find the sum of the integers between 100 and 200 that are divisible by 9.

Find the sum of all integers between 100 and 550, which are divisible by 9.

The integers between 100 and 550 that are divisible by 9 are: 108, 117...549 Here, we have: a= 108 d = 9an = 549⇒108+(n-1)(9)=549⇒9n-9=441⇒9n=450⇒n=50Sn = n22a+(n-1)d⇒S50 = 5022×108+(50-1)×9⇒S50 = 25657 = 16425