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Question

(A) The integers between $100$ and $200$ that are divisible by $9$ form an arithmetic progression.The first integer greater than $100$ divisible by $9

Vedclass · Accepted Answer

$1683$

Vedclass · Answer

(A) The integers between $100$ and $200$ that are divisible by $9$ form an arithmetic progression.The first integer greater than $100$ divisible by $9$ is $108$ $(9 	imes 12 = 108)$.The last integer less than $200$ divisible by $9$ is $198$ $(9 	imes 22 = 198)$.Thus,the sequence is $108, 117, 126, \ldots, 198$.Here,the first term $a = 108$,the common difference $d = 9$,and the last term $l = a_n = 198$.Using the formula for the $n^{th}$ term: $a_n = a + (n - 1)d$.$198 = 108 + (n - 1)9$.$198 - 108 = (n - 1)9$.$90 = (n - 1)9$.$n - 1 = 10 \implies n = 11$.Now,find the sum $S_n$ using the formula $S_n = \frac{n}{2}(a + l)$.$S_{11} = \frac{11}{2}(108 + 198)$.$S_{11} = \frac{11}{2}(306)$.$S_{11} = 11 	imes 153 = 1683$.Therefore,the sum of the integers between $100$ and $200$ that are divisible by $9$ is $1683$.

Find the sum of the integers between $100$ and $200$ that are divisible by $9$.

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