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(A) The numbers between $100$ and $1000$ that are divisible by $9$ form an Arithmetic Progression $(A.P.)$.The first number greater than $100$ divisib

Vedclass · Accepted Answer

$55350$

Vedclass · Answer

(A) The numbers between $100$ and $1000$ that are divisible by $9$ form an Arithmetic Progression $(A.P.)$.The first number greater than $100$ divisible by $9$ is $108$ $(a = 108)$.The last number less than $1000$ divisible by $9$ is $999$ $(l = 999)$.The common difference $d = 9$.Using the formula for the $n^{th}$ term of an $A.P.$,$l = a + (n - 1)d$:$999 = 108 + (n - 1)9$$891 = (n - 1)9$$n - 1 = 99$$n = 100$.The sum of an $A.P.$ is given by $S_n = \frac{n}{2}(a + l)$:$S_{100} = \frac{100}{2}(108 + 999)$$S_{100} = 50 	imes 1107$$S_{100} = 55350$.

The sum of the numbers between $100$ and $1000$ which are divisible by $9$ is:

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