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(A) The natural numbers lying between $100$ and $1000$ which are multiples of $5$ are $105, 110, \dots, 995.$This forms an arithmetic progression wher

Vedclass · Accepted Answer

$98450$

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(A) The natural numbers lying between $100$ and $1000$ which are multiples of $5$ are $105, 110, \dots, 995.$This forms an arithmetic progression where the first term $a = 105,$ common difference $d = 5,$ and the last term $l = 995.$Using the formula for the $n^{th}$ term: $a_n = a + (n - 1)d.$$995 = 105 + (n - 1)5.$$890 = (n - 1)5.$$n - 1 = 178 \Rightarrow n = 179.$Now,the sum $S_n = \frac{n}{2}(a + l).$$S_{179} = \frac{179}{2}(105 + 995).$$S_{179} = \frac{179}{2}(1100).$$S_{179} = 179 	imes 550 = 98450.$Thus,the sum is $98450.$

Find the sum of all natural numbers lying between $100$ and $1000$ which are multiples of $5.$

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