The sum of numbers from 250 to 1000 which are | vedclass

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(D) The numbers divisible by $3$ between $250$ and $1000$ form an arithmetic progression: $252, 255, \dots, 999$.Here,the first term $a = 252$,the las

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$156375$

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(D) The numbers divisible by $3$ between $250$ and $1000$ form an arithmetic progression: $252, 255, \dots, 999$.Here,the first term $a = 252$,the last term $l = 999$,and the common difference $d = 3$.Using the formula for the $n^{th}$ term: $l = a + (n - 1)d$.$999 = 252 + (n - 1)3$.$747 = (n - 1)3$.$n - 1 = 249$.$n = 250$.The sum $S_n$ of an arithmetic progression is given by $S_n = \frac{n}{2}(a + l)$.$S_{250} = \frac{250}{2}(252 + 999)$.$S_{250} = 125 	imes 1251$.$S_{250} = 156375$.

The sum of numbers from $250$ to $1000$ which are divisible by $3$ is

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