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(A) The multiples of $3$ lying between $250$ and $1000$ form an arithmetic progression ($A$.$P$.).The first multiple of $3$ after $250$ is $252$,and t

Vedclass · Accepted Answer

$1,56,375$

Vedclass · Answer

(A) The multiples of $3$ lying between $250$ and $1000$ form an arithmetic progression ($A$.$P$.).The first multiple of $3$ after $250$ is $252$,and the last multiple of $3$ before $1000$ is $999$.So,the $A$.$P$. is $252, 255, 258, ..., 999$.For this $A$.$P$.,the first term $a = 252$,the common difference $d = 3$,and the last term $l = 999$.Using the formula for the $n^{th}$ term: $a_n = a + (n - 1)d$.$999 = 252 + (n - 1)3$$999 - 252 = (n - 1)3$$747 = (n - 1)3$$n - 1 = 249$$n = 250$.Now,the sum of $n$ terms is given by $S_n = \frac{n}{2}(a + l)$.$S_{250} = \frac{250}{2}(252 + 999)$$S_{250} = 125 	imes 1251$$S_{250} = 1,56,375$.Thus,the sum of the multiples of $3$ lying between $250$ and $1000$ is $1,56,375$.

Find the sum of the multiples of $3$ lying between $250$ and $1000$.

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