Find the sum of the multiples of 3 lying betw | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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 $1000$.

Similar Questions

Find the $45^{th}$ term of the $A.P.$ $13, 8, 3, -2, \ldots$

Find the $12^{th}$ term and the $24^{th}$ term of the $A.P.$ $9, 13, 17, \ldots$

For an $A.P.$,$a=11$ and $d=7$. Find the sum of the first $40$ terms of the $A.P.$

The sum of the first $n$ terms of an $A.P.$ is given by $S_{n} = 7n^{2} - 3n$. Find the $n^{th}$ term of the $A.P.$

The sum of four numbers in $A.P.$ is $50$. The greatest number is four times the smallest number. Find those numbers.

Vedclass Test Series

Exam Paper Generator

Online Exam Module