Question

) Find the sum of all numbers between 250 and 1000 which are exactly divisible by 3.​

Answer 1

Answer:

Hello mate............⚘

Explanation:

The first number after 250 divisible by 3 is 252. and the number just before 1000 divisible by 3 is 999.

252 = 84 x 3 and 999 = 333 x 3

So there are 333–84+1 = 250 numbers

between 250 and 1000 divisible by 3.

Their sum = (n/2)(a+l) = (250/2)(252+999) = 156375.

Answer 2

Answer:

156375.

Explanation:

Given question is related to arithmetic progression.

Where a(first term) = 252. and Tₙ (nth term) = 999, and common difference = 3. where n(no.of terms) = ?

First you have to find no.of multiples from 0-250 and then, 0-1000.

Then after, you have to subtract one from other.

0-250:

largest multiple= 249.

249/3 = 83........................ multiples.

0-1000

Largest multiple = 999

999/3 = 333.......................... multiples.

So, we have, 333 multiples between ''0 and 1000''. And 83 multiples  between "0 and 250".

Subtract 83 from 333 and you will get the no.of multiples b/w 250 and 1000.

333 - 83 = 250. no.s

From 250, the first terms is 252 and the last term is 999

Sₙ= n/2 (a+tₙ)

Sum = 250/2 (252+999)

Sum = 125 (1251)

Sum = 156375.