Answer

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**Hint:**Use the formula for the nth term of an A. P. series:Substitute the values to get the required answer.

**Formula used**

${T_n} = a + (n - 1)d$

Take $a = 252$,$d = 3$,${T_n} = 999$

**Complete step by step solution:**

In this question, we are asked to find the number of natural numbers between 250 and 1000 which are exactly divisible by 3.

That is, we have to count those numbers between 250 and 1000 which when divided by 3 give the remainder as 0.

Now, let us consider the natural numbers between 250 and 1000 which are exactly divisible by 3 as a series of natural numbers.

Then this series is a series of finitely many numbers which are in arithmetic progression.

The first term of the series is 252 as it is the smallest number between 250 and 1000 which is divisible by 3.

We know that the difference between two consecutive multiples of 3 is 3.

The last term or the nth term of this series is 999.

Thus, we have

First term$ = a = 252$

Common difference$ = d = 3$

Last term$ = {T_n} = 999$

Now, the formula for computing the nth term of a series in A. P. is

${T_n} = a + (n - 1)d$where n denotes the number of terms in the series.

Therefore, we have

$

999 = 252 + 3(n - 1) \\

\Rightarrow 999 - 252 = 3(n - 1) \\

\Rightarrow 747 = 3(n - 1) \\

\Rightarrow n - 1 = 747 \div 3 = 249 \\

\Rightarrow n = 249 + 1 = 250 \\

$

**Hence there are 250 natural numbers between 250 and 1000 which are exactly divisible by 3.**

**Note:**As the common difference must be considered for successive terms, the order of the terms is important while solving problems related to A.P. Do not change the order randomly when solving questions related to AP

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