Answer

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Hint: Here we go through by first finding the smallest three digit number which is divisible by 7 and the largest three digit number which is divisible by 7 and make this term as AP then find out the total number of terms by the rule of AP.

Complete step-by-step answer:

As we know the smallest 3 digit number which is divisible by 7 is 105 i.e.$7 \times 15$.

And the largest 3 digit number can be found by dividing the largest 3 digit number by 7 and subtract the remaining remainder from the largest 3 digit number. i.e. $999 \div 7 = 7 \times 142 + 5$ here 5 is the remainder so subtract 5 from 999 we get, the largest number which is divisible by 7 is 994.

The first three-digit number which is divisible by 7 is 105 and the last three-digit number which is divisible by 7 is 994.

Now form an AP with first term 105 an with the common difference of 7 because by adding 7 in first the number comes out is also divisible by 7 by continuing the step we form an AP

This is an A.P in which a=105, d=7 and l=994.

Let the number of terms be n then ${T_n} = 994$.

We know that nth term of AP is written as ${T_n} = a + (n - 1)d$

By putting the values we get,

$\

\Rightarrow 994 = 105 + (n - 1)7 \\

\Rightarrow 994 - 105 = (n - 1)7 \\

\Rightarrow (n - 1) = \dfrac{{889}}{7} = 127 \\

\Rightarrow n = 127 + 1 = 128 \\

$

$\therefore n = 128$

Therefore there are 128 three-digit numbers which are divisible by 7.

Note: Whenever we face such type of question the key concept for solving the question is first find out the smallest 3 digit number which is divisible by 7 and then find out the largest 3 digit number which is divisible by 7 and then make the series of AP, 7 as a common difference and then find out the numbs in that series to get the answer

Complete step-by-step answer:

As we know the smallest 3 digit number which is divisible by 7 is 105 i.e.$7 \times 15$.

And the largest 3 digit number can be found by dividing the largest 3 digit number by 7 and subtract the remaining remainder from the largest 3 digit number. i.e. $999 \div 7 = 7 \times 142 + 5$ here 5 is the remainder so subtract 5 from 999 we get, the largest number which is divisible by 7 is 994.

The first three-digit number which is divisible by 7 is 105 and the last three-digit number which is divisible by 7 is 994.

Now form an AP with first term 105 an with the common difference of 7 because by adding 7 in first the number comes out is also divisible by 7 by continuing the step we form an AP

This is an A.P in which a=105, d=7 and l=994.

Let the number of terms be n then ${T_n} = 994$.

We know that nth term of AP is written as ${T_n} = a + (n - 1)d$

By putting the values we get,

$\

\Rightarrow 994 = 105 + (n - 1)7 \\

\Rightarrow 994 - 105 = (n - 1)7 \\

\Rightarrow (n - 1) = \dfrac{{889}}{7} = 127 \\

\Rightarrow n = 127 + 1 = 128 \\

$

$\therefore n = 128$

Therefore there are 128 three-digit numbers which are divisible by 7.

Note: Whenever we face such type of question the key concept for solving the question is first find out the smallest 3 digit number which is divisible by 7 and then find out the largest 3 digit number which is divisible by 7 and then make the series of AP, 7 as a common difference and then find out the numbs in that series to get the answer

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