Question

4. How many three-digit numbers are there
divisible by 3? Find their sum.​

Answer 1

$\bf\purple{QuestioN:-}$

How many three-digit numbers are there  divisible by 3? Find their sum.​

$\bf\green{AnsweR:-}$

Generally the numbers whose sum of the digits are divisible by “3” are divisible by 3 .(i.e. the numbers are divisible by 3).  

So the first number of 3 digit divisible by 3  is 102 and the last number divisible by 3, of three digit number is 999.

Then if it formed an AP,

$\pink\bf{a=102}$

$\pink\bf{a_n=999}$

$\pink\bf{d=3}$

So there is an A.P. whose common difference is 3 and where the first number is 102 and the n th  number is 999.

Now the formula becomes:

$\red:\implies\sf{a_n=a+(n-1)\times{d}}$

$\red:\implies\sf{999=102+(n-1)\times{3}}$

$\red:\implies\sf{(n-1)\times{d}=a_n-a}$

$\red:\implies\sf{(n-1)\times{3}=999-102}$

$\red:\implies\sf{(n-1)\times{3}=897}$

$\red:\implies\sf{(n-1)=\dfrac{897}{3}}$

$\red:\implies\sf{(n-1)=299}$

$\red:\implies\sf{n=299+1=300}$

Next we want to see the sum of these terms, ie, S₃₀₀.

Formula applied :

$\Large\blue:\implies\pmb{S_n=\dfrac{n}{2}[\:\:2a+(n-1)\times{d}\:\:\:]}$

$\blue:\implies\bf{S_{300}=\dfrac{300}{2}[\:\:2\times102+(300-1)\times{3}\:\:\:]}$

$\blue:\implies\bf{S_{300}=150[204+(299)\times{3}]}$

$\blue:\implies\bf{S_{300}=150[204+897]}$

$\blue:\implies\bf{S_{300}=150\times1,101}$

$\blue:\implies\bf{S_{300}=1,65,150}$

Hence, S₃₀₀ = 1,65,150.

Happy Learning!! ☺