Question

The sum of numbers from $250$ to $1000$ which are divisible by $3$ ?

Answer 1

Hint: For these kinds of questions, we need to use the concept of arithmetic progression. Arithmetic progression(AP) or arithmetic sequence is a sequence of numbers such that the difference between the consecutive terms is constant, which is often referred to as the common difference, usually denoted by the letter , $d$. In this question we have to find the sum of numbers which are between $250 $ and $1000$ which are divisible by $3$.

Complete step by step solution:
So this question is nothing but the sum of terms in an arithmetic progression whose common difference is $3$.
The starting term is an AP and is often denoted by the letter $a$. It is the first term.
Our first term in this AP would be the first number we find starting from $250$ which is divisible by $3$.$250,251$ are not divisible by $3$. But $252$ is divisible by $3$. So that would be the first term of our AP.
Now, we have to look for our last term in this AP. Let us come from the back and find out the biggest number which is less than or equal to $1000$ which is divisible by $3$.
$1000$ is the last number in this question. So let us start from there. $1000$ is clearly not divisible by $3$ since it leaves a reminder of $1$. But $999$ is exactly divisible by $3$. So $999$ would be our last number.
Now we have to find the number of terms in this AP. We use the following formula :
$\Rightarrow {{a}_{n}}=a+\left( n-1 \right)d$
We just have to know which number of terms $999$ is. Since it is the last term which is divisible by $3$, we would get the number of terms present in this AP.
We know our $a,d$and ${{a}_{n}}$ .
Let us substitute and find the number of terms.
$\begin{align}
  & \Rightarrow {{a}_{n}}=a+\left( n-1 \right)d \\
 & \Rightarrow 999=252+\left( n-1 \right)3 \\
\end{align}$
Upon solving we, get the following :
$\begin{align}
  & \Rightarrow {{a}_{n}}=a+\left( n-1 \right)d \\
 & \Rightarrow 999=252+\left( n-1 \right)3 \\
 & \Rightarrow 999=252+3n-3 \\
 & \Rightarrow \dfrac{750}{3}=n \\
 & \Rightarrow n=250 \\
\end{align}$
There are $250$ terms in this AP.
Now, let us find the sum of all these terms. We use the following formula :
$\Rightarrow {{S}_{n}}=\dfrac{n}{2}\left[ a+l \right]$ , where ${{S}_{n}} $ denotes the sum of $n$ terms of an AP.
$l$ denotes the last term in an AP.
Let us substitute the value and get the sum.
$\begin{align}
  & \Rightarrow {{S}_{n}}=\dfrac{n}{2}\left[ a+l \right] \\
 & \Rightarrow {{S}_{n}}=\dfrac{250}{2}\left[ 252+999 \right] \\
 & \Rightarrow {{S}_{n}}=125\left[ 1251 \right]=156375 \\
\end{align}$

$\therefore $ Hence, the sum of numbers from $250$ to $1000$ which are divisible by $3$ is $156375$.

Note: We should be very careful while solving the question as there is a lot of scope for calculation mistakes. We should remember all the formulae regarding Arithmetic Progression to solve a question. There are also some short-cuts for problems involving Arithmetic Progression. These can be used to solve a question quickly. We should also remember the formulae of Geometric Progression and Harmonic Progression since all these three concepts can be clubbed to form a single question.