Question

Find the sum of numbers between $250$ and $500$ that are divisible by $5$.

Answer 1

Here, the numbers that are divisible by 5 form an arithmetic progression. We will then find the first term, last term and common difference using the given information. We will substitute these values in the formula of ${n^{th}}$ term of an arithmetic progression to find the number of terms. Then using these values and the formula of Sum of $n$ terms in an arithmetic progression, we will find the required sum.

Formula used:
We will use the following formulas:

1. The ${n^{th}}$ term of an arithmetic progression is given by ${a_n} = a + (n - 1)d$, where $a$ is the first term, $n$ is the number of terms, ${a_n}$ is the ${n^{th}}$ term, and $d$ is the common difference.
2. Sum of $n$ terms in an arithmetic progression, ${S_n} = na + \dfrac{{n(n - 1)}}{2}d$

Complete step by step solution:
We are supposed to find the sum of numbers between $250$ and $500$ that are divisible by $5$. The first term that is divisible by $5$ is $250$. The last term that is divisible by $5$ is $500$.
We observe that all the numbers that are divisible by $5$ form an arithmetic progression. We need to find the number of terms whose sum we require.
Substituting $a = 250,d = 5$ and ${a_n} = 500$ in equation ${a_n} = a + (n - 1)d$ we get
$500 = 250 + \left( {n - 1} \right)5$
Subtracting both sides 250 we get
$\Rightarrow 5\left( {n - 1} \right) = 500 - 250 \\\ \Rightarrow 5\left( {n - 1} \right) = 250 \\\$
Dividing both sides by 5, we have
\Rightarrow \dfrac{{5\left( {n - 1} \right)}}{5} = \dfrac{{250}}{5} \\\ \Rightarrow \left( {n - 1} \right) = 50 \\\
Adding 1 on both sides, we get
$\Rightarrow n = 50 + 1 = 51$
So, there are 51 terms between $250$ and $500$ that are divisible by $5$.
Now, we have to find the sum of these terms.
Substituting $n = 51,a = 250$ and $d = 5$ in the above formula ${S_n} = na + \dfrac{{n(n - 1)}}{2}d$, we get
${S_{51}} = 51 \times 250 + \dfrac{{51(51 - 1)}}{2} \times 5$
Simplifying the expression, we get
$\Rightarrow {S_{51}} = 12750 + \dfrac{{51 \times 50}}{2} \times 5$
Multiplying the terms, we get
$\Rightarrow {S_{51}} = 12750 + 6375$
Adding the terms, we get
$\Rightarrow {S_{51}} = 19125$

Therefore, the sum of terms between $250$ and $500$ that are divisible by $5$ is 19125.

Note:
Here, the terms that are divisible by 5 forms an AP such that the common difference is 5. An arithmetic progression is a sequence or series where there is a common difference between consecutive terms. A real-life example of AP is when we add a fixed amount in our money bank every week. Similarly, when we ride a taxi, we pay an amount for the initial kilometer and pay a fixed amount for all the further kilometers, this also turns out to be an AP.