Question

What is the sum of even numbers between $500\,\& \,600$ ?
$a)\,26950 \\\ b)\,27500 \\\ c)\,27950 \\\ d)\,26500 \\\$

Answer 1

So all even numbers between $500\,\& \,600$ must be in AP and the sum of AP is given by $\mathop S\nolimits_n = \dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right)$ here $n$ is the number of terms, $d$ is the common difference and $a$ is the first term.

Complete step-by-step answer:
So according to question we need to find the sum of all even numbers between $500\,\& \,600$
So first of all you should know what is even number, the number which are divisible by $2$ are called even and the set of even number between $500\,\& \,600$ is
\left\\{ {502,504,506,...........................,598} \right\\}
Here $\mathop a\nolimits_1 = 502,\mathop a\nolimits_2 = 504,\mathop a\nolimits_3 = 506$ and so on
Now $\mathop a\nolimits_2 - \mathop a\nolimits_1 = 504 - 502 = 2 \\\ \mathop a\nolimits_3 - \mathop a\nolimits_2 = 506 - 504 = 2 \\\$
Hence we can say that common difference $d = 2$
Hence \left\\{ {502,504,506,...........................,598} \right\\} are in AP
Now we are to find the sum of even numbers between $500\,\& \,600$ that means \left\\{ {502,504,506,...........................,598} \right\\}
So there are in AP where first term is $502$ i.e. $a = 502$
And common difference $d = 2$
$\mathop a\nolimits_n = a + \left( {n - 1} \right)d$
So we need to find out total number of terms i.e. $n$
So last term it was $598$. So
$598 = a + \left( {n - 1} \right)d \\\ 598 = 502 + \left( {n - 1} \right)2 \\\ 598 - 502 = \left( {n - 1} \right)2 \\\ \dfrac{{96}}{2} = \left( {n - 1} \right) \\\ n = 48 + 1 = 49 \\\$
So there are $49$ terms between $500\,\& \,600$. So we have to find the sum of 49 terms that is given by formula
$\mathop s\nolimits_n = \dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right)$
We know $n = 49\,\,a = 502\,\,d = 2$
Sum of even terms is $\therefore$
$= \dfrac{{49}}{2}\left( {2(502) + (49 - 1)2} \right) \\\ = \dfrac{{49}}{2}\left( {1004 + 48 \times 2} \right) \\\ = \dfrac{{49}}{2}\left( {1004 + 96} \right) \\\ = \dfrac{{49 \times 1100}}{2} \\\ = 26950 \\\$

So, the correct answer is “Option A”.

Note: Here it is said that to find the sum of even numbers between $500\,\& \,600$. So we don’t include $500\,\& \,600$. Even no. in between $500\,\& \,600$ are\left\\{ {502,504,506,...........................,598} \right\\} and as the common difference are equal so they are Arithmetic Progression and we know the formula for sum of $n$ terms in an AP i.e. $\mathop S\nolimits_n = \dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right)$.