The sum of all even numbers between 21 and 51 is
(a) 518
(b) 540
(c) 560
(d) 596

Hint: Arithmetic Progression (AP) is a sequence of numbers in order that the difference of any two successive numbers is a constant value. We will use the nth term formula
\[T=a+(n-1)d\] to calculate the number of even numbers between 21 and 51 and the sum of n terms formula \[\dfrac{n}{2}\left[ 2a+(n-1)d \right]\] to solve this question

Complete step-by-step answer:
Before proceeding with the question we should understand the definition of even numbers.

An even number is a number which on being divided by 2 gives remainder as 0.

Series of even numbers between 21 and 51 is 22, 24, 26, 28………50.

So here the first term(a) is 22 and the last term(T) is 50. And the common difference(d) is 2.

We know that the nth term formula is \[T=a+(n-1)d.......(1)\]

So substituting all these in equation (1) we get,

\[50=22+(n-1)\times 2.......(2)\]

Now solving equation (2) to get the number of terms, we get,

\[\begin{align}

  & \Rightarrow 50-22=2n-2 \\

 & \Rightarrow 2n=30 \\

 & \Rightarrow n=\dfrac{30}{2}=15.......(3) \\

\end{align}\]

We also know that the sum of n terms of an AP is \[\dfrac{n}{2}\left[ 2a+(n-1)d \right]\]. So
substituting the value of n from equation (3) in this formula and also all the other values we get,

Sum of even numbers between 21 and 51 \[=\dfrac{15}{2}\left[ 2\times 22+(15-1)\times 2

\right]=\dfrac{15}{2}\left[ 44+28 \right]=\dfrac{15}{2}\times 72=540\].

Hence the correct answer is option (b).


Note: Remembering the formula of the nth term of an arithmetic progression and the formula of the sum of n terms of an A.P is the key here. We in a hurry can make a mistake in solving equation (2) by putting value of a in place of d or value of T in place of a and hence we need to be careful while substituting the values in the formulas.