Question

Find the sum of even numbers between 11 and 51.
A) 600
B) 610
C) 620
D) 630

Answer 1

Hint- We will firstly write the even numbers between 11 and 51. We get 12, 14, 16, 18, 20, ….50. We can see that it is in arithmetic progression. So, we can easily find out the sum of this series using this concept.
Complete step-by-step answer:
We have been asked to find the sum of even numbers between 11 and 51 which are 12, 14, 16, 18, …. 50. It is in arithmetic progression, so we will find the sum by using the summation formula of arithmetic progression.
We know that the sum of first $n$ terms of arithmetic series is given by,
$\Rightarrow Sn = \dfrac{n}{2}[2a + (n - 1)d] = \dfrac{n}{2}[{T_1} + {T_n}]$
We have $a = 12 = {T_1}$(initial/first term), $d = 2$ (common difference of A.P.) and last term ${T_n} = 50$. We need to calculate $n$, so
$\Rightarrow {T_n} = 50 = a + (n - 1)d$
$\Rightarrow {T_n} = 50 = 12 + (n - 1)2$
$\Rightarrow 50 = 12 + 2n - 2$
$\Rightarrow 50 - 12 + 2 = 2n$
$\Rightarrow 40 = 2n$
$\Rightarrow n = 20$
Now calculating sum of even numbers
$\Rightarrow S20 = \dfrac{{20}}{2}[2 \times 12 + (20 - 1)2]$
$\Rightarrow 10[24 + (19)2]$
$\Rightarrow 10[24 + 38]$ $\Rightarrow 10[62]$
$\Rightarrow 620$
So, the sum of even numbers between 11 and 51 is 620. Hence, option (C) is correct.
Note- We can calculate it by simple calculation but that is time consuming and complex as well. As it is a series in arithmetic progression, so we can use the summation formula for an A.P. mentioned below:
$\Rightarrow Sn = \dfrac{n}{2}[2a + (n - 1)d] = \dfrac{n}{2}[{T_1} + {T_n}]$. We can find the sum of even numbers as difference of sum of 11 to 51 numbers and sum of odd numbers from 11 to 51.