Question

Find the sum of the integers between 301 and 400 inclusive that are multiples of 4.

Answer 1

Short Answer 8800

Step One
Find the first term after 300 that is divisible by 4.
300 / 4 = 75. It is divisible by 4, but it is not in the series of numbers. The next number that is divisible by 4 is 304
304 / 4 =76 

Step Two
Solve for the number of terms in the series.

Givens
a = 304
L = 400
n = ??
d = 4

Formula
L = a + (n - 1)*d

Substitute and solve
400 = 304 + (n -1)*4   Subtract 304 from both sides.
400 - 304 = (n - 1)*4
96 = (n - 1)*4              Divide by 4
96/4 = (n-1)  
24 = n - 1                    Add 1 to both sides.
24 + 1 = n
n = 25

Step 3
Find the sum

Formula
Sum = (a + L )*n/2

Givens
a = 304
L = 400
n = 25

Sub and solve
Sum = (304 + 400)*25/2
Sum = 704*25/2
Sum = 8800

Answer 2

First we will find the number of integers between 301 and 400 that are multiples of 4. We can see that the first such number is 304 being a multiple of 4. So the required numbers are 304, 308, 312,...,400 which form an AP.

To find the number of such integers use,

$a_n=a_1+(n-1)d$. Here $a_n$ is the nth term, $d$ is the common difference of the AP. Here its is $d=4$. So

$400=304+(n-1)4\\ (n-1)4=96\\ n=25$

The required sum of the AP is

$S_n=\frac{n}{2} (a_1+a_n)\\ S_n=\frac{25}{2} (304+400)\\ S_n=8,800$