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What is the sum of first 50 even numbers?

  • Last Updated : 03 Sep, 2021

Arithmetic is a part of mathematics that works with different types of numbers, fractions, applied different operations on numbers like addition, multiplication, etc. The word Arithmetic comes from the Greek word arithmos, which means number. Also Arithmetic involves exponentiation, the calculation of percentages, finding the value of number series, logarithmic functions, and square roots, etc.

There is series in arithmetic called Arithmetic Progression (AP),  this is a sequence of numbers, where the difference between any two consecutive terms is always the same. Let’s say, a series is 2,4,6,8,10,12,….., in this series, the difference between any two consecutive numbers is 2. If we add this 2 with the previous number then we get the next number in the series, similarly, if we subtract 2 from the next number, we get the previous number.

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In this article, we are finding the sum of the first 50 even numbers. To work with this series there are some formulas available, that are,



Let’s say a series A consist of some element a1, a2, a3, a4, a5, a6,…an

A = {a1, a2, a3, a4, a5, a6,…an}

  • Common difference between two terms (d) =  (a1-a2)
  • Sum of the series(S) = (n/2)[2a + (n – 1)d]
  • First term = a
  • 2nd term = a+d
  • 3rd term = a+2d
  • Similarly, Nth term = a+(n-1)d

Now let’s take a series that has 5 terms, then the sum of the series is as below,

1st term, a1 = a

2nd term, a2 = a+d

3rd term, a3 = a+2d

4th term, a4 = a+3d

5th term, a5 = a+4d



so, sum of this series, S5= (a1 +a2 +a3 +a4+a5) = (a + a+d + a+2d + a+3d + a+4d)

It can also be written as,

=> S5 = a+a+a+a+a + d+2d+3d+4d

=> S5 = 5a+10d

=> S5 = 5×(a + 2d)     [take 5 as common] ——————-(1)

we can this equation as,

2×S5 = 5 × 2 × [a + 2d]   [multiply 2 in both side]

=> 2×S5 = 5×[2a+4d]

=> S5 = (5/2)×[2a + (5-1)d]     Here n = 5,

Now let’s take a series that has 4 terms, then the sum of the series is as below,

1st term, a1 = a

2nd term, a2 = a+d

3rd term, a3 = a+2d

4th term, a4 = a+3d

so, sum of this series, S4 = (a + a+d + a+2d + a+3d)

It can also be written as,

=> S4 = a+a+a+a + d+2d+3d

=> S4 = 4a+6d

=> S4 = 2×(2a + 3d)     [take 2 as common]

we can this equation as,



S4 = (4/2) × [2a + (4-1)*d]     Here n = 4, 

So, for finding the sum of series which has N term, the formula looks like

Sn = (n/2)×[2a + (n-1)×d]

What is the sum of the first 50 even numbers?

First of all, we have to find the series of even numbers, let’s find that

The first even number is 2,

2nd even number is 4

3rd even number is 6 and so on….. so the last term will be 50 × 2 = 100

Secondly, find the common difference between them, d = 4 – 2 = 2 or 6 – 4 = 2 

So, d = 2,

Thirdly, find the sum of the first 50 even terms,

Use the formula for for finding the sum of series, that is Sn = (n/2)×[2a + (n-1)×d]

here, n=50, a=2, d=2

therefore, S50 = (50/2)×[2×2 + (50-1)×2]

=> S50 = 25×[4+ 49×2] 

=> S50 = 25×102

=> S50 = 2550 

The sum of first 50 even number is 2550

Another Method of Solving:

To find the sum of even numbers up to N we can use another formula, that is, N×(N+1)

This formula only works with consecutive even numbers, that is, 2,4,6,8,10,12,14,16…… like this.



Let’s say a series has the first 5 even numbers, 2,4,6,8,10

so the sum of this series,

S5 = 5×(5+1)   [N =5]

S5 = 5×6 = 30

To check if the sum is correct or not we can sum up the numbers, 2+4+6+8+10 = 30, so the formula works fine

Here we are finding the sum of first 50 even numbers,

The sum, S50 = N*(N+1)

S50 = 50*(50+1) = 50*51

S50 = 2550

The sum of first 50 even number is 2550

Another Method of Solving:

For this method, we need to find the last term of the series.

To find the last term we a formula,

Tn = a+(n-1)d

=> T50 = 2+(50-1)×2    [Here, a=2, d=2. n=50]

=> T50 = 2+49×2 = 2+98

=> T50 = 100 ———–(2)

Now to find the sum, the formula is, Sn=(N/2) × (a + Tn), This formula is correct for all series, 

let’s say a series has 4 terms 1,5,9,13. the sum of this series according to this formula will be,

S4 = (4/2)×(1+13) = 2 × 14  = 28

Check if it is correct or not, 1+5+9+13 = 28, so the formula is correct.

Now find the sum of the first 50 even numbers,

S50 = (50/2) × (2 + 100)   [Here, a=2, Tn=100 — from (2)]

S50 = 25 × 102

S50 = 2550

The sum of first 50 even number is 2550

Similar Questions

Question 1. Find the sum of the first 30 even numbers.

Solution: 

To find this sum we can use any previously defined method,

Let’s use the 2nd method, which is N×(N+1)



S30 = 30×(30+1)

=> S30 = 30×31

=>S30 = 930

The sum of the first 30 even numbers is 930

Question 2. Find the sum of a series whose first term is 6 and the common difference is 4 and the number of terms in the series is also 6.

Solution: 

To solve this problem we can use the 3rd and 1st methods.

Using 1st method:

Given, d=4,a=6,N=6

put all the values in this formula Sn = (n/2)×[2a + (n-1)×d]

S6 = (6/2)×[2×6+(6-1)×4] = 3×(12+20)

S6 = 3×32

S6 = 96

Using 3rd method:

Given, d=4,a=6,N=6

Find the nth term,

Tn=a+(n-1)d

T6 = 6 + (6-1)×4

T6 = 6+ 20

T6 = 26——–nth term

Now put all the values in this formula Sn=(N/2) × (a + Tn)

S6 = (6/2) × (6+26)

S6 = 3×32

S6 = 96

Now find the series, first term is 6 and common difference is 4

so the series will be,

a1 = 6, a2 = 6+4 = 10, a3 = 10+4 =14,

a4 = 14+4 = 18, a5= 18+4= 22, a6= 22+4 = 26

the series is 6,10,14,18, 22, 26

the sum is 6+10+14+18+22+26 = 96

So the solution is correct. and the sum of this series is 96

Question 3. Find the series whose sum is 147, the last term is 33 and the number of terms in the series is 7.

Solution: 

Given, Sn =147, Tn = 33, N=7

we can use this Sn=(N/2) × (a + Tn) formula here to find the first term in that series

Putting all the given values, we get

147 = (7/2)×(a + 33)

=> 147×2 = 7×a + 7×33

=> 294 = 7a + 231

=> 7a = 294-231



=>7a = 63

=> a = 63/7

=>a = 9

So the first term is 9.

Now find the common difference d, to find d we can use any formula which contains d

Here we are using the formula for finding the Nth term, that is  Tn=a+(n-1)d

Putting all the values, we get,

33 = 9 + (7-1)×d       [Given Tn = 33, a=9, n=7]

=> 33-9 = 6×d

=>6d = 24

=>d=24/6

=> d=4

the common difference is 4

Now find the series,

a1 = 9, a2 = 9+4 = 13, a3 = 13+4 = 17, a4 = 17+4= 21, a5 = 21+4=25, a6 = 25+4 = 29, a7 = 29+4 = 33

So the final series is 9, 13, 17, 21, 25, 29, 33




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