# What is the sum of first 50 even numbers?

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 a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, a_{6},…a_{n}

A = {a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, a_{6},…a_{n}}

- Common difference between two terms (d) = (a
_{1}-a_{2}) - 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, a

_{1}= a2nd term, a

_{2}= a+d3rd term, a

_{3}= a+2d4th term, a

_{4}= a+3d5th term, a

_{5}= a+4dso, sum of this series, S

_{5}= (a_{1}+a_{2}+a_{3}+a_{4}+a_{5}) = (a + a+d + a+2d + a+3d + a+4d)It can also be written as,

=> S

_{5}= a+a+a+a+a + d+2d+3d+4d=> S

_{5}= 5a+10d=> S

_{5}= 5×(a + 2d) [take 5 as common] ——————-(1)we can this equation as,

2×S

_{5}= 5 × 2 × [a + 2d] [multiply 2 in both side]=> 2×S

_{5}= 5×[2a+4d]=>

S×_{5}= (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, a

_{1}= a2nd term, a

_{2}= a+d3rd term, a

_{3}= a+2d4th term, a

_{4}= a+3dso, sum of this series, S

_{4 }= (a + a+d + a+2d + a+3d)It can also be written as,

=> S

_{4}= a+a+a+a + d+2d+3d=> S

_{4}= 4a+6d=> S

_{4}= 2×(2a + 3d) [take 2 as common]we can this equation as,

SHere n = 4,_{4}= (4/2) × [2a + (4-1)*d]

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

S_{n}= (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

S_{n}= (n/2)×[2a + (n-1)×d]here, n=50, a=2, d=2

therefore, S

_{50}= (50/2)×[2×2 + (50-1)×2]=> S

_{50}= 25×[4+ 49×2]=> S

_{50}= 25×102=>

S_{50}= 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,

S

_{5}= 5×(5+1) [N =5]S

_{5}= 5×6 = 30To 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, S

_{50}= N*(N+1)S

_{50}= 50*(50+1) = 50*51

S_{50}= 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,

T

_{n}= a+(n-1)d=> T

_{50}= 2+(50-1)×2 [Here, a=2, d=2. n=50]=> T

_{50}= 2+49×2 = 2+98

=> T_{50}= 100 ———–(2)

Now to find the sum,the formula is, SThis formula is correct for all series,_{n}=(N/2) × (a + T_{n}),let’s say a series has 4 terms 1,5,9,13. the sum of this series according to this formula will be,

S

_{4}= (4/2)×(1+13) = 2 × 14 = 28Check 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,

S

_{50}= (50/2) × (2 + 100)[Here, a=2, Tn=100 — from (2)]S

_{50}= 25 × 102

S_{50}= 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)S

_{30}= 30×(30+1)=> S

_{30}= 30×31=>

S_{30}= 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 S

_{n}= (n/2)×[2a + (n-1)×d]S

_{6}= (6/2)×[2×6+(6-1)×4] = 3×(12+20)S

_{6}= 3×32

S_{6}= 96

Using 3rd method:Given, d=4,a=6,N=6

Find the nth term,

T

_{n}=a+(n-1)dT

_{6}= 6 + (6-1)×4T

_{6}= 6+ 20

T——–nth term_{6}= 26Now put all the values in this formula

S_{n}=(N/2) × (a + T_{n})S

_{6}= (6/2) × (6+26)S

_{6}= 3×32

S_{6}= 96

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

a

_{1}= 6, a_{2}= 6+4 = 10, a_{3}= 10+4 =14,a

_{4}= 14+4 = 18, a_{5}= 18+4= 22, a_{6}= 22+4 = 26the series is 6,10,14,18, 22, 26

the sum

is 6+10+14+18+22+26 = 96So 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, S

_{n}=147, T_{n}= 33, N=7we can use this

Sformula here to find the first term in that series_{n}=(N/2) × (a + T_{n})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 = 9So 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

T_{n}=a+(n-1)dPutting all the values, we get,

33 = 9 + (7-1)×d [Given T

_{n}= 33, a=9, n=7]=> 33-9 = 6×d

=>6d = 24

=>d=24/6

=>

d=4the common difference is 4

Now find the series,

a= 9,_{1}a= 9+4 = 13,_{2 }a= 13+4 = 17,_{3}a= 17+4= 21,_{4}a= 21+4=25,_{5 }a= 25+4 = 29,_{6 }a= 29+4 = 33_{7 }So

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