Sum of Arithmetic Progression

If someone asked you to find out the sum of 1 to 100 then you will be confused but if you know the concept of arithmetic progressions it will become easy for you to solve this in a few seconds. This is actually what the great German mathematician Carl Friedrich Gauss did when he was asked to do so. In this article, we will learn the formula to find out the sum of any arithmetic series. This is one of the basic and most important concepts which will help you to increase your calculation speed and will ultimately boost your confidence. Before understanding the formula of the sum of arithmetic progression, let's understand what is an arithmetic progression which is mentioned below.

Arithmetic Progression Sum Formula

Arithmetic progression is one of the important mathematical concepts which defines the numbers in a sequence where the difference between all the consecutive terms is constant and equal. To know the sum of the series formula, we need to check first whether we have nth term or not or we can the last term. The formula of summation lies on this factor. Thus, there are two cases where you know the last term and when you do not know the last term.

• Sum of arithmetic series when the nth term or last term is known, can be calculated with the help of the following formula:

S =n/2 (a + L) or  n/2( a1 + an )

• Sum of arithmetic sequence when the nth term or last term is not known, can be calculated with the help of the following formula:

S = n/2 {2a + (n − 1) d}

Here, 

S = The sum of the arithmetic sequence

a = First term

d = Common difference between the terms

n = The total number of terms in the sequence 

an or L= The last term of the sequence.

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Other Related Formulas 

Some of the other sum of arithmetic sequence formula are given below which you must know:

• Sum of series formula of 'n' Natural numbers

\[\frac{n(n+1)}{2}\]

• Sum of series formula which is infinite 

If d > 0 then Sn = + ∞

If d < 0 then Sn = - ∞

Examples

Let's check out some of the solved examples to understand the summation of series formula:

Problem 1. Find out the sum of the series

10, 15, 20, 25, 30, _ _ _ _ _, 100

Solution:

10, 15, 20, 25, 30, _ _ _ _ _, 100

Here,  a=10,  l=100 & d= 5

(We know the value of last term here, so we will use the related formula)

Let's calculate 'n' first:

L = a+ (n-1)d

100 = 10+(n-1)5

90 = (n-1)5

90 = 5n-5

90+5 = 5n

95/5= n

n = 19

Using arithmetic sum formula:

S_n =n/2 (a + L) 

S_n =19/2(10+100)

S_n = \[\frac{19X110}{2}\]

S_n = 1045

Therefore, the sum of the given series is 1045.

Problem 2. Calculate the sum of first 10 terms

11,17, 23, 29, 35, ____

Solution:

11,17, 23, 29, 35, ____

Here, a= 11, d= 6 & n=10

Using the formula of the sum of arithmetic progression: 

S_n =n/2 {2a + (n − 1) d}

S₁₀ =  10/2(2X 11+ (10-1)6)

S₁₀ =  5( 22+9X6)

S₁₀ =  5(22+54)

S₁₀ = 5(76)

S₁₀ = 380

Therefore, the sum of the first 10 numbers is 380.

Problem 3. Find the sum of the first 35 terms of an A.P. whose 3rd term is 7 & 7^th term is two more than thrice of its 3^rd term.

Solution:

Let us assume: The first term is ‘a’ as well as the common difference is ‘d’. 

According to the Question:

Given:  3^rd term = 7

We also can write in following way:

a + (3 - 1)d = 7

a + 2d = 7          _______ ( 1 )

Given: 7^th term is two more than thrice of its  3^rd term.

7^th term = 3 ×  3^rd term + 2

We can write it in the following way:

a + (7 - 1)d = 3 × [a + (3 - 1)d] + 2

a + 6d = 3 × [a + 2d] + 2

Put: ( a + 2d = 7 )

a + 6d = 3 × 7 + 2

a + 6d = 21 + 2

a + 6d = 23      ________ ( 2 )

Subtract the equation ( 1 ) from ( 2 ) 

4d = 16

d = 164

d = 4

Put d = 4 in the equation ( 1 )

a + 2 × 4 = 7

a + 8 = 7

a = 7 - 8

a = -1

Therefore, now we know both the terms that we assumed in the beginning i.e first term 'a' which is -1 & the common difference 'd' which is 4. 

Now, using the sum of sequence formula:

S_n = n/2{2a + (n − 1) d}

Put all the values

= 35/2[-2 + 34 × 4]

=35/2 [-2 + 136]

= 35/2[134]

= 35 × 67

= 2345

Therefore, the sum of the series is 2345.

Problem 4. Find the sum of 26 terms, if the 5^th term & 12^th term of an A.P. are 30 & 65 respectively.

Solution: 

Let us assume: the first term is ‘a’ and the common difference is ‘d’. 

According to the Question:

Given: 5^th term = 30

It can be written in the following way:

a + (5 - 1)d = 30

a + 4d = 30      _______  ( 1 )

Given: 12^th term = 65

It can be written in the following form:

a + (12 - 1)d = 65

a + 11d = 65    ________  ( 2 )

Subtract the Eqn ( 1 ) from ( 2 ) 

7d = 35

d = 357

d = 5

Put d = 5 in eqⁿ first

a + 4 × 5 = 30

a + 20 = 30

a = 30 - 20

a = 10

Therefore, we have calculated both the required values that we assumed in the beginning i.e first term 'a' which is 10 & the common difference 'd' which is 5. Now we can calculate the sum of the series. 

Using the arithmetic progression sum formula: 

S_n = n/2{2a + (n − 1) d}

=26/2{ 2×10 + ( 26 - 1 ) × 5 }

= 13 {20 + 25 × 5}

= 13 {20 + 125}

= 13 {145}

= 1885

Therefore, the sum of the series is 1885.

Conclusion

To summarise in the end, we can say that calculating the summation of any number of series is possible with the help of this formula of arithmetic progression. Whether we know the last term or not, we can find out the sum of the series of numbers. It helps in calculating the summation of series very quickly which leads to an increase in the speed of calculations. In this article, we have covered a short intro of arithmetic series, arithmetic progression sum formula along with solved examples, etc. which will help you to understand this particular topic through which you can use this particular formula in your real lives.