Sum of the First n Terms of an Arithmetic Progression

We will learn how to find the sum of first n terms of an Arithmetic Progression.

Prove that the sum S\(_{n}\) of n terms of an Arithmetic Progress (A.P.) whose first term ‘a’ and common difference ‘d’ is

S = \(\frac{n}{2}\)[2a + (n - 1)d]

Or, S = \(\frac{n}{2}\)[a + l], where l = last term = a + (n - 1)d

Proof:

Suppose, a\(_{1}\), a\(_{2}\), a\(_{3}\), ……….. be a\(_{n}\)  Arithmetic Progression whose first term is a and common difference is d.

Then,

a\(_{1}\) = a

a\(_{2}\) = a + d

a\(_{3}\) = a + 2d

a\(_{4}\) = a + 3d

………..

………..

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

Now,

S = a\(_{1}\) + a\(_{2}\) + a\(_{3}\) + ………….. + a\(_{n -1}\) + a\(_{n}\)

S = a + (a + d) + (a + 2d) + (a + 3d) + ……….. + {a + (n - 2)d} + {a + (n - 1)d} ……………….. (i)

By writing the terms of S in the reverse order, we get,

S = {a + (n - 1)d} + {a + (n - 2)d} + {a + (n - 3)d} + ……….. + (a + 3d) + (a + 2d) + (a + d) + a

Adding the corresponding terms of (i) and (ii), we get

2S = {2a + (n - 1)d} + {2a + (n - 1)d} + {2a + (n - 1)d} + ………. + {a + (n - 2)d}

2S = n[2a + (n -1)d

S = \(\frac{n}{2}\)[2a + (n - 1)d]

Now, l = last term = nth term = a + (n - 1)d

Therefore, S = \(\frac{n}{2}\)[2a + (n - 1)d] = \(\frac{n}{2}\)[a {a + (n - 1)d}] = \(\frac{n}{2}\)[a + l].

 

We can also find find the sum of first n terms of a\(_{n}\) Arithmetic Progression according to the process below.

Suppose, S denote the sum of the first n terms of the Arithmetic Progression {a, a + d, a + 2d, a + 3d, a + 4d, a + 5d ……………...}.

Now nth term of the given Arithmetic Progression is a + (n - 1)d

Let the nth term of the given Arithmetic Progression = l

Therefore, a + (n - 1)d = l

Hence, the term preceding the last term is l – d.

The term preceding the term (l - d) is l - 2d and so on.

Therefore, S = a + (a + d) + (a + 2d) + (a + 3d) + …………………….. to n tems

Or, S = a + (a + d) + (a + 2d) + (a + 3d) + …………………….. + (l - 2d) + (l - d) + l ……………… (i)

Writing the above series in reverse order, we get

S = l + (l - d) + (l - 2d) + ……………. + (a + 2d) + (a + d) + a………………(ii) 

Adding the corresponding terms of (i) and (ii), we get

2S = (a + l) + (a + l) + (a + l) + ……………………. to n terms

2S = n(a + l)

S = \(\frac{n}{2}\)(a + l)

⇒ S = \(\frac{Number of terms}{2}\) × (First term + Last term) …………(iii)

⇒ S = \(\frac{n}{2}\)[a + a + (n - 1)d], Since last term l = a + (n - 1)d

⇒ S = \(\frac{n}{2}\)[2a + (n - 1)d]

Solved examples to find the sum of first n terms of an Arithmetic Progression:

1. Find the sum of the following Arithmetic series:

1 + 8 + 15 + 22 + 29 + 36 + ………………… to 17 terms

Solution:

First term of the given arithmetic series = 1

Second term of the given arithmetic series = 8

Third term of the given arithmetic series = 15

Fourth term of the given arithmetic series = 22

Fifth term of the given arithmetic series = 29

Now, Second term - First term = 8 - 1 = 7

Third term - Second term = 15 - 8 = 7

Fourth term - Third term = 22 - 15 = 7

Therefore, common difference of the given arithmetic series is 7.

The number of terms of the given A. P. series (n) = 17

We know that the sum of first n terms of the Arithmetic Progress, whose first term = a and common difference = d is

S = \(\frac{n}{2}\)[2a + (n - 1)d]

Therefore, the required sum of first 20 terms of the series = \(\frac{17}{2}\)[2 ∙ 1 + (17 - 1) ∙ 7]

\(\frac{17}{2}\)[2 + 16 ∙ 7]

\(\frac{17}{2}\)[2 + 112]

\(\frac{17}{2}\) × 114

= 17 × 57

= 969

 

2. Find the sum of the series: 7 + 15 + 23 + 31 + 39 + 47 + ……….. + 255

Solution:

First term of the given arithmetic series = 7

Second term of the given arithmetic series = 15

Third term of the given arithmetic series = 23

Fourth term of the given arithmetic series = 31

Fifth term of the given arithmetic series = 39

Now, Second term - First term = 15 - 7 = 8

Third term - Second term = 23 - 15 = 8

Fourth term - Third term = 31 - 23 = 8

Therefore, the given sequence is a\(_{n}\) arithmetic series with the common difference 8.

Let there be n terms in the given arithmetic series. Then

a\(_{n}\) = 255

⇒ a + (n - 1)d = 255

⇒ 7 + (n - 1) × 8 = 255

⇒ 7 + 8n - 8 = 255

⇒ 8n - 1 = 255

⇒ 8n = 256

⇒ n = 32

Therefore, the required sum of the series = \(\frac{32}{2}\)[2 ∙ 7 + (32 - 1) ∙ 8]

= 16 [14 + 31 ∙ 8]

= 16 [14 + 248]

= 16 × 262

= 4192

 

Note:

1. We know the formula to find the sum of first n terms of a\(_{n}\) Arithmetic Progression is S = \(\frac{n}{2}\)[2a + (n - 1)d]. In the formula there are four quantities. They are S, a, n and d. If any three quantities  are known, the fourth quantity can be determined.

Suppose when two quantities are given then, the remaining two quantities are provided by some other relation.

2. When the sum S\(_{n}\) of n terms of an Arithmetic Progression is given, then the nth term a_n of the Arithmetic Progression cann be determined by the formula a\(_{n}\) = S\(_{n}\) - S\(_{n -1}\).

Arithmetic Progression






11 and 12 Grade Math

From Sum of the First n Terms of an Arithmetic Progression to HOME PAGE




Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need.



New! Comments

Have your say about what you just read! Leave me a comment in the box below. Ask a Question or Answer a Question.




Share this page: What’s this?

Recent Articles

  1. Formation of Square and Rectangle | Construction of Square & Rectangle

    Jul 16, 25 02:45 AM

    Construction of a Square
    In formation of square and rectangle we will learn how to construct square and rectangle. Construction of a Square: We follow the method given below. Step I: We draw a line segment AB of the required…

    Read More

  2. Perimeter of a Figure | Perimeter of a Simple Closed Figure | Examples

    Jul 16, 25 02:33 AM

    Perimeter of a Figure
    Perimeter of a figure is explained here. Perimeter is the total length of the boundary of a closed figure. The perimeter of a simple closed figure is the sum of the measures of line-segments which hav…

    Read More

  3. Formation of Numbers | Smallest and Greatest Number| Number Formation

    Jul 15, 25 11:46 AM

    In formation of numbers we will learn the numbers having different numbers of digits. We know that: (i) Greatest number of one digit = 9,

    Read More

  4. 5th Grade Quadrilaterals | Square | Rectangle | Parallelogram |Rhombus

    Jul 15, 25 02:01 AM

    Square
    Quadrilaterals are known as four sided polygon.What is a quadrilateral? A closed figure made of our line segments is called a quadrilateral. For example:

    Read More

  5. 5th Grade Geometry Practice Test | Angle | Triangle | Circle |Free Ans

    Jul 14, 25 01:53 AM

    Name the Angles
    In 5th grade geometry practice test you will get different types of practice questions on lines, types of angle, triangles, properties of triangles, classification of triangles, construction of triang…

    Read More