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

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