**Arithmetic Series**

The series associated with an arithmetic sequence is known as an
**arithmetic series**. For example: 3 +
7 + 11 + 15 + 19 + 23 + 27 + 31 is the arithmetic series associated with the
arithmetic sequence: 3, 7, 11, 15, 19, 23, 27, 31.

**Sum of Arithmetic
Series**

To derive the formula for the **sum of arithmetic series**, let us suppose an arithmetic series where
a be the first term, d be the common difference, n be the number of terms, l be
the last term and S_{n} be the sum of the n terms of an arithmetic
series (AS), then,

S_{n} = a + (a + d) + (a + 2d) + … … … + (l – 2d) + (l –
d) + l ………… (i)

Writing in the reverse order, we have,

S_{n} = l + (l – d) + (l – 2d) + … … … + (a + 2d) + (a +
d) + a ………… (ii)

Adding relations (i) and (ii), we get

2S_{n} = (a + l) + (a + l) + (a + l) + … … … + (a + l) +
(a + l) + (a + l).

or, 2S_{n} = n(a + l)

Again, the last term, l = a + (n – 1)d

Therefore, the sum of the arithmetic series can be found by using the following formulas:

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**Worked Out Examples**

**Example 1:** Find the sum of the series 2 + 8 + 14 + 20 + … … … 20 terms.

**Solution:**

Here,

First term (a) = 2

Common difference (d) = 8 – 2 = 6

Number of terms (n) = 20

Sum (S_{n}) = ?

We Know,

**Solution:**

Here,

First term (a) = 22

Number of terms (n) = ?

Sum (S_{n}) = ?

We know,

l = a + (n – 1)d

∴ n = 7

Now,

Hence, the required sum is 266.

**Solution:**

Given,

or, (12 – m)+ (15 – m)+ (18 – m)+ (21 – m) = 58

or, 12 – m + 15 – m + 18 – m + 21 – m = 58

or, 66 – 4m = 58

or, 4m = 66 – 58

or, 4m = 8

∴ m = 2

**Solution:**

Given,

**Example 5:** How many terms of an AS 17 + 15 + 13 + … … are needed to give the
sum 80?

**Solution:**

Here,

First term (a) = 17

Common difference (d) = -2

Sum (S_{n}) = 80

Number of terms (n) = ?

By using formula,

or, 2n^{2} – 36n + 160 = 0

or, n^{2} – 18n + 80 = 0

or, n^{2} – 10n – 8n + 80 = 0

or, (n – 10)(n – 8) = 0

∴ n = 8 or 10

**Example 6:** If the sum of the first 16 terms of an AS is 280 and the 6^{th}
term of the series is 14. Find the sum of the first 21 terms.

**Solution:**

Let a be the first term and d be the common difference of the
given series.

Sum of 16 terms (S_{16}) = 280

6^{th} term (t6) = 14

By using formula, we have

or, 2a + 15d = 35 …………… (i)

Again,

t_{n} = a + (n – 1)d

or, t_{4} = a + (6 – 1)d

or, a + 5d = 14 …………… (ii)

Solving equations (i) and (ii), we get

a + 5 × 7/5 = 14

or, a + 7 = 14

or, a = 7

Now,

_{21}= 441

Hence, the sum of 21 terms of the series is 441.

**Example 7:** Find the sum of all natural numbers less than 100 which are exactly
divisible by 7.

**Solution:**

Here,

First term (a) = 7

Common difference (d) = 7

Last term (l) =98

Sum (S_{n}) = ?

Number of terms (n) = ?

By using formula,

l = a + (n – 1)d

or, 98 = 7 + (n – 1)7

or, 98 = 7 + 7n – 7

or, 7n = 98

or, n = 14

Hence by using formula,

_{14}= 7 × 105

∴ S_{14} = 735

∴ The required sum is 735.

**Sum of the First n Natural
Numbers**

The first n natural numbers are 1, 2, 3, 4, … … … n.

Let S_{n} = 1 + 2 + 3 … … … + n

Here,

First term (a) = 1

Common difference (d) = 2 – 1 = 1

Number of terms (n) = n

Sum of first n terms (S_{n}) = ?

By using formula,

**Sum of the First n Odd
Natural Numbers**

The first n odd natural number are 1, 3, 5, … … …, (2n – 1)

Let S_{n} = 1 + 3 + 5 + … … + (2n – 1)

Here,

First term (a) = 1

Common difference (d) = 3 – 1 = 2

Number of terms (n) = n

Sum of n terms (S_{n}) = ?

By using formula,

**The Sum of the First n Even
Natural Numbers**

The first n even natural numbers are 2, 4, 6, … …, 2n.

Let S_{n} = 2 + 4 + 6 + … … + 2n

Here,

First term (a) = 2

Common difference (d) = 4 – 2 = 2

Number of terms (n) = n

Sum of n terms (S_{n}) = ?

By using formula,

**Sum of the Squares of the
First n Natural Numbers**

The squares of the first n natural numbers are 1^{2}, 2^{2},
3^{2}, … …, n^{2}

Let S_{n} = 1^{2} + 2^{2} + 3^{2}
+ … … + n^{2}

Consider the identity,

r^{3} – (r – 1)^{3} = r^{3} – (r^{3}
– 3r^{2} + 3r – 1)

or, r^{3} – (r – 1)^{3} = 3r^{2} – 3r +
1

Putting r = 1, 2, 3, 4, … …, (n – 1), n

1^{3} – 0^{3} = 3.1^{2} – 3.1 + 1

2^{3} – 1^{3} = 3.2^{2} – 3.2 + 1

3^{3} – 2^{3} = 3.3^{2} – 3.3 + 1

4^{3} – 3^{3} = 3.4^{2} – 3.4 + 1

… … … … … … … …

(n – 1)^{3} – (n – 2)^{3} = 3(n – 1)^{2}
– 3(n – 1) + 1

n^{3} – (n – 1)^{3} = 3n^{2} – 3n + 1

Adding above all identities, we have

n^{3} = 3(1^{2}
+ 2^{2} + 3^{2} + … … + n^{2}) – 3(1 + 2 + 3 + … … + n)
+ n

**Sum of the Cubes of the
First n Natural Numbers**

The cubes of the first n natural numbers are 1^{3}, 2^{3},
3^{3}, … …, n^{3}.

Let S_{n} = 1^{3} + 2^{3} + 3^{3}
+ … … + n^{3}

Consider the identity,

k^{4} – (k – 1)^{4} = k^{4} – (k^{4}
– 4k^{3} + 6k^{2} – 4k + 1)

or, k^{4} – (k – 1)^{4} = 4k^{3} – 6k^{2}
+ 4k – 1

Putting k = 1, 2, 3, 4, … …, (n – 1), n.

1^{4} – 0^{4} = 4.1^{3} – 6.1^{2}
+ 4.1 – 1

2^{4} – 1^{4} = 4.2^{3} – 6.2^{2}
+ 4.2 – 1

3^{4} – 2^{4} = 4.3^{3} – 6.3^{2}
+ 4.3 – 1

4^{4} – 3^{4} = 4.4^{3} – 6.4^{2}
+ 4.4 – 1

… … … … … … … … … …

(n – 1)^{4} – (n – 2)^{4} = 4.(n – 1)^{3}
– 6.(n – 1)^{2} + 4.(n – 1) – 1

n^{4} – (n – 1)^{4} = 4.n^{3} – 6.n^{2}
+ 4.n – 1

Adding all above identities, we have,

n^{4} = 4(1^{3} + 2^{3} + 3^{3}
+ … … + n^{3}) – 6(1^{2} + 2^{2} + 3^{2} + … …
+ n^{2}) + 4(1 + 2 + 3 + … … + n) – n

_{n}= n

^{4}+ n + n(n + 1)(2n + 1) – 2n(n + 1)

or, 4S_{n} = n(n + 1)(n^{2} – n + 1) + n(n + 1)(2n
+ 1) – 2n(n + 1)

or, 4S_{n} = n(n + 1)(n^{2} – n + 1 + 2n + 1 –
2)

**More Examples**

**Example 8:** Find the sum of the following series:

a. 2 + 4 + 6 + … … … 12 terms

b. 1^{3} + 2^{3} + 3^{3} + … … 10 terms

**Solution:**

a. The given series is 2 + 4 + 6 + … … … 12 terms

This is the sum of the first 12 even natural numbers.

Here, n = 12

∴ S_{n} = n(n + 1) = 12(12 + 1) = 156

b. The given series is 1^{3} + 2^{3} + 3^{3}
+ … … 10 terms

This is the sum of the cubes of the first 10 natural numbers

Here, n = 10

**Example 9:** The sum of three numbers in AP is 27 and the sum of their squares
is 293. Find the numbers.

**Solution:**

Let the three numbers in AP be a – d, a and a + d.

Then,

(a – d) + a + (a + d) = 27

i.e. a = 9

And,

(a – d)^{2} + a^{2} + (a + d)^{2} = 293

or, a^{2} – 2ad + d^{2} + a^{2} + a^{2}
+ 2ad + d^{2} = 293

or, 3a^{2} + 2d^{2} = 293

or, 3 × 9^{2} + 2d^{2} = 293

or, d^{2} = 25

or, d = 5

Hence, the required numbers are a – d, a and a + d.

i.e. 4, 9 and 14.

**Example 10:** Find n^{th} term and the sum of n terms of the series 2×3 +
3×4 + 4×5 + … … n terms.

**Solution:**

The given series is 2×3 + 3×4 + 4×5 + … … n terms.

n^{th} term of 2, 3, 4, … … = 2 + (n – 1)1 = n + 1

n^{th} term of 3, 4, 5, … … = 3 + (n – 1)1 = n + 2

The n^{th} term of the given series = (n + 1)(n + 2) = n^{2}
+ 3n + 2

Sum of the series of n terms,

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