**How to Find Sum of Arithmetic Series :**

Here we are going to see, how to find the sum of arithmetic series.

A series whose terms are in Arithmetic progression is called Arithmetic series.

To find sum of arithmetic series, we use one of the formulas given below.

Sn = (n/2) [a + l]

Sn = (n/2) [2a + (n - 1)d]

a = first term,

n = number of terms of the series,

d = common difference and l = last term

**Question 1 :**

Find the sum of the following

(i) 3, 7, 11,… up to 40 terms.

**Solution :**

Number of terms (n) = 40

First term (a) = 3

Common difference (d) = 7 - 3 = 4

Sn = (n/2) [2a + (n - 1)d]

= (40/2) [2(3) + (40 - 1)4]

= 20 [6 + 156]

= 3240

Hence the sum of 40 terms of the given series is 3240.

(ii) 102, 97, 92,… up to 27 terms.

**Solution :**

Number of terms (n) = 27

First term (a) = 102

Common difference (d) = 97 - 102 = -5

Sn = (n/2) [2a + (n - 1)d]

= (27/2) [2(102) + (27 - 1)(-5)]

= (27/2) [204 + 26(-5)]

= (27/2) [204 - 130]

= (27/2) (74)

= 27 (37)

= 999

Hence the sum of 27 terms of the given series is 999.

(iii) 6 + 13 + 20 +...........+ 97

**Solution :**

l = 97, a = 6, d = 13 - 6 = 7

From this, we have to find the number of terms.

n = [(l - a)/d] + 1

n = [(97 - 6)/7] + 1

= (91/7) + 1

n = 13 + 1

n = 14

Sn = (n/2) [a + l]

= (14/2)[6 + 97]

= 7[103]

= 721

Hence the sum of given series is 721.

**Question 2 :**

How many consecutive odd integers beginning with 5 will sum to 480?

**Solution :**

Consecutive integers starting from 5 are 5, 7, 9, ............

5 + 7 + 9 + 11 + ............. = 480

S_{n} = 480

(n/2)[2a + (n - 1)d] = 480

a = 5, d = 7 - 5 = 2

(n/2)[2(5) + (n - 1)(2)] = 480

(n/2)[10 + 2n - 2] = 480

(n/2)[2n + 8] = 480

2n^{2} + 8n = 960

Dividing by 2, we get

n^{2} + 4n - 480 = 0

(n + 24) (n - 20) = 0

n = -24 or n = 20

Hence by finding sum of 20 terms we will get 480.

After having gone through the stuff given above, we hope that the students would have understood, "How to Find Sum of Arithmetic Series".

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