A series whose terms are in arithmetic sequence is called arithmetic series.

To find sum of terms in arithmetic sequence, we can 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.

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