**How to Find the Sum of Arithmetic Series ? **

To find the sum of an arithmetic series, we use the formula give below.

S_{n} = (n/2) [a + l] (or)

S_{n} = (n/2) [2a + (n - 1)d]

a = first term, d = common difference, n = number of terms and l = last term.

** Question 1 :**

Find the sum of the following APs

(i) 2 , 7 , 12,....... to 10 terms

**Solution :**

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

n = 10 a = 2 d = 7 - 2 = 5

S_{10 }= (10/2) [2(2) + (10-1)5]

= 5 [4 + 9(5)]

= 5 [4 + 45]

= 5 [49]

= 245

(ii) -37,-33,-29,..................... to 12 terms

**Solution :**

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

n = 12 a = -37 d = -33 - (-37)

= -33 + 37

= 4

S_{12 }= (12/2) [ 2 (-37) + (12 - 1) 4 ]

= 6 [ -74 + 11 (4) ]

= 6 [ -74 + 44 ]

= 6 [-30]

= -180

(iii) 0.6, 1.7, 2.8,............... to 100 terms

**Solution :**

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

n = 100 a = 0.6 d = 1.7 - 0.6

= 1.1

S_{100 }= (100/2) [ 2 (0.6) + (100 - 1) 1.1 ]

= 50 [ 1.2 + 99 (1.1) ]

= 50 [1.2 + 108.9 ]

= 50 [110.1]

= 5505

(iv) 1/15, 1/12, 1/10,................... to 11 terms

**Solution :**

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

n = 11 a = 1/15 d = (1/12) - (1/15)

= (5 - 4)/60

= 1/60

S_{11 }= (11/2) [ 2 (1/15) + (11 - 1) (1/60) ]

= (11/2) [(2/15) + (10/60)]

= (11/2) [(2/15) + (1/6)]

= (11/2) [ (4 + 5)/30 ]

= (11/2) [ 9/30 ]

= (11/2) [ 3/10 ]

= 33/20

**Question 2 :**

Find the sums given below

(i) 7 + 10 1/2 + 14 + ...............+ 84

**Solution :**

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

a = 7 d = (21/2) - (7) l = 84

= (21 - 14)/2

= 7/2

a_{n} = a + (n - 1) d

84 = 7 + (n - 1) (7/2)

84 - 7 = (n - 1) (7/2)

77 x (2/7) = n - 1

11 x 2 = n - 1

n - 1 = 22

n = 22 + 1 = 23

So, the total number of terms in the given series is 23.

S_{23} = (23/2) [7 + 84]

S_{23} = (23/2) [91]

= 2093/2

After having gone through the stuff given above, we hope that the students would have understood, how to find the sum of arithmetic series.

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