HOW TO FIND THE SUM OF ARITHMETIC SERIES

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

 S₁₀  =  (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

 S₁₂  =  (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

S₁₀₀  =  (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

S₁₁  =  (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.

S₂₃ = (23/2) [7 + 84]

 S₂₃ = (23/2) [91]

        = 2093/2 

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