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 :

Sn = (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

7 + 10   1/2 + 14 + ...............+ 84

Solution :

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

 a = 7,  d = (21/2) - (7) and 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

Question 3 :

In an arithmetic series a₁ = 7 and a₁₂ = 29. Find the sum of first 12 terms.

Solution :

From the given information, it is clear that first term and last terms are given.

To find, sum of 12 terms since we know the last term, we use the formula.

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

n = 12, a = 7 and l = 29

= (12/2) [7 + 29]

= 6(36)

= 216

So, the sum of first 12 terms is 216.

Question 4 :

In an arithmetic series the first term is 8 the common difference is 1/2 and the sum of the first n terms is 220.5. Find the value of n.

Solution :

First term = 8

Common difference = 1/2

Sn = (n/2) [2a + (n - 1) d]

220.5 = (n/2)[2(8) + (n - 1)1/2]

220.5 = (n/2) [16 + n/2 - 1/2]

220.5 = (n/2) [31/2 + n/2]

220.5 = (n/2) [(31 + n)/2]

220.5(4) = n(31 + n)

882 = 31n + n²

n² + 31n - 882 = 0

(n + 49)(n - 18) = 0

n = -49 and n = 18

So, the possible value of n is 18.

Question 5 :

How many terms of the arithmetic series 25 + 19 + 13 + ........ are required to give a sum of -20 ?

Solution :

25 + 19 + 13 + ........

Let n be the required number of terms to be added to get the sum which is -20.

(n/2) [2a + (n - 1) d] = -20

a = 25, d = 19 - 25 ==> -6

(n/2) [2(25) + (n - 1) (-6)] = -20

(n/2) [50 - 6n + 6] = -20

(n/2) [56 - 6n] = -20

56n - 6n² = -40

6n² - 56n - 40 = 0

3n² - 28n - 20 = 0

3n² - 30n + 2n - 20 = 0

3n(n - 10) + 2(n - 10) = 0

(3n + 2) (n - 10) = 0

n = 10 and n = -2/3

So, the possible value of n is 10.

Question 6 :

The first 3 terms of an arithmetic progression are 1, 4, 7. Find the 47th term and the sum of first 47 terms.

Solution :

a_n = a + (n - 1)d

a = 1, d = 4 - 1 ==> 3 and n = 47

a₄₇ = 1 + (47 - 1)(3)

= 1 + 46(3)

= 1 + 138

= 139

Sum of 47 terms :

S_n = (n/2)(a + l)

= (47/2)(1 + 139)

= 47(70)

= 3290

Question 7 :

Find the sum of first 31 terms of the sequence 15, 13, 11, .......

Solution :

15, 13, 11, .......

Here a = 15, d = 13 - 15 ==> -2 and n = 31

S_n = (n/2) [2a + (n - 1) d]

= (31/2) [2(15) + (31 - 1) (-2)]

= (31/2)[30 + 30(-2)]

= 31/2 (-30)

= 31(-15)

= -465

So, the required sum is -465.

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