How to Find the Sum of Arithmetic Series ?
To find the sum of an arithmetic series, we use the formula give below.
Sn = (n/2) [a + l] (or)
Sn = (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
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
an = 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
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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