An arithmetic series is a series whose terms form an arithmetic sequence.
We use the one of the formulas given below to find the sum of arithmetic series.
Sn = (n/2) [2a+(n-1)d]
Sn = (n/2) [a+l]
n = Total number terms, a = First term,
d = Common difference, l = Last term
Problem 1 :
Find the sum of first 75 positive integers
Solution :
1 + 2 + 3 + ..........+ 75
Since we know the last term, we can use the second formula to find the sum of the arithmetic series.
Sn = (n/2) (a+l)
= (75/2) (1+75)
= (75/2) (76)
= 75 x 38
= 2850
Problem 2 :
Find the sum of first 125 natural numbers
Solution :
1 + 2 + 3 + ..........+ 125
Total number of terms in the series is 125 so n = 125
Sn = (n/2) (a+L)
= (125/2) (1+125)
= (125/2) (126)
= 125 ⋅ 63
= 7875
So, sum of first 125 natural numbers is 7875.
Problem 3 :
Find the sum of 6 + 5 ¼ + 4 ½ + ......... 25 terms
Solution :
In this series we have 25 terms so n = 25
a = 6 d = t₂ - t₁ ==> = 5 ¼ - 6 ==> -3/4
Sn = (n/2) [2a + (n-1) d]
S_{25} = (25/2) [2(6) + (25-1) (-3/4)]
= (25/2) [12 + (24) (-3/4)]
= (25/2) [12 + (6) (-3)]
= (25/2) [12-18 ]
= (25/2) (-6)
= -75
So, sum of 25 terms of the given series is -75.
Problem 4 :
Find the sum of each arithmetic series described
a = 5, n = 30 and L = 121
Solution :
S_{n} = (n/2) [a+L]
= (30/2) (5+121)
= (15) [126]
S_{30} = 1890
So, the sum of 30 terms is 1890.
Problem 5 :
Find the sum of each arithmetic series described
a = 50, n = 25 and d = -4
Solution :
S_{n} = (n/2) [2a+(n-1)d]
S_{25} = (25/2) [2(50)+(25-1)(-4)]
= (25/2) [100+(24)(-4)]
= (25/2) (100 - 96)
= (25/2)(4)
S_{25 }= 50
So, sum of 25 terms is 50.
Problem 6 :
Find the sum of first 40 terms of the series
1^{2} - 2^{2} + 3^{2} - 4^{2} + ........
Solution :
= 1^{2} - 2^{2} + 3^{2} - 4^{2} + ........ 40 terms
= (1-4) + (9-16) + ........ 20 terms
= -3 -7 + ........ 20 terms
here a = -3, d = -7 - (-3) ==> -4 and n = 20
S_{n} = (n/2) [2a+(n-1)d]
= (20/2)[2(-3)+(20-1)(-4)]
= 10 [-6+19(-4)]
= 10[-6+19(-4)]
= 10(-6-76)
= 10 (-82)
= -820
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