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 first n terms of an arithmetic series.
S_{n} = (n/2)[2a_{1}+ (n-1)d]
S_{n} = (n/2)[a_{1} + l]
n ---> number terms
a_{1} ----> first term
d ----> common difference
l ----> last term
When we do not know the last term, the first formula can be used to find the sum of first n terms of an arithmetic series.
When we know the last term, the second formula can be used to find the sum of first n terms of an arithmetic series.
Example 1 :
Find the sum of first 75 positive integers.
Solution :
1 + 2 + 3 + ..........+ 75
The above series is an arithmetic series with last term 75.
Since we know the last term, the second formula can be used to find the required sum.
S_{n} = (n/2)(a_{1} + l)
S_{75 }= (75/2)(1 + 75)
= (75/2)(76)
= 75 x 38
= 2850
Example 2 :
Find the sum of first 125 natural numbers.
Solution :
1 + 2 + 3 + ..........+ 125
The above series is an arithmetic series with last term 125.
Since we know the last term, the second formula can be used to find the required sum.
S_{n} = (n/2)(a_{1} + l)
S_{125} = (125/2)(1 + 125)
= (125/2)(1 + 125)
= (125/2)(126)
= 125 x 63
= 7875
Example 3 :
Find the sum of the terms in following arithmetic series.
6 + 5¼ + 4½ + ......... to 25 terms
Solution :
In the given arithmetic series a_{1} = 6 and n = 25.
Common difference :
d = a_{2} - a_{1}
= 5¼ - 6
= 21/4 - 6
= (21 - 24)/4
= -3/4
Since we do not know the last term, the first formula can be used to find the required sum.
Sn = (n/2)[2a_{1} + (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]
= (25)[-3]
= -75
Example 4 :
In an arithmetic sequence 60, 56, 52, 48, ....... starting from the first term, how many terms are needed so that their sum is 368?
Solution :
Let n be the number of terms required in the given arithmetic sequence to get the sum 368.
S_{n} = 368
(n/2)[2a_{1} + (n - 1)d] = 368 ----(1)
60, 56, 52, 48, .......
a_{1} = 60,
d = a_{2} - a_{1}
= 60 - 56
= -4
Substitute a_{1} = 60 and d = -4 in (1).
(n/2)[2(60) + (n - 1)(-4)] = 368
(n/2)[120 - 4n + 4] = 368
(n/2)[124 - 4n] = 368
Multiply both sides by 2.
n[124 - 4n] = 736
124n - 4n^{2} = 736
4n^{2} - 124n + 736 = 0
Divide both sides by 4.
n^{2} - 31n + 184 = 0
(n - 23)(n - 8) = 0
n = 23 or 8
Sum of 8 or 23 terms of the given arithmetic sequence is 368.
Example 5 :
Find the sum of all 3 digit natural numbers, which are divisible by 9.
Solution :
The first 3 digit number is 100 and the last 3 digit number is 9999.
3 digit numbers which are divisible by 9 :
108, 117, 126, ............ 999
3 digit numbers which are divisible by 9 form an arithmetic sequence.
a_{1} = 108
d = a_{2} - a_{1}
= 117 - 108
= 9
n = [(l - a_{1})/d] + 1
= [(999 - 108)/9] + 1
= [891/9] + 1
= 99 + 1
= 100
S_{n} = (n/2)(a + l)
= (100/2)[108 + 999]
= 50[1107]
= 55350
Example 6 :
In an arithmetic series, the sum of first 11 terms is 44 and the that of the next 11 terms is 55. Find the arithmetic series.
Solution :
Sum of first 11 terms = 44
S_{11} = 44
(11/2)[2a_{1} + (11 - 1)d] = 44
2a_{1} + 10d = (44 x 2)/11
2a_{1} + 10d = 4 x 2
2a_{1} + 10d = 8 ----(1)
Sum of next 11 terms = 55
Sum of first 11 terms + Sum of next 11 terms = 44 + 55
Sum of first 22 terms = 99
S_{22} = 99
(22/2)[2a_{1} + (22 - 1)d] = 99
11[2a_{1} + 21d = 99
2a_{1} + 21d = 9
2a_{1} + 21d = 9 ----(2)
(2) - (1) :
11d = 1
d = 1/11
Substitute d = 1/11 in (1).
2a_{1} + 10(1/11) = 8
2a_{1} + 10/11 = 8
2a_{1} = 8 - 10/11
2a_{1} = (88 - 10)/11
2a_{1} = 78/11
a_{1} = 39/11
a_{2} = 39/11 + 1/11 = 40/11
a_{3} = 40/11 + 1/11 = 41/11
Arithmetic series :
39/11 + 40/11 + 41/11 + ............
Example 7 :
Solve for x :
1 + 6 + 11 + 16 + .......... + x = 148
Solution :
The above series is an arithmetic series with
a_{1} = 1, d = 5 and l = x
n = [(l - a)/d] + 1
= [(x - 1)/5] + 1
= (x - 1 + 5)/5
= (x + 4)/5
S_{n} = 148
(n/2)[a_{1} + l] = 148
(1/2)(n)[a_{1} + l] = 148
(1/2)[(x + 4)/5](1 + x) = 148
[(x + 4)/10](1 + x) = 148
(x + 4)(x + 1) = 1480
x^{2} + x + 4x + 4 = 1480
x^{2} + 5x + 4 - 1480 = 0
x^{2} + 5x - 1476 = 0
(x - 36)(x + 41) = 0
x = 36 or -41
The given arithmetic series is an increasing series with positive values. So, the last term can never be a negative value.
Therefore the last term of the series is 36.
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