## HOW TO FIND SUM OF AN ARITHMETIC SERIES

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

• We can use the first formula to find sum of n terms, if we don't have the last term of the given arithmetic series.
• We can use the second formula to find sum of n terms, if we have the last term of the given arithmetic series.

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]

S25 = (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 :

Sn = (n/2) [a+L]

=  (30/2) (5+121)

=  (15) [126]

S30  =  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 :

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

S25  =  (25/2) [2(50)+(25-1)(-4)]

=  (25/2) [100+(24)(-4)]

=  (25/2) (100 - 96)

=  (25/2)(4)

S25  =  50

So, sum of 25 terms is 50.

Problem 6 :

Find the sum of first 40 terms of the series

12 - 22 + 32 - 42  + ........

Solution :

=  12 - 22 + 32 - 42  + ........ 40 terms

=  (1-4) + (9-16)  + ........ 20 terms

=  -3 -7 + ........ 20 terms

here a = -3, d = -7 - (-3) ==> -4 and n = 20

Sn  =  (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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