SUM OF ARITHMETIC PROGRESSION FORMULA

Sum of arithmetic progression formula :

An arithmetic series is a series whose terms form an arithmetic sequence.

We use the one of the formula given below to find the sum of arithmetic series.

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

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

  • "n" stands for the total number terms
  • "a" stands for the first term
  • "d" stands for common difference
  • "l" stands for the 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.

Sum of arithmetic progression formula - Examples

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 :

(ii) 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 x 63 

    = 7875

Problem 3 :

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 ]

  =  (25) [ -3 ]

  = -75

Problem 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 :

In this problem we have to find the number of terms are needed so that  their sum is 368

S n = 368

60,56,52,48,.......

 a = 60  d = 56 - 60 = -6

(n/2) [ 2a + (n-1) d ]  =  368

  (n/2) [ 2(60) + (n-1)(-4) ]  =  368           

  (n/2) [ 120 - 4 n + 4 ]  =  368           

  (n/2) [ 124 - 4 n ]  =  368           

  n  [ 124 - 4 n ]  =  368 x 2           

  n  [ 124 - 4 n ]  =  368 x 2       

  124 n - 4 n²  =  736           

  4 n² - 124 n + 736 = 0

÷ by 4 = > n² - 31 n + 184 = 0

   (n-23) (n-8) = 0 

   n = 23,8

Sum of 8 or 23 terms of arithmetic sequence is  368.

Problem 5 :

Find the sum of all 3 digit natural numbers,which are divisible by 9.

Solution :

3 digit number starts from 100 and ends with 999. From this sequence we have to find number of terms which are divisible by 9 and also we have to find their sum.

  108, 117, 126,........................999

The first number which is divisible by 9 from this sequence is 108, the second number will be 117 and the last number will be 999.

108 + 117+ 126 + ........... + 999

a = 108    d = 117 - 108 = 9  and L = 999      

to find number of terms, we have to use the formula for (n)

  n = [ (L-a)/d ]+ 1

  n = [(999-108)/9] + 1

  n = [891/9] + 1

  n = 99 + 1

  n = 100

 S n = (n/2) [a+l]

      = (100/2) [108 + 999]

      = 50 [1107]

      = 55350

Let us see the next example on "Sum of arithmetic progression formula"

Problem 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

                         S₁₁ = 44

     (11/2)[2a + (11-1) d] = 44

  2a + 10 d = (44 x 2)/11

  2a + 10 d = 4 x 2

  2 a + 10 d = 8   ----- (1)

Sum of next 11 term = 55

  S₂₂ = S₁₁ + 55

  S₂₂ = 44 + 55

  S₂₂ = 99

(22/2)[2a + (22-1) d] = 99

2a + 21 d = (99 x 2)/22

2a + 21 d = 9

2 a + 21 d = 9   ----- (2)

Solving (1) and (2) we get

   -11 d = -1

  d = 1/11

Substitute d = 1/11 in the first  equation we get

  2 a + 10 (1/11) = 8

  2 a + 10/11 = 8

  2 a  = 8 - (10/11) ==> 78/11 ==> a = 39/11

Therefore the series is (39/11) + (40/11) + (41/11) + ............

Let us see the next example on "Sum of arithmetic progression formula"

Problem 7 :

Solve 1 + 6 + 11 + 16 + ..........  + x = 148

Solution :

148 represents sum of x number of terms starting from 1.

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

a = 1  d = 5    

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

  =  [(x - 1)/5] + 1 ==> (x - 1 + 5)/5 ==>  (x + 4)/5  

148 = [(x + 4)/5] /2 [1+x]

148 = (x+4)/10 [1+x]

148 x 10 = (x + 4) (1 + x)

1480 = x + x² + 4 + 4x

1480 = x² + 4 + 5 x

x² + 5 x + 4 - 1480 = 0

x² + 5 x - 1476 = 0

(x - 36) (x + 41) = 0

x = 36,-41

Therefore the last term of the series is 36.

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