Arithmetic Series Worksheet Solution3





In the page arithmetic series worksheet solution3 you are going to see solution of each questions from the arithmetic series worksheet.

(7) 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

              = -4

           (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.


(8) 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     and L = 999      

              d = 9      

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


(9) Find the sum of first 20 terms of the arithmetic series in which 3 rd term is 7 and 7th term is 2 more than three times its 3rd term.

Solution:

 3rd term = 7

          t₃ = 7

          t₇ = 3(t₃) + 2

          t₇ = 3(7) + 2

          t₇ = 21 + 2

          t₇ = 23

      a + 2d = 7   -----(1)

      a + 6d = 23 -----(2)

Solving (1) and (2) we get

  - 4 d = -16

       d = 16/4

       d = 4

Substitute d = 4 in the first equation

      a + 2 (4) = 7

      a + 8 = 7

            a = 7 - 8 

           a = -1

Now we need to find sum of 20 terms

   Sn = (n/2) [ 2a + (n-1) d ]

  S₂₀ = (20/2) [ 2(-1) + (20-1) (4) ]

       = 10 [ - 2 + 19 (4) ]

       = 10 [ - 2 + 76 ]

       = 10 [ 74 ]

       = 740



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arithmetic series worksheet solution3