10th CBSE math solution for exercise 5.2 part 5

This page 10th CBSE math solution for exercise 5.2 part 5 is going to provide you solution for every problems that you find in the exercise no 5.2 part 5

10th CBSE math solution for exercise 5.2 part 5

(7) Find the 31st term of an AP whose 11th term is 38 and 16th term is 73

Solution:

a = 38

a + 10 d = 38 ----- (1)

a= 73

a + 15 d = 73 ----- (2)

(1) - (2)

            a + 10 d = 38

            a + 15 d = 73

         (-)   (-)   (-)

        --------------

            - 5 d = -35

                  d = 35/5

                   d = 7

Substitute d = 7 in the first equation

 a + 10 (7) = 38

  a + 70 = 38

        a = 38 - 70

        a = -32

a₃₁ = a + 30 d

     = -32 + 30 (7)

     = -32 + 210

     = 178


(8) An AP consists of 50 terms of which 3rd term is 12 and the last term is 73.

Solution:

   a₃ = 12

    l = 73

               = 6 

a n = 205

a + (n -1) d = 205

 7 + (n - 1) 6 = 205

        (n - 1) 6 = 205 - 7

         (n - 1) 6 = 198

           n - 1 = 198/6

            n - 1 = 33

              n = 33 + 1

             n = 34

Therefore total number of terms is 34


In this topic 10th CBSE math solution for exercise 5.2 part 5 we are going to see solution of 9th question

(9) If the 3rd and 9th terms of an AP are 4 and -8 respectively,which term of this AP is zero?

Solution:

   a₃ = 4

   a = -8

 a + 2 d = 4  ------- (1)

 a + 8 d = -8  ------- (2)

 (1) - (2)

             a + 2 d = 4 

            a + 8 d = -8

               (-)    (-)      (+)

              --------------

                  - 6 d = 12

                        d = -2

Substitute d = -2 in first equation

         a + 2 (-2) = 4

          a - 4 = 4

            a = 4 + 4

              a = 8 

 let us assume nth term is zero

        a n = 0

      a + (n - 1) d = 0

     8 + (n - 1) (-2) = 0

            (n - 1) (-2) = -8

               n - 1 = 4

                 n = 5

So fifth term of the sequence is zero.   




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