10th CBSE math solution for exercise 5.3 part 6

This page 10th CBSE math solution for exercise 5.3 part 6 is going to provide you solution for every problems that you find in the exercise no 5.3 part 6

10th CBSE math solution for exercise 5.3 part 6

(6) The first and last term of an AP are 17 and 350 respectively.If the common difference is 9, how many terms are there and what is their sum?

Solution:

a = 17       l = 350   d = 9  

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

S n= (n/2) [2 (17) + (n - 1) 9]

     = (n/2) [34 + 9 n - 9]

     = (n/2) [25 + 9 n]  ----(1)

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

    = (n/2)[17 + 350] ------(2)

(1) = (2)

(n/2) [25 + 9 n] = (n/2)[17 + 350]

        25 + 9 n = 17 + 350

                9 n = 367 - 25

                 9 n = 342

                    n = 342/9

                    n = 38 

38 terms are needed.

S₃₈ = (38/2)[17 + 350]

      = 19 [367]

      = 6973


(7) The sum of first 22 terms of an AP in which d = 7and 22nd term is 149.

Solution:

n = 22  d = 7 

a₂₂ = a + 21 d

     = a + 21 (7)

 149 = a + 147

 149 -147 = a

  a = 2

 S n = (n/2) [2a + (n-1)d]

      = (22/2) [2(2) + (22-1)7]

      = 11 [4 + 21(7)]

      = 11 [4 + 147]

      = 11 [151]

      = 1661


In the page 10th CBSE math solution for exercise 5.3 part 6 next we are going to see the solution of next problem.

(8) The sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.

Solution:

a₂ = a + 1d

14 = a + d

 a + d = 14 ---(1)

 a= a + 2d

 18 = a + 2d 

 a + 2 d = 18 ------(2)

 (1)-(2)

       a + d = 14

       a + 2 d = 18

     (-)   (-)   (-)

     --------------

            - d = -4

              d = 4

 Substitute d = 4 in the first equation 

        a + 4 = 14

            a = 14 - 4

             a = 10

 S n = (n/2) [2a + (n - 1) d]

 n = 51

S n  = (51/2)[2(10) + (51-1) 4]

       = (51/2)[20 + 50(4)]

       = (51/2) [20 + 200]

       = (51/2) [220]

       = 51 (110)

        = 5610        back to exercise




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