10th CBSE math solution for exercise 5.3 part 3

This page 10th CBSE math solution for exercise 5.3 part 3 is going to provide you solution for every problems that you find in the exercise no 5.3 part 3

10th CBSE math solution for exercise 5.3 part 3

(ii) Given a = 7, a₁₃ = 35 find d and S₁₃

Solution:

a n = a + (n - 1) d

a₁₃ = 7 + (13 - 1) d

35 - 7 = 12 d

28/12 = d

d = 7/3

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

S₁₃ = (13/2) [7 + 35]

     = (13/2) (42)

     = 13 (21)          

     = 273


(iii) Given a₁₂ = 37 , d = 3 find a and S₁₂

Solution:

a n = a + (n - 1) d

a₁₂ = a + (12 - 1) 3

37 =  a + 11 (3)

37 = a + 33

37 - 33 = a

a = 4

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

S₁₂ = (12/2) [4 + 37]

     = 6 (41)

     = 246


In the page 10th CBSE math solution for exercise 5.3 part 3 next we are going to see the solution of next problem.

(iv) Given a= 15 , S₁₀ = 125 find d and a₁₀

Solution:

a n = a + (n - 1) d

a = a + (3 - 1) d

15 =  a + 2 d

a + 2 d = 15 -----(1)

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

S₁₀ = (10/2) [ 2 a + (10-1) d]

125 = 5 [2a + 9 d]

 125/5 = 2 a + 9 d

2 a + 9 d = 25  -----(2)

 (1) x 2 => 2 a + 4 d = 30

                2 a + 9 d = 25

                (-)     (-)     (-)

                ----------------

                       - 5 d = 5

                             d = - 1

  Substitute d = -1 in the first equation

   a + 2 (-1) = 15

          a = 15 + 2

          a = 17

a₁₀ = a + 9 d

    = 17 + 9 (-1)

    = 17 - 9

    = 8


(v) Given d = 5 , S₉ = 75 find a and a

Solution:

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

    S₉ = (9/2) [ 2a + (9 - 1) 5]

     75 = (9/2) [ 2 a + 8 (5) ]

     75 x (2/9) = 2 a + 40

      (50/3) -40 = 2 a

     2 a = (50 - 120)/3

         2 a = -70/3

           a = - 35/3

   a= a + 8 d

      = (-35/3) + 8 (5)

      = (-35/3) +  40

     = (-35 + 120)/3

     = 85/3




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