10th CBSE math solution for exercise 5.2 part 8

This page 10th CBSE math solution for exercise 5.2 part 8 is going to provide you solution for every problems that you find in the exercise no 5.2 part 6

10th CBSE math solution for exercise 5.2 part 8

(16) Determine the AP whose third term is 16 and 7th term exceeds the 5th term by 12.

Solution:

a₃ = 16

a₇ = a₅ + 12

a + 2 d = 16

a + 6 d = a + 4 d + 12

 a - a + 6d - 4d = 12

    2 d = 12

      d = 12/2

      d = 6

 a + 2 (6) = 16

 a + 12 = 16

  a = 16 - 12

  a = 4

 4, 4 + 6, 4 + 2(6),.........

 4,10, 16,.................


(17) Find the 20th term from the last term of the AP: 3,8,13,.........253

Solution:

Common difference of the given sequence is 5

Now we need to find the 20th term from the last term of the sequence.So, we have to consider 253 as first term of the sequence 

253,248,243,...........,3

a = 253  d = 248 - 253    n = 20

              = -5  

a n = a + (n - 1) d

a ₀ = 253 + (20 - 1) (-5)

     = 253 - 19 (5)

     = 253 - 95

     = 158  


In this topic 10th CBSE math solution for exercise 5.2 part 8 we are going to see solution of 18th question

(18) The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms of the AP.

Solution:

 a₄ + a= 24

 a₆ +  a ₁₀= 44  

a + 3 d + a + 7 d = 24

 2 a + 10 d = 24

 2(a + 5 d) = 24

   a + 5 d = 24/2

    a + 5 d = 12 -------(1)

 a + 5 d + a + 9 d = 44

 2 a + 14 d = 44

 2(a + 7d) = 44

  a + 7 d = 44/2

    a + 7 d = 22 -------(2)

(1) - (2)

     a + 5 d = 12

     a + 7 d = 22

     (-)   (-)  (-)

     ------------

         - 2 d = -10

             d = 5

 Substitute d = 5 in the first equation

 a + 5(5) = 12

 a + 25 = 12

  a = 12 - 25

 a = -13

 a,a + d,a + 2d ,.........

 -13,-13 + 5,-13 + 2(5),...........

  -13,-8,-13 + 10,.......

  - 13,-8,-3,.........




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