10th CBSE math solution for exercise 5.3 part 2

This page 10th CBSE math solution for exercise 5.3 part 2 is going to provide you solution for every problems that you find in the exercise no 5.3 part 2

10th CBSE math solution for exercise 5.3 part 2

(ii) 34 + 32 + 30 + .................. + 10

Solution:

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

 a = 34    d = 32 - 34         l = 10 = a n

                 = -2

a n = a + (n - 1) d

10 = 34 + (n - 1) (-2)

10 - 34 = (n - 1) (-2)

-24 = (n - 1) (-2)

 -24/(-2) = (n - 1)

   12 = n - 1

   n = 12 + 1

   n = 13

There are 13 terms in the above sequence

S₃ = (13/2) [34 + 10]

 S = (13/2) [ 44]

        = 13 (22)

      = 286


(iii)  - 5 + (-8) + (-11) + .............+ (-230)

Solution:

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

 a = -5    d = -8 - (-5)         l = -230 = a n

                 = -8 + 5

                 = -3

a n = a + (n - 1) d

-230 = -5 + (n - 1) (-3)

-230 + 5 = (n - 1) (-3)

-225 = (n - 1) (-3)

 225/3 = (n - 1)

   75 = n - 1

   n = 75 + 1

   n = 76

There are 76 terms in the above sequence

S₇₆ = (76/2) [-8 + (-5)]

 S₇₆ = 38 [ -8 - 5]

        = 38 (-13)

      = -494


In the page 10th CBSE math solution for exercise 5.3 part 2 next we are going to see the solution of third problem.

(3) In an AP

(i) Given a = 5, d = 3, a n= 50 find n and Sn

Solution:

a n = a + (n - 1) d

50 = 5 + (n - 1) (3)

50 - 5 = (n - 1) (3)

45 = (n - 1) (3)

45/3 = n - 1

 n - 1 = 15

 n = 15 + 1

 n = 16

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

S₁₆ = (16/2) [5 + 50]

     = 8 (55)

     = 440                         back to exercise




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