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

(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

S₁₃ = (13/2) [34 + 10]

S₁₃ = (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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