This page 10th CBSE math solution for exercise 5.3 part 1 is going to provide you solution for every problems that you find in the exercise no 5.3 part 1

(1) Find the sum of the following APs

(i) 2 , 7 , 12,....... to 10 terms

Solution:

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

n = 10 a = 2 d = 7 - 2

= 5

S ₁₀ = (10/2) [ 2 (2) + (10 - 1) 5 ]

= 5 [ 4 + 9 (5) ]

= 5 [ 4 + 45 ]

= 5 [49]

= 245

(ii) -37,-33,-29,..................... to 12 terms

Solution:

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

n = 12 a = -37 d = -33 - (-37)

= -33 + 37

= 4

S ₁₂ = (12/2) [ 2 (-37) + (12 - 1) 4 ]

= 6 [ -74 + 11 (4) ]

= 6 [ -74 + 44 ]

= 6 [-30]

= -180

(iii) 0.6,1.7,2.8,............... to 100 terms

Solution:

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

n = 100 a = 0.6 d = 1.7 - 0.6

= 1.1

S ₁₀₀ = (100/2) [ 2 (0.6) + (100 - 1) 1.1 ]

= 50 [ 1.2 + 99 (1.1) ]

= 50 [1.2 + 108.9 ]

= 50 [110.1]

= 5505

(iv) 1/15,1/12,1/10,................... to 11 terms

Solution:

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

n = 11 a = 1/15 d = (1/12) - (1/15)

= (5 - 4)/60

= 1/60

S ₁₁ = (11/2) [ 2 (1/15) + (11 - 1) (1/60) ]

= (11/2) [ (2/15) + (10/60) ]

= (11/2) [ (2/15) + (1/6) ]

= (11/2) [ (4 + 5)/30 ]

= (11/2) [ 9/30 ]

= (11/2) [ 3/10 ]

= 33/20

In the page 10th CBSE math solution for exercise 5.3 part 1 next we are going to see the solution of second problem.

(2) Find the sums given below

(i) 7 + 10 (1/2) + 14 + ...............+ 84

Solution:

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

a = 7 d = (21/2) - (7) l = 84

= (21 - 14)/2

= 7/2

a n = a + (n - 1) d

84 = 7 + (n - 1) (7/2)

84 - 7 = (n - 1) (7/2)

77 x (2/7) = n - 1

11 x 2 = n - 1

n - 1 = 22

n = 22 + 1 = 23

There are 23 terms in the above sequence

S₂₃ = (23/2) [7 + 84]

S₂₃ = (23/2) [ 91]

= 2093/2 back to exercise

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