This page 10th CBSE math solution for exercise 5.2 part 6 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 6

(10) The 17th term of an AP exceeds its 10th term by 7. Find the common difference.

Solution:

a₁₁ = 38

a + 10 d = 38 ----- (1)

a₁₆ = 73

a + 15 d = 73 ----- (2)

(1) - (2)

a + 10 d = 38

a + 15 d = 73

(-) (-) (-)

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- 5 d = -35

d = 35/5

d = 7

Substitute d = 7 in the first equation

a + 10 (7) = 38

a + 70 = 38

a = 38 - 70

a = -32

a₃₁ = a + 30 d

= -32 + 30 (7)

= -32 + 210

= 178

(11) Which term of the AP:3,15,27,39,.......... will be 132 more than its 54th term?

Solution:

a n = 132 + a 54

a + (n - 1) d = 132 + a + 53 d

a = 3 d = 15 - 3

= 12

3 + (n - 1) 12 = 132 + 3 + 53 (12)

3 + 12 n - 12 = 135 + 636

-9 + 12 n = 771

12 n = 771 + 9

12 n = 780

n = 780/12

n = 65

In this topic 10th CBSE math solution for exercise 5.2 part 6 we are going to see solution of 12th question

(12)
Two APs have the same common difference. The difference between their
100th term is 100, What is the difference between their 1000th terms?

Solution:

Let the first two terms of two APs a₁ and a₂ respectively and the common difference of these two A.Ps be d

for first A.P

a₁₀₀ = a₁ + (100 - 1) d

= a₁ + 99 d

a₁₀₀₀ = a₁ + (1000 - 1) d

= a₁ + 999 d

For second AP

a₁₀₀ = a₂ + (100 - 1) d

= a₂ + 99 d

a₁₀₀₀ = a₂ + (1000 - 1) d

= a₂ + 999 d

given that,difference between two 100th term of two APs = 100