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

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

Solution:

**a****₁****₁** = 38

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

a₁₆ = 73

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

(1) - (2)

a + 10 d = 38

a + 15 d = 73

(-) (-) (-)

--------------

- 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?**

**Let the first two terms of two APs a**₁ 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

(a₁ + 99 d) - (a₂ + 99 d) = 100

a₁ + 99 d - a₂ - 99 d = 100

a₁ - a₂ = 100

difference between 1000th term of two APs

= (a₁ + 999 d) - (a₂ + 999 d)

= a₁ + 999 d - a₂ - 999 d

= a₁ - a₂

= 100

difference between 1000th term of two APs = 100

HTML Comment Box is loading comments...