Class 10 NCERT Maths Solution Exercise-5.2, Arithmetic Progression(A.P).

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NCERT MATHS CLASS - X 

Arithmetic Progressions

Exercise 5.2

Question 9:

If the 3^rd and the 9^th terms of an A.P. are 4 and − 8 respectively. Which term of this A.P. is zero.

Given , a₃ = 4, a₉ = −8
Since, a_n = a + (n − 1) d
a₃ = a + (3 − 1) d
4 = a + 2d ................(I)
a₉ = a + (9 − 1) d
−8 = a + 8d .............(II)

On subtracting equation (I) from (II),
−12 = 6d
d = −2

From equation (I),

4 = a + 2 (−2)
4 = a − 4
a = 8

Let n^th term of this A.P. be zero.

a_n = a + (n − 1) d
0 = 8 + (n − 1) (−2)
0 = 8 − 2n + 2
2n = 10
n = 5
Hence, 5^th term of this A.P. is 0.

Question 10:

If 17^th term of an A.P. exceeds its 10^th term by 7. Find the common difference.

In A.P. n^th term is given by

a_n = a + (n − 1) d

a₁₇ = a + (17 − 1) d

a₁₇ = a + 16d

Similarly, a₁₀ = a + 9d

It is given that

a₁₇ − a₁₀ = 7

(a + 16d) − (a + 9d) = 7

7d = 7

d = 1

Therefore, the common difference is 1.

Question 11:

Which term of the A.P. 3, 15, 27, 39,… will be 132 more than its 54^th term?

A.P. is 3, 15, 27, 39,…

Given
a = 3
d = a₂ − a₁ = 15 − 3 = 12

In A.P. n^th term is given by 

a_n = a + (n − 1) d
a₅₄ = a + (54 − 1) d
= 3 + (53) (12)
= 3 + 636 = 639
a₅₄ = 639
 
The term 132 more than its 54^th term = 639 + 132 = 771.

Let n^th term be 771.
a_n = a + (n − 1) d
771 = 3 + (n − 1) 12
768 = (n − 1) 12
(n − 1) = 64
n = 65
Therefore, 65^th term was 132 more than 54^th term.

Question 12:

Two APs have the same common difference. The difference between their 100^th term is 100, what is the difference between their 1000^th terms?

Let the first term of these A.P.s be a₁ and a₂ respectively and the common difference of these A.P.s be d.

For first A.P.,

a₁₀₀ = a₁ + (100 − 1) d

= a₁ + 99d

a₁₀₀₀ = a₁ + (1000 − 1) d

a₁₀₀₀ = a₁ + 999d

For second A.P.,

a₁₀₀ = a₂ + (100 − 1) d

= a₂ + 99d

a₁₀₀₀ = a₂ + (1000 − 1) d

= a₂ + 999d

Given that, difference between 

100^th term of these A.P.s = 100

Therefore, (a₁ + 99d) − (a₂ + 99d) = 100

a₁ − a₂ = 100 .............(1)

Difference between 1000^th terms of these A.P.s

(a₁ + 999d) − (a₂ + 999d) = a₁ − a₂

From equation (1),

This difference, a₁ − a₂ = 100

Hence, the difference between 1000^th terms of these A.P. will be 100.  

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