161-NCERT-10-5-Arithmetic Progressions - Ex-5.2

NCERT

10th Mathematics

Exercise 5.2

Topic: 5 Arithmetic Progressions

Click here for ⇨ NCERT-10-5-Arithmetic Progressions - Ex- 5.1

EXERCISE 5.2

Q1. Fill in the blanks in the following table, given that a is the first term, d is the common difference, and a_n the nth term of the AP: 

	a	d	n	an
i	7	3	8	-----
ii	-18	-----	10	0
iii	-----	-3	18	-5
iv	-18.9	2.5	-----	3.6
v	3.5	0	105	-----

Explanation:

1) The nth term a_n of an AP with the first term 'a' and common difference 'd' is given

by a_n = a + (n – 1) d.

2) In general, a_n is known as the nth term of an AP, and a_r is known as the general

term or the rth term of an AP.

3) Sometimes a_n is also called the last term of an AP and is denoted by 'l'.

Solution:

i) a = 7, d = 3, n = 8, find a_n. 

1) According to the problem,

a = 7, d = 3, n = 8,

a_n = a + (n – 1) d

a_n = 7 + 3(8 – 1)

a_n = 7 + 3(8 – 1)

a_n = 7 + 3(7)

a_n = 7 + 21

a_n = 28

2) So, here a_n = 28.

ii) a = - 18, a_n = 0, n = 10, find d_.

1) According to the problem,

a = - 18, a_n = 0, n = 10,

a_n = a + (n – 1) d

0 = - 18 + d(10 – 1)

0 = - 18 + 9d

9d = 18

d = 18/9

d = 2.

2) So, here d = 2.

iii) d = - 3, a_n = - 5, n = 18, find a_.

1) According to the problem,

d = - 3, a_n = - 5, n = 18,

a_n = a + (n – 1) d

- 5 = a + (- 3)(18 – 1)

- 5 = a + (- 3)(17)

- 5 = a - 51

a = 51 - 5

a = 46.

2) So, here a = 46.

iv) a = - 18.9, d = 2.5, a_n = 3.6, find n_. 

1) According to the problem,

a = - 18.9, d = 2.5, a_n = 3.6,

a_n = a + (n – 1) d

3.6 = - 18.9 + (2.5)(n – 1)

(2.5)(n – 1) = 3.6 + 18.9

(2.5)(n – 1) = 22.5

(n – 1) = 22.5/2.5

(n – 1) = 225/25

(n – 1) = 9

n = 9 + 1

n = 10.

2) So, here n = 10.

v) a = 3.5, d = 0, n = 105, find a_n. 

1) According to the problem,

a = 3.5, d = 0, n = 105,

a_n = a + (n – 1) d

a_n = 3.5 + 0(105 – 1)

a_n = 3.5 + 0(104)

a_n = 3.5 + 0

a_n = 3.5

2) So, here a_n = 3.5.

Q2. Choose the correct choice in the following and justify :

(i) 30th term of the AP: 10, 7, 4, . . . , is

(A) 97 (B) 77 (C) –77 (D) – 87

(ii) 11th term of the AP: – 3, - 1/2, 2, . . ., is

(A) 28 (B) 22 (C) –38 (D) – 48 and 1/2

Explanation:

1) The nth term a_n of an AP with the first term 'a' and common difference 'd' is given

by a_n = a + (n – 1) d.

2) In general, a_n is known as the nth term of an AP, and a_r is known as the general

term or the rth term of an AP.

3) Sometimes a_n is also called the last term of an AP and is denoted by 'l'.

Solution:

(i) 30th term of the AP: 10, 7, 4, . . . , is

(A) 97 (B) 77 (C) –77 (D) – 87

1) According to the problem,

d = a₂ - a₁

d = 7 - 10
d = - 3

 2) So here,

a = 10, d = - 3, n = 30,

a_n = a + (n – 1) d 

a₃₀ = 10 + (30 – 1) (- 3)

a₃₀ = 10 - 3(29)

a₃₀ = 10 - 87

a₃₀ = - 77

3) So, the answer is B i.e. a₃₀ = - 77.

(ii) 11th term of the AP: - 3, - 1/2, 2, . . ., is

(A) 28 (B) 22 (C) –38 (D) – 48 and 1/2

1) According to the problem,

d = a₂ - a₁

d = (- 1/2) - (- 3)

d = - 1/2 + 3

d = (6 - 1)/2

d = 5/2

 2) So here,

a = - 3, d = 5/2, n = 11,

a_n = a + (n – 1) d 

a₁₁ = - 3 + (11 – 1) (5/2)

a₁₁ = - 3 + (5/2)(10)

a₁₁ = - 3 + 25

a₁₁ = 22

3) So, the answer is B i.e. a₁₁ = 22.

Q3. In the following APs, find the missing terms in the boxes :

(i) 2, 囗, 26

(ii) 囗, 13, 囗, 3

(iii) 5, 囗, 囗, 9½

(iv) - 4, 囗, 囗, 囗, 囗, 6

(v) 囗, 38, 囗, 囗, 囗, - 22

 

Explanation:

1) The nth term a_n of an AP with the first term 'a' and common difference 'd' is given

by a_n = a + (n – 1) d.

2) In general, a_n is known as the nth term of an AP, and a_r is known as the general

term or the rth term of an AP.

3) Sometimes a_n is also called the last term of an AP and is denoted by 'l'.

Solution:

(i) 2, 囗, 26

1) Here,

a₁ = 2, a₂ = ?, a₃ = 26

2) Using the formula a_n = a + (n – 1) d, we can find the value of d and all other terms. 

a₁ = a = 2 --------- equation 1

a₃ = a + (3 - 1)(d)

a₃ = 2 + d(2)

26 = 2 + 2d

2d = 26 - 2

2d = 24

d = 24/2

d = 12 --------- equation 2

3) From equation 1 and equation 2, we have a = 2, d = 12.

4) Using the formula a_n = a + (n – 1) d, we can find the value of a_2.

a_n = a + (n – 1) d

a₂ = 2 + (2 – 1) (12)

a₂ = 2 + 12(1)
a₂ = 14 

5) So, the missing term a₂ is 14.

(ii) 囗, 13, 囗, 3

1) Here,

a₁ = ? (say a), a₂ = 13, a₃ = ?, a₄ = 3

2) Using the formula a_n = a + (n – 1) d, we can find the value of d and all other terms. 

a₁ = a --------- equation 1

a₂ = a + (2 - 1)(d)

a₂ = a + d

13 = a + d

a + d = 13 --------- equation 2

3) Using formula a_n = a + (n – 1) d, and a₄ = 3 we have,

a_n = a + (n – 1) d

a₄ = a + (4 - 1)(d)

a₄ = a + 3d

3 = a + 3d

a + 3d = 3 --------- equation 3

4) Subtract equation 2 from equation 3, and we get,

a + 3d =   3

a +   d = 13

  ( - ) ( - )     ( - )  

  --------------------------- 

      2d = - 10 

d = - 10/2

d = - 5 --------- equation 4

5) Put d = - 5 from equation 4 in equation 2, we get,

a + d = 13

a + (- 5) = 13
a - 5 = 13
a = 13 + 5
a = 18 --------- equation 5

6) Now we will find a_3.

a_n = a + (n – 1) d

a₃ = a + (3 - 1)(d)

a₃ = 18 + 2 (- 5)

a₃ = 18 - 10

a₃ = 8

7) So, the missing term a₁ = 18, and a₃ = 8

(iii) 5, 囗, 囗, 9½

1) Here,

a₁ = a = 5, a₂ = ?, a₃ = ?, a₄ = 9½

2) Using the formula a_n = a + (n – 1) d, we can find the value of d and all other terms. 

a₁ = a = 5 --------- equation 1

a_n = a + (n – 1) d 

a₄ = 5 + (4 - 1)(d)

a₄ = 5 + 3d

9½ = 5 + 3d

3d = 9½ - 5

3d = 19/2 - 5

3d = (19 - 10)/2
3d = 9/2
d = 3/2 --------- equation 2

3) Using formula a_n = a + (n – 1) d, a = 5, d = 3/2 we have,

a_n = a + (n – 1) d

a₂ = a + (2 - 1)(d)

a₂ = a + d 

a₂ = 5 + 3/2

a₂ = (10 + 3)/2

a₂ = 13/2 --------- equation 3

4) Now we will find a₂,

a_n = a + (n – 1) d

a₃ = a + (3 - 1)(d)

a₃ = a + 2d 

a₃ = 5 + 2(3/2)
a₃ = 5 + 3
a₃ = 8 --------- equation 4

5) So, the missing term a₂ = 13/2, and a₃ = 8.

(iv) - 4, 囗, 囗, 囗, 囗, 6

1) Here,

a₁ = a = - 4, a₂ = ?, a₃ = ?, a₄ = ?, a₅ = ?, a₆ = 6.

2) Using the formula a_n = a + (n – 1) d, we can find the value of d and all other

terms. 

a₁ = a = - 4 --------- equation 1

a_n = a + (n – 1) d 

a₆ = - 4 + (6 - 1)(d)

a₆ = - 4 + 5d

6 = - 4 + 5d

5d = 6 + 4

5d = 10

d = 10/5
d = 2 --------- equation 2

3) Using formula a_n = a + (n – 1) d, a = - 4, d = 2 we have,

a_n = a + (n – 1) d

a₂ = a + (2 - 1)(d)

a₂ = a + d 

a₂ = - 4 + 2

a₂ = - 2 --------- equation 3

4) Now we will find a₃,

a_n = a + (n – 1) d

a₃ = a + (3 - 1)(d)

a₃ = a + 2d 

a₃ = - 4 + 2(2)
a₃ = - 4 + 4
a₃ = 0 --------- equation 4

5) Now we will find a₄,

a_n = a + (n – 1) d

a₄ = a + (4 - 1)(d)

a₄ = a + 3d 

a₄ = - 4 + 3(2)
a₄ = - 4 + 6

a₄ = 2 --------- equation 5

6) Now we will find a₅,

a_n = a + (n – 1) d

a₅ = a + (5 - 1)(d)

a₅ = a + 4d 

a₅ = - 4 + 4(2)
a₅ = - 4 + 8

a₅ = 4 --------- equation 6

7) So, the missing term a₂ = - 2, a₃ = 0, a₄ = 2, a₅ = 4.

(v) 囗, 38, 囗, 囗, 囗, - 22

1) Here,

a₁ = a = ?, a₂ = 38, a₃ = ?, a₄ = ?, a₅ = ?, a₆ = - 22.

2) Using the formula a_n = a + (n – 1) d, we can find the value of d and all other

terms. 

a₁ = a --------- equation 1

a_n = a + (n – 1) d 

a₂ = a + (2 - 1)(d)

a₂ = a + d

38 = a + d

a + d = 38 --------- equation 2

3) Using formula a_n = a + (n – 1) d, a₆ = - 22 we have,

a_n = a + (n – 1) d

a₆ = a + (6 - 1)(d)

a₆ = a + 5d 

- 22 = a + 5d

a + 5d = - 22 --------- equation 3

4) Subtract equation 2 from equation 3, and we get,

a + 5d = - 22

a +   d =   38

  ( - ) ( - )     ( - )  

  --------------------------- 

      4d = - 60 

d = - 60/4

d = - 15 --------- equation 4

5) Put d = - 15 from equation 4 in equation 2, we get,

a + d = 38

a + (- 15) = 38
a - 15 = 38
a = 38 + 15
a = 53 --------- equation 5 

6) Now we will find a₃,

a_n = a + (n – 1) d

a₃ = a + (3 - 1)(d)

a₃ = a + 2d 

a₃ = 53 + 2(- 15)
a₃ = 53 - 30
a₃ = 23 --------- equation 6

7) Now we will find a₄,

a_n = a + (n – 1) d

a₄ = a + (4 - 1)(d)

a₄ = a + 3d 

a₄ = 53 + 3(- 15)
a₄ = 53 - 45

a₄ = 8 --------- equation 7

8) Now we will find a₅,

a_n = a + (n – 1) d

a₅ = a + (5 - 1)(d)

a₅ = a + 4d 

a₅ = 53 + 4(- 15)
a₅ = 53 - 60

a₅ = - 7 --------- equation 6

7) So, the missing term a₁ = 53, a₃ = 23, a₄ = 8, a₅ = - 7.

Q4. Which term of the AP : 3, 8, 13, 18, . . . , is 78?

Solution:

1) Here,

a₁ = a = 3, a₂ = 8, a₃ = 13, a₄ = 18.

2) Here.

d = a₂ - a₁

d = 8 - 3
d = 5 --------- equation 1

3) Here.

d = a₃ - a₂

d = 13 - 8
d = 5 --------- equation 2

4) From equation 1 and equation 2 as d = a₂ - a₁ = a₃ - a₂ = 5.

5) Using the formula a_n = a + (n – 1) d,

a_n = a + (n – 1) d 

78 = 3 + (n - 1)(d)

78 = 3 + 5(n - 1)

5(n - 1) = 78 - 3

5(n - 1) = 75

(n - 1) = 75/5

(n - 1) = 15

n = 15 + 1

n = 16 --------- equation 3

6) So, the 16th term of the given AP is 78.

Q5. Find the number of terms in each of the following APs :

(i) 7, 13, 19, . . . , 205     (ii) 18, 15½, 13, . . . , – 47

Explanation:

1) The nth term a_n of an AP with the first term 'a' and common difference 'd' is given

by a_n = a + (n – 1) d.

2) In general, a_n is known as the nth term of an AP, and a_r is known as the general

term or the rth term of an AP.

3) Sometimes a_n is also called the last term of an AP and is denoted by 'l'.

Solution:

(i) 7, 13, 19, . . . , 205

1) Here,

a₁ = a = 7, a₂ = 13, a₃ = 19, a_n = 205.

2) Here.

d = a₂ - a₁

d = 13 - 7
d = 6 --------- equation 1

3) Here.

d = a₃ - a₂

d = 19 - 13
d = 6 --------- equation 2

4) From equation 1 and equation 2 as d = a₂ - a₁ = a₃ - a₂ = 6. 

5) Using the formula a_n = a + (n – 1) d,

a_n = a + (n – 1) d 

205 = 7 + (n - 1)(6)

205 = 7 + 6(n - 1)

6(n - 1) = 205 - 7

6(n - 1) = 198

(n - 1) = 198/6

(n - 1) = 33

n = 33 + 1

n = 34

6) The number of terms in the given AP is 34.

(ii) 18, 15½, 13, . . . , – 47

1) Here,

a₁ = a = 18, a₂ = 15½, a₃ = 13, a_n = - 47.

2) Here.

d = a₂ - a₁

d = 15½ - 18

d = 31/2 - 18

d = (31 - 36)/2

d = - 5/2 --------- equation 1

3) Here.

d = a₃ - a₂

d = 13 - 15½

d = 13 - 31/2
d = (26 - 31)/2

d = - 5/2 --------- equation 2

4) From equation 1 and equation 2 as d = a₂ - a₁ = a₃ - a₂ = - 5/2. 

5) Using the formula a_n = a + (n – 1) d,

a_n = a + (n – 1) d 

- 47 = 18 + (n - 1)(- 5/2)

- 47 = 18 - 5(n - 1)/2

5(n - 1)/2 = 18 + 47

5(n - 1)/2 = 65

5(n - 1) = 2(65)

(n - 1) = 2(65)/5

(n - 1) = 2(13)

(n - 1) = 26

n = 26 + 1

n = 27

6) The number of terms in the given AP is 27.

 

Q6. Check whether – 150 is a term of the AP: 11, 8, 5, 2 . . .

1) Here,

a₁ = a = 11, a₂ = 8, a₃ = 5, a₄ = 2.

2) Here.

d = a₂ - a₁

d = 8 - 11

d = - 3

3) So d = a₂ - a₁ = - 3. 

4) Using the formula a_n = a + (n – 1) d,

a_n = a + (n – 1) d 

- 150 = 11 + (n - 1)(- 3)

- 150 = 11 - 3(n - 1)

3(n - 1) = 11 + 150

3(n - 1) = 161

(n - 1) = 161/3

n = (161/3) + 1

n = (161 + 3)/3

n = (164)/3 

5) As, n = 164/3 is not an integer, (- 150) is not a term of this AP.

Q7. Find the 31st term of an AP whose 11th term is 38 and the 16th term is 73.

Explanation:

1) The nth term a_n of an AP with the first term 'a' and common difference 'd' is given

by a_n = a + (n – 1) d.

2) In general, a_n is known as the nth term of an AP, and a_r is known as the general

term or the rth term of an AP.

3) Sometimes a_n is also called the last term of an AP and is denoted by 'l'.

Solution:

1) Here,

a₁₁ = 38, a₁₆ = 73, find a₃₁.

2) Let the first term be a and the common difference be d.

a_n = a + (n – 1) d 

a₁₁ = a + (11 - 1)(d)

a₁₁ = a + 10d

38 = a + 10d

a + 10d = 38 --------- equation 1

3) Using formula a_n = a + (n – 1) d, a₁₆ = 73 we have,

a_n = a + (n – 1) d

a₁₆ = a + (16 - 1)(d)

a₁₆ = a + 15d 

73 = a + 15d

a + 15d = 73 --------- equation 2

4) Subtract equation 1 from equation 2, and we get,

a + 15d = 73

a + 10d = 38

  ( - )  ( - )     ( - )  

  --------------------------- 

      5d = 35 

d = 35/5

d = 7 --------- equation 3

5) Put d = 7 from equation 3 in equation 1, we get,

a + 10d = 38

a + 10(7) = 38
a + 70 = 38
a = 38 - 70
a = - 32 --------- equation 4 

6) Now we will find a₃₁,

a_n = a + (n – 1) d

a₃₁ = a + (31 - 1)(d)

a₃₁ = a + 30d 

a₃₁ = - 32 + 30(7)
a₃₁ = - 32 + 210
a₃₁ = 178 --------- equation 5

7) So, here a₃₁ = 178.

Q8. An AP consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term.

Solution:

1) Here,

n = 50, a₃ = 12, a₅₀ = 106, find a₂₉.

2) Let the first term be "a" and the common difference be "d".

a_n = a + (n – 1) d 

a₃ = a + (3 - 1)(d)

a₃ = a + 2d

12 = a + 2d

a + 2d = 12 --------- equation 1

3) Using formula a_n = a + (n – 1) d, a₅₀ = 106 we have,

a_n = a + (n – 1) d

a₅₀ = a + (50 - 1)(d)

a₅₀ = a + 49d 

106 = a + 49d

a + 49d = 106 --------- equation 2

4) Subtract equation 1 from equation 2, and we get,

a + 49d = 106

a +  2d =    12

  ( - )  ( - )     ( - )  

  --------------------------- 

      47d = 94 

d = 94/47

d = 2 --------- equation 3

5) Put d = 2 from equation 3 in equation 1, we get,

a + 2d = 12

a + 2(2) = 12
a + 4 = 12
a = 12 - 4
a = 8 --------- equation 4 

6) Now we will find a₂₉,

a_n = a + (n – 1) d

a₂₉ = a + (29 - 1)(d)

a₂₉ = a + 28d 

a₂₉ = 8 + 28(2)
a₂₉ = 8 + 56
a₂₉ = 64--------- equation 5

7) So, here a₂₉ = 64.

Q9. If the 3rd and the 9th terms of an AP are 4 and – 8 respectively, which term of this AP is zero?

Solution:

1) Here,

a₃ = 4, a₉ = - 8, find which term is 0.

2) Let the first term be "a" and the common difference be "d".

a_n = a + (n – 1) d 

a₃ = a + (3 - 1)(d)

a₃ = a + 2d

4 = a + 2d

a + 2d = 4 --------- equation 1

3) Using formula a_n = a + (n – 1) d, a₉ = - 8 we have,

a_n = a + (n – 1) d

a₉ = a + (9 - 1)(d)

a₉ = a + 8d 

- 8 = a + 8d

a + 8d = - 8 --------- equation 2

4) Subtract equation 1 from equation 2, and we get,

a + 8d = - 8

a + 2d =   4

  ( - )  ( - )     ( - )  

  --------------------------- 

      6d = - 12 

d = - 12/6

d = - 2 --------- equation 3

5) Put d = - 2 from equation 3 in equation 1, we get,

a + 2d = 4

a + 2(- 2) = 4
a - 4 = 4
a = 4 + 4
a = 8 --------- equation 4 

6) Now we will find n when a_n = 0.

a_n = a + (n – 1) d

0 = 8 + (n - 1)(- 2)

0 = 8 - 2(n - 1) 

2(n - 1) = 8

(n - 1) = 8/2

(n - 1) = 4 

n = 5 --------- equation 5

7) So, here the 5th term is 0.

Q10. The 17th term of an AP exceeds its 10th term by 7. Find the common difference.

Solution:

1) Here,

a₁₇ = a₁₀ + 7 --------- equation 1.

2) Let the first term be "a" and the common difference be "d".

a_n = a + (n – 1) d 

a₁₇ = a + (17 - 1)(d)

a₁₇ = a + 16d --------- equation 2

3) Using formula a_n = a + (n – 1) d, a₉ = - 8 we have,

a_n = a + (n – 1) d

a₁₀ = a + (10 - 1)(d)

a₁₀ = a + 9d --------- equation 3

4) From equations 1, 2, and 3 we have,

a₁₇ = a₁₀ + 7

a + 16d = a + 9d + 7

a + 16d - a - 9d = 7

16d - 9d = 7 

7d = 7 

d = 7/7

d = 1 --------- equation 4

5) So, here the common difference is 1.

Q11. Which term of the AP : 3, 15, 27, 39, . . . will be 132 more than its 54th term?

Solution:

1) Here,

a₁ = a = 3, a₂ = 15, a₃ = 27, a₄ = 39.

2) Here.

d = a₂ - a₁

d = 15 - 3
d = 12 --------- equation 1

3) Using the formula a_n = a + (n – 1) d,

a_n = a + (n – 1) d 

a₅₄ = 3 + 12 (54 – 1)

a₅₄ = 3 + 12 (53)

a₅₄ = 3 + 636

a₅₄ = 639 --------- equation 2

4) Now we will find n for the term more by 132 than the 54th term.

a_n = a₅₄ + 132

a_n = 639 + 132
a_n = 771 --------- equation 3

5) Now we will find the value of n. 

a_n = a + (n – 1) d 

771 = 3 + 12 (n – 1)

12 (n – 1) = 771 - 3

12 (n – 1) = 768

(n – 1) = 768/12

(n – 1) = 64

n = 64 + 1

n = 65 --------- equation 4

6) So, 65th term is 132 more than 54th term.

7) Second method: our term is 132 more than the 54th term. As the difference is 12,

132 will give us 132/12 = 11. So our term is 54 + 11 = 65. So, 65th term is 132 more than 54th term.

Q12. Two APs have the same common difference. The difference between their 100th terms is 100, what is the difference between their 1000th terms?

Solution:

1) Let the first terms of two APs be "x"  and "y", and the common difference be "d".

2) The 100th term first AP of which the first term is x:

a_n = a + (n – 1) d 

a₁₀₀ = x + (100 - 1)(d)

a₁₀₀ = x + 99d --------- equation 1

3) The 100th term of the second AP of which the first term is y:

a_n = a + (n – 1) d 

a₁₀₀ = y + (100 - 1)(d)

a₁₀₀ = y + 99d --------- equation 2

4) As the difference between their 100th term 100, we have,

(x + 99d) - (y + 99d) = 100

(x - y) = 100 --------- equation 3

5) The 1000th term first AP of which the first term is x:

a_n = a + (n – 1) d 

a₁₀₀₀ = x + (1000 - 1)(d)

a₁₀₀₀ = x + 999d --------- equation 4

6) The 1000th term second AP of which the first term is y:

a_n = a + (n – 1) d 

a₁₀₀₀ = y + (1000 - 1)(d)

a₁₀₀₀ = y + 999d --------- equation 5

7) Now we find the difference between their 1000th terms,

The difference = (x + 999d) - (y + 999d)

The difference = (x - y) --------- equation 6

8) From equations 3 and 6, we have, 

The difference = 100

9) So, the difference between their 1000th terms is 100.

Q13. How many three-digit numbers are divisible by 7?

Solution:

1) Here the first 3-digit number which is divisible by 7 is a = 105 and the common

difference d = 7.

2) The greatest 3-digit number is 999. When 7 divides 999, the remainder is 5. 

So 999-5 divisible by 7. i.e 994 is divisible by 7. So here a_n = 997.

3) Using the formula a_n = a + (n – 1) d, we have,

a_n = a + (n – 1) d 

994 = 105 + 7 (n – 1)

7 (n – 1) = 994 - 105

7 (n – 1) = 889

(n – 1) = 889/7

(n – 1) = 127

n = 127 + 1

n = 128

4) So, there are 128 3-digit numbers that are divisible by 7.

Q14. How many multiples of 4 lie between 10 and 250?

Solution:

1) Here the first number between 10 and 250 which is divisible by 4 is a = 12 and

the common difference d = 4.

2) When 4 divides 250, the remainder 2. So 250-2 is divisible by 4. i.e. 248 is

divisible by 4. So here a_n = 248.

3) Using the formula a_n = a + (n – 1) d, we have,

a_n = a + (n – 1) d 

248 = 12 + 4 (n – 1)

4 (n – 1) = 248 - 12
4 (n – 1) = 236
(n – 1) = 236/4

(n – 1) = 59

n = 59 + 1

n = 60

4) So, there are 60 numbers that are divisible by 4 between 10 and 250.

Q15. For what value of n, are the nth terms of two APs: 63, 65, 67, . . . and 3, 10, 17, . . . equal?

Solution:

1) For first AP, a₁ = a = 63, a₂ = 65, a₃ = 67.

2) Here.

d = a₂ - a₁

d = 65 - 63
d = 2

2) The nth term first AP will be,

a_n = a + (n – 1) d 

a_n = 63 + (n - 1)(2)

a_n = 63 + 2(n - 1) --------- equation 1

3) For first AP, a₁ = a = 3, a₂ = 10, a₃ = 17.

4) Here.

d = a₂ - a₁

d = 10 - 3
d = 7

5) The nth term second AP will be,

a_n = a + (n – 1) d 

a_n = 3 + (n - 1)(7)

a_n = 3 + 7(n - 1) --------- equation 2

6) As nth term of both the APs are the same, from equations 1 and 2 we have,

63 + 2(n - 1) = 3 + 7(n - 1)

7(n - 1) - 2(n - 1) = 63 - 3

7n - 7 - 2n + 2 = 60
5n - 5 = 60
5n = 60 + 5
5n = 65
n = 65/5
n = 13

7) Therefore, the 13th terms of both these A.P.s are equal.

Q16. Determine the AP whose third term is 16 and the 7th term exceeds the 5th term by 12.

Solution:

1) Let the first term of an AP be "a" and the common difference be "d".

2) Using the formula a_n = a + (n – 1) d,

a_n = a + (n – 1) d 

a₃ = a + (3 – 1) d

16 = a + 2d

a + 2d = 16 --------- equation 1

3) Now we will find the 5th term.

a_n = a + (n – 1) d

a₅ = a + (5 – 1) d 
a₅ = a + 4d --------- equation 2

4) Now we will find the 7th term.

a_n = a + (n – 1) d

a₇ = a + (7 – 1) d 
a₇ = a + 6d --------- equation 3 

5) As the 7th term exceeds the 5th term by 12, we have,

a₇ = a₅ + 12 --------- equation 4

6) So, from equations 2, 3, and 4, we have,

a₇ = a₅ + 12

a + 6d = a + 4d + 12
a + 6d - a - 4d = 12
6d - 4d = 12
2d = 12
d = 12/2
d = 6 --------- equation 5 

7) Put d = 6 from equation 5 in equation 1, and we get.

a + 2d = 16

a + 2(6) = 16

a + 12 = 16

a = 16 - 12

a = 4 --------- equation 6

8) So, the AP will be 4, 10, 16, 22...

Q17. Find the 20th term from the last term of the AP : 3, 8, 13, . . ., 253.

Solution:

1) Here,

a₁ = a = 3, a₂ = 8, a₃ = 13, a_n = 253.

2) Here.

d = a₂ - a₁

d = 8 - 3

d = 5 --------- equation 1 

3) Using the formula a_n = a + (n – 1) d,

a_n = a + (n – 1) d 

253 = 3 + (n - 1)(5)

253 = 3 + 5(n - 1)

5(n - 1) = 253 - 3

5(n - 1)/2 = 65

5(n - 1) = 2(65)

(n - 1) = 2(65)/5

(n - 1) = 2(13)

(n - 1) = 26

n = 26 + 1

n = 27

6) The number of terms in the given AP is 27.

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