## NCERT10th MathematicsExercise 5.3Topic: 5 Arithmetic Progressions

## Click here for ⇨ NCERT-10-5-Arithmetic Progressions - Ex- 5.2

### EXERCISE 5.3

Q1. Find the sum of the following APs:

(i) 2, 7, 12, . . ., to 10 terms. (ii) –37, –33, –29, . . ., to 12 terms.

(iii) 0.6, 1.7, 2.8, . . ., to 100 terms. (iv) 1/15, 1/12, 1/10, . . ., to 11 terms.

### Explanation:

1) The nth term a

_{n}of an AP with the first term 'a' and common difference 'd' is givenby a_{n}= a + (n – 1) d.

2) The sum of the first n terms of an AP is given by:

S = (n/2)[2a + (n – 1) d]

3) If a

_{n}= l is the last term of an AP, thenS = (n/2)[2a + (n – 1) d]

S = (n/2)[a + (a + (n – 1) d)]

S = (n/2)[a + a_{n}]

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

### Solution:

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

1) Here, a = 2, d = 7 - 2 = 5, and n = 10

2) We know that the sum of the first n terms of an AP is given by:

S = (n/2)[2a + (n – 1) d]

S = (10/2)[2(2) + (10 – 1) (5)]

S = (5)[4 + 5(9)]3) So, the sum of the given AP is 245.

S = (5)[4 + 45]

S = (5)(49)

S = 245

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

1) Here, a = - 37, d = - 33 + 37 = 4, and n = 12

2) We know that the sum of the first n terms of an AP is given by:

S = (n/2)[2a + (n – 1) d]

S = (12/2)[2(- 37) + (12 – 1) (4)]

S = (6)[- 74 + 4(11)]3) So, the sum of the given AP is - 180.

S = (6)[- 74 + 44]

S = (6)(- 30)

S = - 180

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

1) Here, a = 0.6, d = 1.7 - 0.6 = 1.1, and n = 100

2) We know that the sum of the first n terms of an AP is given by:

S = (n/2)[2a + (n – 1) d]

S = (100/2)[2(0.6) + (100 – 1) (1.1)]

S = (50)[1.2 + 1.1(99)]3) So, the sum of the given AP is 5505.

S = (50)[1.2 + 108.9]

S = (50)(110.1)

S = 5505

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

1) Here, a = 1/15,

d = 1/12 - 1/15

d = (15 - 12)/(12x15)d = (3)/(12x15)

d = 1/(4x15)

d = 1/60 and

n = 11

2) We know that the sum of the first n terms of an AP is given by:

S = (n/2)[2a + (n – 1) d]

S = (11/2)[2(1/15) + (11 – 1) (1/60)]

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

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

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

S = (11/2)/(4 + 5)/30

S = (11/2)/(9/30)

S = (11/2)/(3/10)

S = 33/20

3) So, the sum of the given AP is 33/20.

**Q2. Find the sums given below :**

(i) 7 + 10½ + 14 + . . . + 84(ii) 34 + 32 + 30 + . . . + 10(iii) –5 + (–8) + (–11) + . . . + (–230)

### Solution:

**(i) 7 + 10½ + 14 + . . . + 84**

1) Here, a = 7, d = 10½ - 7 = 3½ = 7/2, nth term is a

_{n}= l = 842) We know that,

a_{n}= a + (n – 1) d

84 = 7 + (n – 1) (7/2)

(n – 1) (7/2) = 84 - 7

(n – 1) (7/2) = 77

(n – 1) = 77(2/7)

(n – 1) = 11(2)

(n – 1) = 22

n = 22 + 1

n = 23

3) We know that the sum of the first n terms of an AP is given by:

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

S = (23/2)[7 + 84]

S = (23/2)[91]4) So, the sum of the given AP is 1046.5.

S = 2093/2

S = 1046.5

**(ii) 34 + 32 + 30 + . . . + 10**

1) Here, a = 34, d = 32 - 34 = - 2, nth term is a

_{n}= 102) We know that,

a_{n}= a + (n – 1) d

10 = 34 + (n – 1) (- 2)

(n – 1) (- 2) = 10 - 34

(n – 1) (- 2) = - 24

(n – 1) = (- 24)/(- 2)

(n – 1) = 12

n = 12 + 1

n = 13

3) We know that the sum of the first n terms of an AP is given by:

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

S = (13/2)[34 + 10]

S = (13/2)[44]

S = (13)[22]

S = 286

4) So, the sum of the given AP is 286.

**(iii) –5 + (–8) + (–11) + . . . + (–230)**

1) Here, a = - 5, d = (- 8) - (- 5) = - 3, nth term is a

_{n}= - 2302) We know that,

a_{n}= a + (n – 1) d

- 230 = - 5 + (n – 1) (- 3)

(n – 1) (- 3) = 5 - 230

(n – 1) (- 3) = - 225

(n – 1) = (- 225)/(- 3)

(n – 1) = 75

n = 75 + 1

n = 76

3) We know that the sum of the first n terms of an AP is given by:

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

S = (76/2)[- 5 + (- 230)]

S = (38)[- 235]

S = - 8930

4) So, the sum of the given AP is - 8930.

**Q3. In an AP:**

(i) given a = 5, d = 3, a_{n}= 50, find n and S_{n}.(ii) given a = 7, a_{13}= 35, find d and S_{13}.(iii) given a_{12}= 37, d = 3, find a and S_{12}.(iv) given a_{3}= 15, S_{10}= 125, find d and a_{10}.(v) given d = 5, S_{n}= 75, find a and a_{9}.(vi) given a = 2, d = 8, S_{n}= 90, find n and a_{n}.(vii) given a = 8, a_{n}= 62, S_{n}= 210, find n and d.(viii) given a_{n}= 4, d = 2, S_{n}= – 14, find n and a.(ix) given a = 3, n = 8, S = 192, find d.(x) given l = 28, S = 144, and there are total 9 terms. Find a.

### Solution:

**(i) given a = 5, d = 3, a**

_{n}= 50, find n and S_{n}.

1) Here, a = 5, d = 3, nth term is a

_{n}= 50.2) We know that,

a_{n}= a + (n – 1) d

50 = 5 + (n – 1) (3)

(n – 1) (3) = 50 - 5

(n – 1) (3) = 45

(n – 1) = 45/3

(n – 1) = 15

n = 15 + 1

n = 16

3) We know that the sum of the first n terms of an AP is given by:

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

S = (16/2)[5 + 50]

S = (8)[55]

S = 440

4) So, S

_{n}= 440, n = 16.

**(ii) given a = 7, a**

_{13}= 35, find d and S_{13}.

1) Here, a = 7, d = ?, 13th term is a

_{13}= 35. Find d and S_{13}2) We know that,

a_{n}= a + (n – 1) d

a_{13}= 7 + (13 – 1) (d)

a_{13}= 7 + 12d

35 = 7 + 12d

12d = 35 - 7

12d = 28

d = 28/12

d = 7/3 --------- equation 1

3) We know that the sum of the first n terms of an AP is given by:

S_{n}= (n/2)[a + l]

S_{13}= (13/2)[7 + 35]

S_{13}= (13/2)[42]

S_{13}= (13)[21]

S_{13}= 273

4) So, S

_{13}= 273, d = 7/3.

**(iii) given a**

_{12}= 37, d = 3, find a and S_{12}.

1) Here, a = ?, d = 3, 12th term is a

_{12}= 37. Find a and S_{12}2) We know that,

a_{n}= a + (n – 1) d

a_{12}= a + (12 – 1) (3)

37 = a + 3(11)

37 = a + 33

a = 37 - 33

a = 4 --------- equation 1

3) We know that the sum of the first n terms of an AP is given by:

S_{n}= (n/2)[a + l]

S_{12}= (12/2)[4 + 37]

S_{12}= (6)[41]

S_{12}= 246

4) So, S

_{12}= 246, a = 4.**(iv) given a**

_{3}= 15, S_{10}= 125, find d and a_{10}.

1) Here, a

_{3}= 15, S_{10}= 125. Find d and a_{10}.2) We know that,

a_{n}= a + (n – 1) d

a_{3}= a + (3 – 1) (d)

15 = a + 2d

a + 2d = 15 --------- equation 1

3) We know that the sum of the first n terms of an AP is given by:

S_{n}= (n/2)[2a + (n - 1) d]

S_{10}= (10/2)[2(a) + (10 - 1) d]

125 = (5)[2(a) + 9d]

[2(a) + 9d] = 125/5

[2(a) + 9d] = 25

2a + 9d = 25 --------- equation 2

4) Subtract double of equation 1 from equation 2

2a + 9d = 25

2a + 4d = 30

( - ) ( - ) ( - )

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

5d = - 5

d = - 5/5

d = - 1 --------- equation 3

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

a + 2d = 15

a + 2(- 1) = 15

a - 2 = 15

a = 15 + 2

a = 17 --------- equation 4

6) We know that,

a_{n}= a + (n – 1) d

a_{10}= 17 + (10 – 1) (- 1)

a_{10}= 17 + 9 (- 1)

a_{10}= 17 - 9

a_{10}= 8

7) So, a

_{10}= 8, d = - 1.**(v) given d = 5, S**

_{9}= 75, find a and a_{9}.

1) Here, d = 5, S

_{9}= 75. Find a and a_{9.}2) We know that,

S_{n}= (n/2)[2a + (n - 1) d]

S_{9}= (9/2)[2a + (9 - 1) (5)]

75 = (9/2)[2a + 8 (5)]

75 = (9/2)[2a + 40]

75 = 9[a + 20]

[a + 20] = 75/9

[a + 20] = 25/3

a = (25/3) - 20

a = (25 - 60)/3

a = - 35/3

3) We know that the

a_{n}= a + (n – 1) d

a_{9}= - 35/3 + (9 – 1) (5)

a_{9}= - 35/3 + (8)(5)

a_{9}= - 35/3 + 40

a_{9}= (- 35 + 120)/3

a_{9}= 85/3

4) So, a

_{9}= 85/3, a = - 35/3.

**(vi) given a = 2, d = 8, S**

_{n}= 90, find n and a_{n}.

1) Here, a = 2, d = 8, S

2) We know that,_{n}= 90. Find n and a_{n.}S_{n}= (n/2)[2a + (n - 1) d]

S_{n}= (n/2)[2(2) + (n - 1)(8)]

90 = (n/2)[4 + 8(n - 1)]

90 = n[2 + 4(n - 1)]

90 = n(2 + 4n - 4)

90 = n(4n - 2)

90 = 4n^{2 }- 2n

45 = 2n^{2 }- n

2n^{2 }- n - 45 = 0

2n^{2 }- 10n + 9n - 45 = 0

2n(n^{ }- 5) + 9(n - 5) = 0

(n^{ }- 5)(2n + 9) = 0

(n^{ }- 5) = 0 or (2n + 9) = 0

n^{ }= 5 or 2n = - 9

n^{ }= 5 or n = - 9/2

3) As n is always a positive integer, ignore n = - 9/2. So we have n = 5.

4) We know that,aa_{n}= a + (n – 1) d_{5}= 2 + (5 – 1) (8)aa_{5}= 2 + (8)(4)_{5}= 2 + 32

a_{5}= 34

5) So, a

_{5}= 34, n = 5.

**(vii) given a = 8, a**

_{n}= 62, S_{n}= 210, find n and d.

1) Here, a = 8, a

_{n}= 62, S_{n}= 210. Find n and d.2) We know that,

S_{n}= (n/2)[a + l]

210 = (n/2)[8 + 62]3) Now we will find d using the formula:

210 = (n/2)[70]

210 = 35n

35n = 210

n = 210/35

n = 6

62 = 8 + 5da62 = 8 + (6 – 1) (d)_{n}= a + (n – 1) d

5d = 62 - 8

5d = 54

d = 54/5

d = 10.8

5) So, d = 10.8, n = 6.

**(viii) given a**

_{n}= 4, d = 2, S_{n}= – 14, find n and a.

1) Here, d = 2, a

_{n}= 4, S_{n}= - 14. Find n and a.2) We know that,

S_{n}= (n/2)[a + l]

- 14 = (n/2)[a + 4]

(n/2)[a + 4] = - 14

n[a + 4] = - 14(2)

n[a + 4] = - 28 --------- equation 1

3) We know that,

a4 = a + (n – 1) (2)_{n}= a + (n – 1) d4 = a + 2n – 2a + 2n = 4 + 2

a + 2n = 6

a = 6 - 2n --------- equation 2

4) Put a = 6 - 2n from equation 2 in equation 1, we get,

n(a + 4) = - 28

n((6 - 2n) + 4) = - 28

n(6 - 2n + 4) = - 28

n(10 - 2n) = - 28

2n(5 - n) = - 28

n(5 - n) = - 14

5n - n^{2 }= - 14

n^{2 }- 5n - 14 = 0

n^{2 }- 7n + 2n - 14 = 0

n(n^{ }- 7) + 2(n - 7) = 0

(n^{ }- 7)(n + 2) = 0

(n^{ }- 7) = 0 or (n + 2) = 0

n^{ }= 7 or n = - 2

5) As n is always a positive integer, ignore n = - 2. So we have n = 7.

6) Put n = 7 in equation 2, and we get,

a = 6 - 2na = 6 - 2(7)

a = 6 - 14

a = - 8

7) So, a = - 8, n = 7.

**(ix) given a = 3, n = 8, S = 192, find d.**

1) Here, a = 3, n = 8, sum of first n terms is S

_{n}= 192, find d.2) We know that the sum of the first n terms of an AP is given by:

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

192 = (8/2)[2(3) + (8 - 1) d]

192 = 4[6 + 7d]

4[6 + 7d] = 192

[6 + 7d] = 192/4

[6 + 7d] = 48

7d = 48 - 6

7d = 42

d = 42/7

d = 6

3) So, d = 6.

**(x) given l = 28, S = 144, and there are total 9 terms. Find a.**

1) Here, l = 28, S

_{n}= 144, n = 9, Find a.2) We know that,

S_{n}= (n/2)[a + l]

144 = (9/2)[a + 28]

(9/2)[a + 28] = 144

[a + 28] = 144(2/9)

[a + 28] = 16 (2)

a + 28 = 32

a = 32 - 28

a = 4

3) So, a = 4.

**Q4. How many terms of the AP: 9, 17, 25, . . . must be taken to give a sum of**

636?

### Solution:

1) Here, a

_{1}= a = 9, a_{2}= 17, a_{3}= 25, sum of first n terms is S_{n}= 636, find n.2) According to the problem,

d = a_{2}- a_{1}

d = 17 - 9

d = 8

3) We know that the sum of the first n terms of an AP is given by:

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

636 = (n/2)[2(9) + (n - 1) (8)]

636 = (n/2)[18 + 8(n - 1)]

636 = n[9 + 4(n - 1)]

636 = n[9 + 4n - 4]

636 = n[4n + 5]

636 = 4n^{2 }+ 5n

4n^{2 }+ 5n - 636 = 0

4n^{2 }+ 53n - 48n - 636 = 0

n(4n^{ }+ 53) - 12(4n + 53) = 0

(4n^{ }+ 53)(n - 12) = 0

(4n^{ }+ 53) = 0 or (n - 12) = 0

n^{ }= - 53/4 or n = 12

4) As n is always a positive integer, ignore n = - 53/4. So we have n = 12.

**Q5. The first term of an AP is 5, the last term is 45 and the sum is 400. Find the number of terms**

**and the common difference.**

### Solution:

1) Here, a

_{1}= a = 5, a_{n}= 45, S_{n}= 400, find n and d.2) We know that the sum of the first n terms of an AP is given by:

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

400 = (n/2)[(5) + 45]

400 = (n/2)[50]

400 = n[25]

n[25] = 400

n = 400/25

n = 163) Now we will find the value of d.

a_{n}= a + (n – 1) d

45 = 5 + d (16 – 1)

15d = 45 - 5

15d = 40

d = 40/15

d = 8/3

4) So, d = 8/3, n = 16.

**Q6. The first and the last terms of an AP are 17 and 350 respectively. If the common difference**

**is 9, how many terms are there, and what is their sum?**

### Solution:

1) Here, a

_{1}= a = 17, a_{n}= 350, d = 9, find n and S_{n}.2) Now we will find the value of d.

a_{n}= a + (n – 1) d

350 = 17 + 9(n – 1)

9(n – 1) = 350 - 17

9(n – 1) = 333

(n – 1) = 333/9

(n – 1) = 37

n = 37 + 1

n = 38

3) We know that the sum of the first n terms of an AP is given by:

S_{n}= (n/2)[a + l]

S_{n}= (38/2)[(17) + 350]

S_{n}= 19[17 + 350]

S_{n}= 19(367)

S_{n}= 6973

4) So, n = 38, S

_{n}= 6973.**Q7. Find the sum of the first 22 terms of an AP in which d = 7 and the 22nd term is 149.**

### Solution:

1) Here, a

_{22}= 149, d = 7, find and S_{22}.2) Now we will find the value of d.

a_{n}= a + (n – 1) d

a_{22}= a + (22 – 1) (7)

149 = a + 7(22 – 1)

a = 149 - 7(21)

a = 149 - 147

a = 2

3) We know that the sum of the first n terms of an AP is given by:

S_{n}= (n/2)[a + l]

S_{22}= (22/2)[(2) + 149]

S_{22}= (11)[151]

S_{22}= 1661

4) So, a = 2, S

_{n}= 1661.**Q8. Find the sum of the first 51 terms of an AP whose second and third terms are 14 and 18**

**respectively.**

### Solution:

1) Here, a

_{2}= 14, a_{3}= 18, find and S_{51}.2) According to the problem,

d = a_{3}- a_{2}

d = 18 - 14

d = 4

3) Now we will find the value of d.

a_{n}= a + (n – 1) d

a_{2}= a + (2 – 1) (4)

14 = a + 4(1)

a = 14 - 4

a = 10

4) We know that the sum of the first n terms of an AP is given by:

S_{n}= (n/2)[2a + (n - 1)d]

S_{51}= (51/2)[2(10) + (51 - 1)(4)]

S_{51}= (51/2)[20 + (50)(4)]

S_{51}= (51/2)[20 + 200]

S_{51}= (51/2)[220]

S_{51}= 51[110]

S_{51}= 5610

5) So, S

_{51}= 5610.**Q9. If the sum of the first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of the**

**first n terms.**

### Solution:

1) Here, S

_{7}= 49, S_{17}= 289, find and S_{n}. Let the first term be 'a' and the commondifference be 'd'.

2) We know that the sum of the first n terms of an AP is given by:

S_{n}= (n/2)[2a + (n - 1)d]

S_{7}= (7/2)[2a + (7 - 1)(d)]

49 = (7/2)[2a + (6)(d)]

49 = 7[a + 3d]

a + 3d = 49/7

a + 3d = 7 --------- equation 1

3) Now we will use S

_{17}= 289 to get another equationS_{n}= (n/2)[2a + (n - 1)d]

S_{17}= (17/2)[2a + (17 - 1)d]SS_{17}= (17/2)[2a + (16)d]_{17}= 17[a + 8d]

289 = 17[a + 8d]

17[a + 8d] = 289

a + 8d = 289/17

a + 8d = 17 --------- equation 2

4) Subtract equation 1 from equation 2

a + 8d = 17

a + 3d = 7

( - ) ( - ) ( - )

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

5d = 10

d = 10/5

d = 2 --------- equation 3

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

a + 3d = 7

a + 3(2) = 7

a + 6 = 7

a = 7 - 6

a = 1

2) We know that the sum of the first n terms of an AP is given by:

S_{n}= (n/2)[2a + (n - 1)d]

S_{n}= (n/2)[2(1) + (n - 1)(2)]

S_{n}= n[1 + (n - 1)]

S_{n}= n[n]

S_{n}= n^{2}

5) So, S

_{n}= n^{2}.**Q10. Show that a**

_{1,}a_{2,}. . ., a_{n}, . . . form an AP where a_{n}is defined as below :(i) a_{n}= 3 + 4n (ii) a_{n}= 9 – 5n.

Also, find the sum of the first 15 terms in each case.

### Solution:

**(i) a**

_{n}= 3 + 4n.1) Put n = 1, 2, 3 in equation a

_{n}= 3 + 4n to get a_{1}, a_{2}, a_{3 }and so on.a) Put n = 1,

a_{n}= 3 + 4n

a_{1}= 3 + 4(1)

a_{1}= 3 + 4a_{1}= 7

b) Put n = 2,a_{n}= 3 + 4na_{2}= 3 + 4(2)

a_{2}= 3 + 8a_{2}= 11

c) Put n = 3,a_{n}= 3 + 4na_{3}= 3 + 4(3)

a_{3}= 3 + 12

a_{3}= 15

2) Here, a

_{1}= 7, a_{2}= 11, a_{3}= 15.a) First difference:

d = a_{2}- a_{1}

d = 11 - 7

d = 4 --------- equation 1

b) Second difference:

d = a_{3}- a_{2}

d = 15 - 11

d = 4 --------- equation 2

3) Here the common difference d = a

_{2}- a_{1 }= a_{3}- a_{2 }= 4, so the equation a_{n}= 3 + 4nform an AP and their terms are 7, 11, 15 . . .

4) Now we will find the sum of the first 15 terms.

5) We know that the sum of the first n terms of an AP is given by:

S_{n}= (n/2)[2a + (n – 1) d]

S_{15}= (15/2)[2(7) + (15 – 1) (4)]

S_{15}= (15/2)[2(7) + 4(14)]

S_{15}= (15)[7 + 2(14)]

S_{15}= (15)[7 + 28]

S_{15}= (15)(35)

S_{15}= 525

6) So, here S

_{n}= 525.**(ii) a**

_{n}= 9 – 5n.

1) Put n = 1, 2, 3 in equation a

_{n}= 9 - 5n to get a_{1}, a_{2}, a_{3 }and so on.a) Put n = 1,

a_{n}= 9 - 5n

a_{1}= 9 - 5(1)

a_{1}= 9 - 5a_{1}= 4

b) Put n = 2,a_{n}= 9 - 5na_{2}= 9 - 5(2)

a_{2}= 9 - 10a_{2}= - 1

c) Put n = 3,

a_{n}= 9 - 5na_{3}= 9 - 5(3)a_{3}= 9 - 15a_{3}= - 6

2) Here, a

_{1}= 4, a_{2}= - 1, a_{3}= - 6.a) First difference:

d = a_{2}- a_{1}

d = - 1 - 4

d = - 5 --------- equation 1

b) Second difference:

d = a_{3}- a_{2}

d = - 6 - (-1)

d = - 6 + 1

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

3) Here the common difference d = a

_{2}- a_{1 }= a_{3}- a_{2 }= - 5,so the equation a_{n}= 9 - 5n form an AP and their terms are 4, - 1, - 6 . . .

4) Now we will find the sum of the first 15 terms.

5) We know that the sum of the first n terms of an AP is given by:

S_{n}= (n/2)[2a + (n – 1) d]

S_{15}= (15/2)[2(4) + (15 – 1) (- 5)]

S_{15}= (15/2)[2(4) - 5(14)]

S_{15}= (15)[4 - 5(7)]

S_{15}= (15)[4 - 35]

S_{15}= (15)[- 31]

S_{15}= - 465

6) So, here S

_{n}= - 465.**Q11. If the sum of the first n terms of an AP is 4n –**

**n**

^{2}**, what is the first term (that is S**

_{1}

**)? What**

**is the sum of the first two terms? What is the second term? Similarly, find the 3rd, the 10th, and**

**the nth terms.**

### Solution:

1) Here S

_{n}= 4n – n^{2}so,a) So, first term S_{1}will be,

S_{n}= 4n – n^{2}

S_{1}= 4(1) – (1)^{2}

S_{1}= 4 – 1S_{1}= 3 = a_{1}= a

b) To get sum of first two terms, put n = 2, in S_{n}= 4n – n^{2}

S_{n}= 4n – n^{2}S_{2}= 4(2) – (2)^{2}S_{2}= 8 – 4S_{2}= 4

2) We know that the second term a

_{2}= S_{2}- S_{1}.a_{2}= S_{2}- S_{1}

a_{2}= 4 - 3

a_{2}= 1

3) The common difference will be:

d = a_{2}- a_{1}

d = 1 - 3

d = - 2

4) Now we will find 3rd term:

a_{n}= a + (n – 1) d

a_{3}= 3 + (3 – 1) (- 2)

a_{3}= 3 + (2) (- 2)

a_{3}= 3 - 4

a_{3}= - 1

5) Now we will find the 10th term:

a_{n}= a + (n – 1) d

a_{10}= 3 + (10 – 1) (- 2)

a_{10}= 3 + (9) (- 2)

a_{10}= 3 - 18

a_{10}= - 15

6) Now we will find the nth term:

a_{n}= a + (n – 1) d

a_{n}= 3 + (n – 1) (- 2)

a_{n}= 3 - 2n + 2

a_{n}= 5 - 2n

7) So here we have:

a) First term S_{1}= a = 3.

b) Sum of first two terms S_{2}= 4.

c) The second term ad) The third term a_{2}= 1_{3}= - 1

e) The 10th term a_{10}= - 15

f) The nth term a_{n}= 5 - 2n

**Q12. Find the sum of the first 40 positive integers divisible by 6.**

### Solution:

1) Here a

_{1}will be 6, and a_{2}will be 12, so here d = 6.So, now we will find sum of first 40 integers,

S_{n}= (n/2)[2a + (n – 1) d

S_{40}= (40/2)[2(6) + (40 – 1) (6)]

S_{40}= 20[12 + 6(39)]S_{40}= 20[12 + 234]

S_{40}= 20[246]

S_{40}= 4920

2) So the sum of the first 40 terms will be 4920.

**Q**

**13. Find the sum of the first 15 multiples of 8.**

### Solution:

1) Here a

_{1}will be 8, and a_{2}will be 16, so here d = 8.So, now we will find sum of first 15 multiples of 8,

S_{n}= (n/2)[2a + (n – 1) d

S_{15}= (15/2)[2(8) + (15 – 1) (8)]

S_{15}= 15[8 + 4(14)]S_{15}= 15[8 + 56]

S_{15}= 15[64]

S_{15}= 960

2) So the sum of the first 15 multiples of 8 will be 960.

**Q14. Find the sum of the odd numbers between 0 and 50.**

### Solution:

1) Here, our odd numbers are 1, 3, 5, . . 47, 49.

2) We have a

_{1}= 1, a_{2}= 3, d = 2 and a_{n}= 49,a_{n}= a + (n – 1) d

49 = 1 + 2(n – 1)

2(n – 1) = 49 - 1

2(n – 1) = 48(n – 1) = 48/2

(n – 1) = 24

n = 25

3) According to the problem,

So, now we will find sum of odd numbers from 0 to 50.

S_{n}= (n/2)[a + l]

S_{25}= (25/2)[1 + 49]

S_{25}= (25/2)[50]S_{25}= 25[25]

S_{25}= 625

4) The sum of all odd numbers between 0 and 50 is 625.

**Q15. A contract on a construction job specifies a penalty for delay of completion beyond a**

**certain date as follows: Rs 200 for the first day, Rs 250 for the second day, Rs 300 for the third**

**day, etc., the penalty for each succeeding day being Rs 50 more than for the preceding day.**

**How much money does the contractor have to pay as a penalty if he has delayed the work by 30**

**days?**

### Solution:

1) Here a

_{1}will be 200, and a_{2}will be 250, so here d = 50.So, now we will find total penalty for the delayed work by 30 days, so we will have to find S_{n}.

S_{n}= (n/2)[2a + (n – 1) d

S_{30}= (30/2)[2(200) + (30 – 1) (50)]

S_{30}= 15[400 + 50(29)]S_{30}= 15[400 + 1450]

S_{30}= 15[1850]

S_{30}= 27750

2) So the total penalty for the delayed work by 30 days is 27750.

**Q16. A sum of Rs 700 is to be used to give seven cash prizes to students of a school for their**

**overall academic performance. If each prize is Rs 20 less than its preceding prize, find the**

**value of each of the prizes.**

### Solution:

1) Let the cost of the first prize be Rs x.

2) According to the problem, the cost of the second prize is (x - 20).

3) The cost of the third prize is (x - 40). So the common difference is d = - 20.

4) As the sum of Rs 700 is to be used give 7 cash prizes, S

_{7}= 700.So, now we will find total penalty for the delayed work by 30 days, so we will have to find S_{n}.

S_{n}= (n/2)[2a + (n – 1) d

S_{7}= (7/2)[2(x) + (7 – 1) (- 20)]

S_{7}= (7)[(x) + (6) (- 10)]700 = (7)[(x) - 60]

100 = x - 60

x = 100 + 60

x = 160

5) So, the value of each of the prizes was (i) Rs 160, (ii) Rs 140, (iii) Rs 120,

(iv) Rs 100, (v) Rs 80, (vi) Rs 60, and (vii) Rs 40.

**Q17. In a school, students thought of planting trees in and around the school to reduce air**

**pollution. It was decided that the number of trees, that each section of each class will**

**plant, will be the same as the class, in which they are studying, e.g., a section of Class I**

**will plant 1 tree, a section of Class II will plant 2 trees and so on till Class XII. There are**

**three sections of each class. How many trees will be planted by the students?**

### Solution:

1) Here a

_{1}will be 1, and a_{2}will be 2, a_{n}will be 12, so here d = 1.So, now we will find total penalty for the delayed work by 30 days, so we will have to find S_{n}.

S_{n}= (n/2)[2a + (n – 1) d

S_{12}= (12/2)[2(1) + (12 – 1) (1)]

S_{12}= 6[2 + 1(11)]S_{12}= 6[2 + 11]

S_{12}= 6[13]

S_{12}= 78

2) So one section of a school will plant 78 trees. So 3 sections will plant

78 x 3 = 234 plants.

**Q18. A spiral is made up of successive semicircles, with centers alternately at A and B, starting with center at A, of radii 0.5 cm, 1.0 cm, 1.5 cm, 2.0 cm, . . . as shown in**

**Fig. What is the total length of such a spiral made up of thirteen consecutive**

**semicircles? (Take p = 22/7)**

a) r = 0.5 so a_{1 }= 0.5 𝞹 = 𝞹/2

b) r = 1 so a_{2 }= 𝞹 = 𝞹

c) r = 1.5 so a_{3 }= 1.5 𝞹 = 3𝞹/2

3) Here a

_{1, }a_{2, }and a_{3}_{, }are the semicircle's perimeter and are in AP.4) Here d = a

_{2 }- a_{1 }= 𝞹 - (𝞹/2) = (𝞹/2)5) We will have to find S

_{13}, so we have,S_{n}= (n/2)[2a + (n – 1) d

S_{13}= (13/2)[2(𝞹/2) + (13 – 1) (𝞹/2)]

S_{13}= (13/2)(𝞹/2)[2 + 12]

S_{13}= (13/2)(𝞹/2)[14]

S_{13}= (13/2)(𝞹)[7]

S_{13}= (13/2)(22/7)[7]

S_{13}= (13/2)(22)

S_{13}= (13)(11)

S_{13}= 143

6) So, the length of such a spiral of thirteen consecutive semicircles is 143 cm.

**Q19. 200 logs are stacked in the following manner: 20 logs in the bottom row, 19 in the next row, 18 in the row next to it, and so on. In how many rows are the 200 logs placed and how many logs are in the top row?**

### Solution:

1) Here a number of logs in the rows are in an AP.

2) According to the problem, a

_{1}_{ }= 20_{, }a_{2}_{ }= 19_{, }a_{3}_{ }= 18. . . and S_{n }= 200, so d = - 1,3) We will have to find n using the above information.

S_{n}= (n/2)[2a + (n – 1) d

S_{n}= (n/2)[2(20) + (n – 1) (- 1)]

200 = (n/2)[40 - (n - 1)]

200 = (n/2)[40 - n + 1)]

400 = n(41 - n)

400 = 41n - n^{2}

n^{2}- 41n + 400 = 0

n^{2}- 16n - 25n + 400 = 0

n(n - 16) - 25(n - 25) = 0

(n - 16)(n - 25) = 0

(n - 16) = 0 or (n - 25) = 04) Now we will find a

n = 16 or n = 25

_{16}and a

_{25}using the formula a

_{n}= a + (n – 1) d.

a) First we will find a_{16}

a_{n}= a + (n – 1) da_{16}= 20 + (- 1) (16 – 1)

a_{16}= 20 - (15)

a_{16}= 5

b) First we will find a_{25}

a_{n}= a + (n – 1) da_{25}= 20 + (- 1) (25 – 1)a_{25}= 20 - (24)a_{25}= - 4

5) As the number of logs can't be negative, a

_{25}= - 4 is impossible. So 200 logscan be placed in 16 rows and there are 5 logs in the 16th log.

**Q20. In a potato race, a bucket is placed at the starting point, which is 5 m from the first potato, and the other potatoes are placed 3 m apart in a straight line. There are ten potatoes in the line.**

**A competitor starts from the bucket, picks up the nearest potato, runs back with it, drops**

**it in the bucket, runs back to pick up the next potato, runs to the bucket to drop it in, and**

**continues in the same way until all the potatoes are in the bucket. What is the total**

**distance the competitor has to run?**

**[Hint: To pick up the first potato and the second potato, the total distance (in meters)**

**run by a competitor is 2 × 5 + 2 × (5 + 3)]**

### Solution:

1) The potatoes are placed at 5, 8, 11, 14, 17 . . . distances from the bucket.

2) These distances form an AP.

3) According to the problem, a competitor travels distances as 10, 16, 22, 28, . . .

4) So a

_{1}_{ }= 10_{, }a_{2}_{ }= 16_{, }a_{3}_{ }= 22, a_{4}_{ }= 28. . . , so d = 6,5) Now we will have to find S

_{10}using the above information.S_{n}= (n/2)[2a + (n – 1) d

S_{10}= (10/2)[2(10) + (10 – 1) (6)]

S_{10}= (10/2)(2)[10 + (9) (3)]

S_{10}= 10[10 + 27]

S_{10}= 10[37]

S_{10}= 370

6) A competitor will have to run a total distance of 370 m.

Need help with math? We're here to help! Our resources include NCERT textbooks, lessons on Arithmetic Progressions, and more. Join our community of students and teachers today! #mathhelp #NCERT #studentsuccess #ArithmeticProgressions #education #learning #students #teachers #math