5 ARITHMETIC PROGRESSIONS - testlabz.com · 5 ARITHMETIC PROGRESSIONS Exercise 5.1 Q.1. In which of the following situations, does the list of numbers involved make an arithmetic

CBSEPracticalSkills.com ©Edulabz International


1

5 ARITHMETIC PROGRESSIONSExercise 5.1

Q.1. In which of the following situations, does the list of numbers involved make an arithmetic progression, and why? (i) The taxi fare after each km when the fare is Rs 15 for the first km and Rs 8

for each additional km.

(ii) The amount of air present in a cylinder when a vacuum pump removes 14

of the air remaining in the cylinder at a time. (iii) The cost of digging a well after every metre of digging, when it costs Rs 150

for the first metre and rises by Rs 50 for each subsequent metre. (iv) The amount of money in the account every year when Rs 10000 is

deposited at compound interest at 8% per annum. Solution.

(i) Taxi fare for 1 km = Rs 15 = a1

Taxi fare for 2 kms 2Rs 15 Rs 8 Rs 23 a= + = = Taxi fare for 3 kms 3Rs 23 Rs 8 Rs 31 a= + = = Taxi fare for 4 kms 4Rs 31 Rs 8 Rs 39 a= + = = and so on. 1 2 1 Rs. 23 Rs. 15 Rs. 8d a a= − = − = 2 3 2 Rs. 31 Rs. 23 Rs. 8d a a= − = − = 3 4 3 Rs. 39 Rs. 31 Rs. 8d a a= − = − =⇒ 1 2d d d= = 3So the given list of numbers form an arithmetic progression with first term a = Rs 15

and common difference Rs. 8d =(ii) Amount of air present in the cylinder x= units 1a= Air present in the cylinder after one time removal of air by the vacuum

pump

34 4x xx= − = units 2a=

Amount of air present in the cylinder after two times removal of air by the vacuum pump

3 1 3 3 3 94 4 4 4 16 16x x x x⎛ ⎞= − = − =⎜ ⎟

⎝ ⎠x units

234

x⎛ ⎞= ⎜ ⎟⎝ ⎠

units 3a=

and so on.

1 2 13 –4 4x xd a a x= − = − = units



2

2

2 3 23 3 – 34 4 16

d a a x x⎛ ⎞= − = − =⎜ ⎟⎝ ⎠

x units

⇒ 1 2d d≠Hence, the given situation does not form an arithmetic progression. (iii) Cost of digging the well after 1 metre of digging 1Rs. 150 a= = Cost of digging the well after 2 metres of digging

2Rs. 150 Rs. 50 Rs. 200 a= + = = Cost of digging the well after 3 metres of digging

3Rs. 200 Rs. 50 Rs. 250 a= + = =and so on. 1 2 1 Rs. 200 Rs. 150 Rs. 50d a a= − = − = 2 3 2 Rs. 250 Rs. 200 Rs. 50d a a= − = − = 3 4 3 Rs. 300 Rs. 250 Rs. 50d a a= − = − =⇒ = d1d d= 2 3

Hence, the given list of numbers forms an arithmetic progression.

(iv) We know amount 1100

nrP ⎛ ⎞= +⎜ ⎟⎝ ⎠

Where Principal, rate, timeP r n= = = (Given) 10000, 8%P r= =

Amount of money after 1 year 1

18 Rs 10000 1

100a⎛ ⎞= + =⎜ ⎟

⎝ ⎠


28 Rs 10000 1

100a⎛ ⎞= + =⎜ ⎟

⎝ ⎠


38 Rs 10000 1

100a⎛ ⎞= + =⎜ ⎟

⎝ ⎠

and so on.

2 1

1 2 1

2

8 810000 1 10000 1100 100

8 810000 1 1 1100 100

8 810000 1100 100

d a a ⎛ ⎞ ⎛ ⎞= − = + − +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎛ ⎞ ⎡ ⎤= + + −⎜ ⎟ ⎢ ⎥⎝ ⎠ ⎣ ⎦

⎛ ⎞ ⎛ ⎞= +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

3 2

2 3 2

2

8 810000 1 10000 1100 100

8 810000 1 1 1100 100

d a a ⎛ ⎞ ⎛ ⎞= − = + − +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎡ ⎤= + + −⎜ ⎟ ⎢ ⎥⎝ ⎠ ⎣ ⎦



3

⇒ 1 2d d≠Hence, the given list of numbers does not form an arithmetic progression.

Q.2. Write first four terms of the AP, when the first term a and the common difference d are given as follows: (i) = 10, = 10a d(ii) = –2, = 0a d(iii) = 4, = – 3a a

(iv) a = – 1, d = 12

(v) = – 1.25, = – 0.25a dSolution.

(i) Given, first term 10a= = Common difference 10d= = ∴ 1 2 110; 10 10 20.a a a a d= = = + = + = a3 = a2 + d = 20 + 10 = 30 a4 = a3 + d = 30 + 10 = 40 Hence, first four terms of the AP are 10, 20, 30, 40... (ii) Given, first term 2a= = − Common difference 0d= = ∴ 1 2a a= = − 2 1 2 0 2a a d= + = − + = − = –2 3 2 2 0a a d= + = − + 4 3 2 0 2a a d= + = − + = −Hence, first four terms of the AP are 2, 2, 2, 2,− − − − … (iii) Given, first term 4a= = Common difference 3d= = − ∴ 1 4a a= = 2 1 4 3 1a a d= + = − = = –2 3 2 1 3a a d= + = − = –5 4 3 2 3a a d= + = − −Hence, first four terms of the AP are 4,1, 2, 5− − … (iv) Given, first term 1a= = −

Common difference 12

d= =

∴ 1 1a a= = −

2 11 112 2

a a d −= + = − + =

3 21 12 2

a a d= + = − + = 0



4

4 3102

a a d= + = + = 12

Hence, first four terms of the AP are : 1 11, ,0,2 2−

− …

(v) Given, first term 1.25a= = − Common difference 0.25d= = −∴ 1 1.25a a= = − 2 1 1.25 0.25 1.50a a d= + = − − = − 3 2 1.50 0.25 1.75a a d= + = − − = − 4 3 1.75 0.25 2a a d= + = − − = −Hence, first four terms of the AP are 1.25, 1.50, 1.75, 2,− − − − …

Q.3. For the following APs, write the first term and the common difference: (i) 3, 1, – 1, – 3,… 3(ii) –5, – 1, 3, 7,…

(iii) 1 5 9 13, , , ,…3 3 3 3

(iv) 0.6,1.7, 2.8, 3.9,…Solution. (i) Given AP is 3 ,1, 1, 3,− − …

Here, 1 2 3 43, 1, 1, 3a a a a= = = − = −First term 1 3a= = 2 1 1 3 2d a a= − = − = − 3 2 1 1 2a a= − = − − = − 4 3 3 1 2a a= − = − + = −Hence, common difference 2= − , first term 3= (ii) Given AP is 5, 1,3,7,− − … Here, 1 2 3 45, 1, 3, 7a a a a= − = − = =First term 1 5a = − 2 1 1 5 4d a a= − = − + = 3 2 3 1 4a a= − = + = 4 3 7 3 4a a= − = − =Hence, common difference 4= , First term 5= − .

(iii) Given AP is 1 5 9 13, , , ,3 3 3 3

…

Here, 1 2 3 41 5 9, , ,3 3 3

a a a a= = = =133

First term 113

a= =



5

2 15 1 5 1 43 3 3 3

d a a −= − = − = =

3 29 5 9 5 43 3 3 3

a a −= − = − = =

4 313 9 13 9 43 3 3 3

a a −= − = − = =

Hence, common difference 43

= , First term 13

=

(iv) Given AP is 0.6,1.7, 2.8,3.9,…Here, 1 2 3 40.6, 1.7, 2.8, 3.9a a a a= = = =First term 1 0.6a= = 2 1 1.7 0.6 1.1d a a= − = − = 3 2 2.8 1.7 1.1a a= − = − = 4 3 3.9 2.8 1.1a a= − = − =Hence, common difference 1.1= , First term 0.6=

Q.4. Which of the following are APs ? If they form an AP, find the common difference d and write three more terms. (i) 2,4,8,16

(ii) 5 72, , 3, ,…2 2

(iii) –1.2, –3.2, –5.2, –7.2,…(iv) –10, –6, –2, 2,…

(v) 3, 3 + 2, 3 + 2 2, 3 + 3 2,… (vi) 0.2,0.22,0.222,0.2222,…(vii) 0, –4, –8, –12,…

(viii) 1 1 1 1– , – , – , – ,…2 2 2 2

(ix) 1, 3,9, 27,…(x) , 2 , 3 ,4 ,…a a a a

(xi) 2 3 4, , , ,…a a a a

(xii) 2, 8, 18, 32,…

(xiii) 3, 6, 9, 12,…

(xiv) 2 2 2 21 ,3 ,5 ,7 ,…

(xv) 2 2 21 ,5 ,7 ,73,…Solution. [ ( 1na a n d= + − ) ]

(i) Given terms are 2, 4,8,16,…



6

3

Here, 1 2 3 42, 4, 8, 16a a a a= = = = 1 2 1 4 2 2d a a= − = − = 2 3 2 8 4 4d a a= − = − = 3 3 2 16 8 8d a a= − = − =∵ 1 2d d d≠ ≠Hence, given terms do not form an AP.

(ii) Given terms are 5 72, ,3, ,2 2

…

Here, 1 2 3 45 72, , 3,2 2

a a a a= = = =

1 2 15 5 422 2

d a a 12

−= − = − = =

2 3 25 6 5 132 2 2

d a a −= − = − = =

3 4 37 7 632 2

d a a 12

−= − = − = =

∵ 1 2 312

d d d= = =

∴ Common difference 12

d= =

Hence, the given terms form an AP.

Now, a5 = a4 + d = 7 1 + 2 2

= 4

a6 = a5 + d = 4 + 12

= 92

a7 = a6 + d = 9 1 + 2 2

= 5

(iii) Given terms are, 1.2, 3.2, 5.2, 7.2,− − − − …Here, 1 2 3 41.2, 3.2, 5.2, 7.2a a a a= − = − = − = − 1 2 1 3.2 1.2 2d a a= − = − + = − 2 3 2 5.2 3.2 2d a a= − = − + = − 3 4 3 7.2 5.2 2d a a= − = − + = −∵ 1 2 3 2d d d= = = −∴ Common difference 2d= = −Hence, the given terms form an AP. Now, 5 4 7.2 2 9.2a a d= + = − − = − 6 5 9.2 2 11.2a a d= + = − − = − 7 6 11.2 2 13.2a a d= + = − − = −(iv) Given terms are, 10, 6, 2, 2,− − − …Here, 1 2 3 410, 6, 2, 2a a a a= − = − = − =



7

1 2 1 6 10 4d a a= − = − + = 2 3 2 2 6 4d a a= − = − + = 3 4 3 2 2 4d a a= − = + =∵ 1 2 3 4d d d= = =∴ Common difference 4d= = Hence, the given terms form an AP. Now, 5 4 2 4 6a a d= + = + = 6 5 6 4 10a a d= + = + = 7 6 10 4 14a a d= + = + =(v) Given terms are 3,3 2,3 2 2 3 3 2,+ + + + … Here, 1 2 3 43, 3 2, 3 2 2, 3 3 2a a a a= = + = + = +

1 2 1 3 2 3d a a= − = + − = 2

2 3 2 3 2 2 (3 2) 3 2 2 3 2 2d a a= − = + − + = + − − =

3 4 3 3 3 2 (3 2 2) 3 3 2 3 2 2 2d a a= − = + − + = + − − =

∵ 1 2 3 2d d d= = =

∴ Common difference 2d= = Hence, the given terms form an AP. Now, 5 4 3 3 2 2 3 4 2a a d= + = + + = +

6 5 3 4 2 2 3 5 2a a d= + = + + = +

7 6 3 5 2 2 3 6 2a a d= + = + + = + (vi) Given terms are 0.2,0.22,0.222,0.2222,…Here, 1 2 3 40.2, 0.22, 0.222, 0.2222a a a a= = = = 1 2 1 0.22 0.2 0.02d a a= − = − = 2 3 2 0.222 0.22 0.002d a a= − = − = 3 4 3 0.2222 0.222 0.0002d a a= − = − =∵ 1 2d d d≠ ≠ 3∴ Given terms do not form an AP. (vii) Given terms are 0, 4, 8, 12,− − − … Here, 1 2 3 40, 4, 8, 12a a a a= = − = − = − 1 2 1 4 0 4d a a= − = − − = − 2 3 2 8 4 4d a a= − = − + = − 3 4 3 12 8 4d a a= − = − + = −∵ 1 2 3 4d d d= = = −∴ Common difference 4d= = − Now, 5 4 12 4 16a a d= + = − − = −



8

6 5 16 4 20a a d= + = − − = − 7 6 20 4 24a a d= + = − − = −

(viii) Given terms are 1 1 1 1, , , ,2 2 2 2

− − − − …

Here, 1 2 3 41 1 1 1, , ,2 2 2 2

a a a a= − = − = − = −

1 2 11 1 02 2

d a a= − = − + =

2 3 21 1 02 2

d a a= − = − + =

3 4 31 1 02 2

d a a= − = − + =

∵ 1 2 3 0d d d= = =∴ Common difference 0d= = Hence, the given terms form an AP.

Now, 5 6 712

a a a= = = − 1 , 02

a d⎡ ⎤= − =⎢ ⎥⎣ ⎦∵

(ix) Given terms are 1, 3,9, 27,…Here, 1 2 3 41, 3, 9, 27a a a a= = = = 1 2 1 3 1 2d a a= − = − = 2 3 2 9 3 6d a a= − = − = 3 4 3 27 9 18d a a= − = − =∵ 1 2d d d≠ ≠ 3∴ Given terms do not form an AP. (x) Given terms are a, 2a, 3a, 4a, … Here, a1 = a, a2 = 2a, a3 = 3a, a4 = 4a d1 = a2 – a1 = 2a – a = a d2 = a3 – a2 = 3a – 2a = a d3 = a4 – a3 = 4a – 3a = a ∵ d1 = d2 = d3 = a ∵ Common difference = d = a Hence, the given terms form an AP. Now, a5 = a4 + d = 4a + a = 5a a6 = a5 + d = 5a + a = 6a a7 = a6 + d = 6a + a = 7a (xi) Given terms are a, a2, a3, a4 … Here, a1 = a, a2 = a2, a3 = a3, a4 = a4

d1 = a2 – a1 = a2 – a = a(a – 1) d2 = a3 – a2 = a3 – a2 = a2 (a – 1) d3 = a4 – a3 = a4 – a3 = a3 (a – 1) ∵ d1 ≠ d2 ≠ d3

∴ The given terms do not form an AP.



9

(xii) Given terms are 2, 8, 18, 32,… 1 2 3 42, 8, 18, 32a a a a= = = =

⇒ 1 2 3 42, 2 2, 3 2, 4 2a a a a= = = =

1 2 1 2 2 2 2d a a= − = − =

2 3 2 3 2 2 2 2d a a= − = − =

3 4 2 3 2 2d = − =

∵ 1 2 3 2d d d= = =

∴ Common difference 2d= = Hence, the given terms form an AP. Now, 5 4 4 2 2 5 2a a d= + = + =

6 5 5 2 2 6 2a a d= + = + =

7 6 6 2 2 7 2a a d= + = + =

(xiii) Given terms are, 3, 6, 9, 12,… 1 2 3 43, 6, 9, 12a a a a= = = =

⇒ 1 2 3 43, 6, 3, 2 3a a a a= = = =

1 2 1 6 3d a a= − = −

2 3 2 3 6d a a= − = − ∵ 1 2d d≠∴ Given terms do not form an AP. (xiv) Given terms are 2 2 2 21 ,3 ,5 ,7 ,…Here, 2 2 2

1 2 3 41 , 3 , 5 , 7a a a a= = = = 2

⇒ 1 2 3 41, 9, 25, 49a a a a= = = = 1 2 1 9 1 8d a a= − = − = 2 3 2 25 9 16d a a= − = − =∵ 1 2d d≠∴ Given term do not form an AP. (xv) Given term are 2 2 21 ,5 ,7 ,73,…Here, 2 2 2

1 2 3 41 , 5 , 7 , 73a a a a= = = =⇒ 1 2 3 41, 25, 49, 73a a a a= = = = 1 2 1 25 1 24d a a= − = − = 2 3 2 49 25 24d a a= − = − = 3 4 3 73 49 24d a a= − = − =∵ 1 2 3 24d d d= = =∴ Common difference 24d= =



10

Hence, the given terms form an AP. 5 4 73 24 97a a d= + = + = 6 5 97 24 121a a d= + = + = 7 6 121 24 145a a d= + = + =

Exercise 5.2 Q.1. Fill in the blanks in the following table, given that a is the first term, d the

common difference and the nth term of the AP: na

(i)

(ii)

(iii)

(iv)

(v)

Solution. (i) Here 7, 3, 8, ?na d n a= = = =∵ nth term of the AP ( 1)na a n d= = + − ⇒ 8 7 (8 1)3 7 21 28a = + − = + =(ii) Here, 18, 10, 0, ?na n a d= − = = =∵ ( 1)na a n d= + −

10 18 (10 1) 0 18 9 9 18 2

a dd

dd

⇒ = − + −

⇒ = − +⇒ =⇒ =

(iii) Here, 3, 18, 5, ?nd n a a= − = = − =∵ ( 1)na a n d= + − – 5 = a + (18 – 1) – 3 – 5 = a – 51 ⇒ a = 51 – 5 = 46 ⇒∴ 3.6 18.9 ( 1)2.5n= − + −

3.6 18.9 ( 1)2.5 ( 1)2.5 22.5

22.5 1 92.5

9 1 10

nn

n

n

⇒ + = −⇒ − =

⇒ − =

⇒ =

=

+ =

(iv) Here, a = –18.9, d = 2.5, an = 3.6, n = ? ∴∴ an = a + (n – 1) d



11

⇒ 3.6 = –18.9 + (n – 1)2.5 ⇒ 3.6 + 18.9 = (n – 1)2.5

⇒ n – 1 = 22.52.5

= 9

⇒ n = 9 + 1 = 10 (v) Here, 3.5, 0, 105, ?na d n a= = = =∵ ( 1)na a n d= + −∴ 3.5 (105 1)0 3.5 0 3.5na = + − = + =

Q.2. 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: 1–3, – ,2,…2

is

(A) 28 (B) 22 (C) –38 (D) 1–482

Solution. (i) Given AP is 10 ,7, 4,… 1 2 310, 7, 4a a a= = = d = 7 – 10 = 4 – 7 = – 3 ∵ 1 ( 1)na a n d= + −Now, 30 10 (30 1)( 3) 10 87 77a = + − − = − = −∴ Option (C) is correct.

(ii) Given AP is 13, , 2,2

− − …

1 2 313, , 22

a a a= = − =

d = 1 1– + 3 = 2 + = 2 2

52

∵ 1 ( 1)na a n d= + −

Now, 115 53 (11 1) 3 10 3 25 222 2

a = − + − = − + × = − + =

∴ Option (B) is correct. Q.3. In the following APs, find the missing terms in the boxes:

(i) 2, , 26

(ii) ,13, , 3

(iii) 15, , ,92

(iv) –4, , , , , 6

(v) , 38, , , , –22



12

6Solution. Let a be the first term and d be the common difference of given AP.

(i) Here and 1 2a a= = 3 1 2 26 2 2 2a a d d= + = ⇒ + = 4

⇒ 2 26 2 2d = − =

⇒24 122

d = =

∴ Missing term 2 1 2 12 14a a d= = + = + =(ii) Here, ...(i) 2 13a a d= + =and ...(ii) 4 3a a d= + = 3Subtracting eq. (ii) from eq. (i), we have

+ 3 = 3 + = 13

– – –2 = – 10

– 10 = = –52

a da d

d

d⇒

Putting the value of d in eq. (i), we have

5 13

13 5 18a

a− =

⇒ = + =∴ 1 18a a= = 3 1 2 18 2( 5) 18 10a a d= + = + − = − = 8 Hence, the missing terms are 18 and 8 respectively.

(iii) Here, and 1 5a a= = 4 113 92

a a d= + =

11932

19 5 32

19 3 52

19 10 9 32 2

9 1 3 2 3 2

a d

d

d

d

d

+ =

⇒ + =

⇒ = −

−⇒ =

⇒ = × =

=

2 13 10 3 1352 2

a a d2

+= + = + = =

3 132 5 2 5 32

a a d ⎛ ⎞= + = + = + =⎜ ⎟⎝ ⎠

8

Hence, the missing terms are 132

and 8 respectively.

(iv) Here, 1 6 14, 5 6a a a a d= = − = + =



13

4 5 6 5 6 4 5 10

10 25

ddd

d

⇒ − + =⇒ =⇒ =

⇒ =

+

=

2

2

2

Now, 2 1 4 2 2a a d= + = − + = − 3 1 2 4 2(2) 4 4 0a a d= + = − + = − + = 4 1 3 4 3(2) 4 6a a d= + = − + = − + = 5 1 4 4 4(2) 4 8 4a a d= + = − + = − + = Hence, the missing terms are –2, 0, 2 and 4 respectively. (v) Here, ...(i) 2 1 38a a d= + =and ...(ii) 6 1 5 2a a d= + = −Subtracting eq. (ii) from eq. (i), we have

1

1

5 238

a da d

+ = −+ =

− − − 4 6d = − 0

⇒

60 154

d = − = −

=


1

1

( 15) 38 38 15 53

aa

+ − =⇒ = +

∴ 1 53a a= = 3 1 2 53 2( 15) 53 30 23a a d= + = + − = − = 4 1 3 53 3( 15) 53 45 8a a d= + = + − = − = 5 1 4 53 4( 15) 53 60 7a a d= + = + − = − = − Hence, the missing terms are 53, 23, 8 and – 7

Q.4. Which term of the AP is 78? 3,8,13,18,…Solution. Given AP is 3,8,13,18,…

1 2 3 43, 8, 13, 18a a a a= = = = d = 8 – 3 = 5 ∵Using, 1 ( 1)na a n d= + −

78 3 ( 1) 5( 1) 78 3

75 1 155

15 1 16

n dn

n

n

⇒ = +⇒ − = −

⇒ − = =

⇒ = + =

−

Hence, 16th term of the given AP is 78.



14

Q.5. Find the number of terms in each of the following APs: (i) 7,13,19,…, 205

(ii) 118,15 ,13,…, –472

Solution. (i) Given AP is 7, 13,19,… 1 2 37, 13, 19a a a= = = d = 13 – 7 = 6 ∵Using formula, ( 1)na a n d= + −

205 7 ( 1)6 ( 1)6 205 7

196 ( 1) 336

33 1 34

nn

n

n

⇒ = +⇒ − = −

⇒ − = =

= + =

−

Hence, 34th term of the AP is 205.

(ii) Given AP is 118,15 ,13,2

…

1 2 31 3118, 15 , 132 2

a a a= = = =

d = ∵ 1 31 15 – 18 = – 18 = 2 2

– 52

Using formula, ( 1)na a n d= + −

5 47 18 ( 1)2

5 ( 1) 47 1825 ( 1) 65

22 1 655

1 26

n

n

n

n

n

−⎛ ⎞⇒ − = + − ⎜ ⎟⎝ ⎠

−⎛ ⎞⇒ − = − −⎜ ⎟⎝ ⎠

−⎛ ⎞⇒ − = −⎜ ⎟⎝ ⎠

⇒ − = − × −

⇒ − =⇒ 26 1 27n = + =

Hence, 27th term of the AP is 47.− Q.6. Check whether – 150 a term of the AP : 11, 8, 5, 2, …

Solution. Here, 1 2 3 411, 8, 5, 2a a a a= = = = d = 8 – 11 = – 3 Let be any term of given AP 150−Then 150na = −



15

1 ( 1) 150 11 ( 1)( 3) 150 ( 1)( 3) 150 11 161

161 13

161 161 3 13 3

164 2 543 3

a n dnn

n

n

n

⇒ + − = −⇒ + − − = −⇒ − − = − − = −

⇒ − =

+⇒ = + =

⇒ = =

Which is not a natural number. Hence, cannot be a term of the given AP. 150−

Q.7. Find the 31st term of an AP whose 11th term is 38 and 16th term is 73. Solution. Let a and d be the first term and common difference of given AP.

Given that 11 38a = ⇒ (11 1) 38 [ ( 1) ]na d a a n+ − = = + −∵ d

1) ]

...(i) ⇒ 10 38a d+ =and 16 73a = ⇒ (16 1) 73a d+ − = [ (na a n d= + −∵ ...(ii) ⇒ 15 73a d+ =Subtracting eq. (i) from eq. (ii), we have

15 7310 38

a da d

+ =+ =

− − − 5 3d = 5

⇒35 75

d = =


10(7) 38

70 38 38 70 32

aa

a

+ =⇒ + =⇒ = − = −

= + −Now, 31 (31 1)a a d 32 30(7) 32 210 178= − + = − + =

Q.8. An AP consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term.

Solution. Let a and d be the first term and common difference of given AP. Given that, 3 12a = ⇒ (3 1) 12a d+ − = [ (na a n d1) ]= + −∵ ...(i) ⇒ 2 1a d+ = 2

1)]Last term 50 106T= = ⇒ (50 1) 106a d+ − = [ (na a n= + −∵



16

...(ii) ⇒ 49 106a d+ =Subtracting eq. (i) from eq. (ii), we have

49 106

2 12a da d+ =

+ = − − − 47 94d =

⇒94 247

d = =

Putting the value of d in (i), we have

2(2) 12

4 12 8

aa

a

+ =⇒ + =⇒ =

Now, 29 (29 1) 8 28(2) 8 56 64a a d= + − = + = + =

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

Solution. Let a and d be the first term and common difference of given AP. Given that: 3 4a = ⇒ (3 1) 4a d+ − = [ (na a n d1) ]= + −∵ ...(i) ⇒ 2a d+ = 4

1) ] 9 8a = ⇒ (9 1) 8a d+ − = [ (na a n d= + −∵ ...(ii) ⇒ 8a d+ = −8Subtracting eq. (i) from eq. (ii), we have

8 82 4

a da d

+ = −+ =

− − − 6 1d = − 2

⇒ 12 26

d −= = −


⇒2( 2) 4

4 44 4 8

aa

a

+ − =− =

= + =Now, (Given) 0na =

( 1) 0 8 ( 1)( 2) 0 2( 1) 8 1 4 4 1 5

a n dn

nn

n

⇒ + − =⇒ + − − =⇒ − − = −⇒ − =⇒ = + =



17

dd

Hence, 5th term of the AP is zero. Q.10. The 17th term of an AP exceeds its 10th term by 7. Find the common

difference. Solution. Let a and d be the first term and common difference of given AP.

17 (17 1) 16a a d a= + − = + 10 (10 1) 9a a d a= + − = +As per condition,

17 10 7 ( 16 ) ( 9 ) 7 16 9 7 7 7

7 17

a aa d a d

a d a dd

d

− =

⇒ + − + =⇒ + − − =⇒ =

⇒ = =

Hence, common difference is 1. Q.11. Which term of the AP will be 132 more than its 54th term? 3,15, 27, 39,…Solution. Let a and d be the first term and common difference of given AP.

Given AP is 3,15, 27,39,… 1 2 3 43, 15, 27, 39a a a a= = = = d = 15 – 3 = 12 Now, 54 (54 1) 3 53(12) 3 636 639a a d= + − = + = + =As per condition,

54 132 ( 1) 639 132 3 ( 1)(12) 771 ( 1)12 771 3 768

768 1 6412

64 1 65

na aa n d

nn

n

n

= +

⇒ + − = +⇒ + − =⇒ − = − =

⇒ − = =

⇒ = + =

Hence, 65th term of the AP is 132 more than its 54th term. Q.12. Two APs have the same common difference. The differnce between their

100th terms is 100, what is the difference between their 1000th terms? Solution. Let a and d be the first term and commond difference of first AP.

Also, ‘A’ and d be the first term and commond difference of second AP. As per condition, of second AP] 100[a − of first AP] 100[a 100= [ (100 1) ] [ (100 1) ] 100A d a d+ − − + − = 99 99 100A d a d+ − − = ...(ii) 100A a− =Now, [ of second AP] 1000a 1000 [a− of first AP] [ (1000 1) ] [ (1000 1) ]A d a d= + − − + −



18

= + − − 999 999A d a d 100A a= − = [Using eq. (i)] Q.13. How many three-digit numbers are divisible by 7? Solution. Three digit numbers which are divisible by 7 are 105 ,112,119, ,994…

Here, and 1 2 3105, 112, 119a a a a= = = = 994na = d = 7 Given than, 994na =

( 1) 994 105 ( 1) 994 ( 1)7 994 105 ( 1)7 889

889 1 1277

127 1 128

a n dn ynn

n

n

⇒ + − =⇒ + − =⇒ − = −⇒ − =

⇒ − = =

⇒ = + =

Hence, 128 three digit numbers are divisible by 7. Q.14. How many multiples of 4 lie between 10 and 250? Solution. Multiples of 4 which lie between 10 and 250 are 12 ,16, 20, 24, , 248…

Here, and 1 2 312, 16, 20a a a a= = = = 248na = d = 4 Given that, 248na =

( 1) 248 12 ( 1)4 248 4( 1) 248 12 236

236 1 594

59 1 60

a n dnn

n

n

+ − =⇒ + − =⇒ − = − =

⇒ − = =

⇒ = + =

Hence, 60 multiples of 4 lie between 10 and 250. Q.15. For what value of n, are the nth terms of two APs 63, and

equal? 65,67,… 3,10,17,…

Solution. Given AP is 63,65,67,…Here, 1 2 363, 65, 67a a a a= = = = d = 65 – 63 = 2 and second AP is 3,10,17,…Here, 1 2 33, 10, 17a a a a= = = = d = 10 – 3 = 7 As per condition, [nth term of first AP] = [nth term of second AP]



19

63 ( 1)2 3 ( 1)7 63 2 2 3 7 7 61 2 7 4 2 7 4 61 5 65

65 135

n nn n

n nn n

n

n

⇒ + − = + −⇒ + − = + −⇒ + = −⇒ − = − −⇒ − = −

⇒ = =

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

Solution. Let a be the first term and d be common difference of the A.P. Given, 3 16a =

6⇒ (3 1) 16a d+ − =

...(i) ⇒ 2 1a d+ =As per condition,

7 5 12 [ (7 1) ] [ (5 1) ] 12 6 4 12 2 12

12 62

a aa d a d

a d a dd

d

− =

⇒ + − − + − =⇒ + − − =⇒ =

⇒ = =

⇒ 16 12 4a = − =

Putting the value of d in eq. (i), we have 2(6) 16a + = Hence, given AP is 4,10,16, 22, 28,…

Q.17. Find the 20th term from the last term of the AP : 3,8,13,…, 253.Solution. Given AP is 3,8,13, , 253…

Here, and 1 2 33, 8, 13a a a a= = = = 253na = d = 8 – 3 = 5 Now, 253na =

3 ( 1)5 253 [ ( 1) ] ( 1)5 250

250 1 505

50 1 51

nn a an

n

n

⇒ + − = = + −

⇒ − =

⇒ − = =

⇒ = + =

∵ n d

20th term from the end of AP (Total number of terms) 20 1= − +

51 20 1 32= − + = nd term ∴ 20th term from the end of AP 32= nd term from the beginning



20

[ ]3 (32 1)5 ( 1)

3 31 53 155 158

na a n d= + − = + −

= + ×= + =

∵

Alternate method nth term from the end = last term + (n – 1) (– d) = 253 + (20 – 1) (– 5) = 253 – 95 = 158 Q.18. The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and

10th terms is 44. Find the first three terms of the AP. Solution. Let a be the first term and d be the common difference of the AP.

Case I: As per condition

4 8 24 (4 1) (8 1) 24 [ ( 1) ] 2 3 7 24 2 10 24

n

a aa d a d a a n

a d da d

d+ =

⇒ + − + + − = = + −⇒ + + =⇒ + =

∵

⇒ 25 1a d+ = ...(i) Case II: As per condition,

6 10 44 (6 1) (10 1) 44 [ ( 1) ] 2 5 9 44 2 14 44

n

a aa d a d a a n

a d da d

d+ =

⇒ + − + + − = = + −⇒ + + =⇒ + =

∵

⇒ 7 2a d 2+ = ...(ii) Subtracting eq. (i) from eq. (ii), we have

7 25 12

a da d

+ =+ =

2

− − − 2 1d = 0

⇒10 52

d = =


5(5) 12

25 12 12 25 13

aa

a

+ =⇒ + =⇒ = − = −

∴ 1 13a a= = − 2 13 5 8a a d= + = − + = − 2 2 13 2(5) 13 10a a d= + = − + = − + = −3Hence, given AP is 13, 8, 4,− − − …

Q.19. Subba Rao started work in 1995 at an annual salary of Rs 5000 and received an increment of Rs 200 each year. In which year did his income reach Rs 7000?

Solution. Subba Rao’s starting salary = Rs 5000



21

Annual increment = Rs 200 Let n denote number of years. ∴ First term = a = Rs 5000 Common difference = d = Rs 200

Rs 7000 5000 ( 1)200 7000 [ ( 1) ]

( 1)200 20002000 1 10200

10 1 11

n

n

an a an

n

n

=n d∴ + − = = + −

⇒ − =

⇒ − = =

⇒ = + =

∵

Hence, in the 11th year his salary will become Rs 7000. Now, in case of year the sequence is 1995, 1996, 1997, 1998, ... Here, and 1995, 1a d= = 11n = Let denote the required year. na∴ 1995 (11 1)1na = + − 1995 10 2005= + = Hence, in 2005, Subba Rao’s salary becomes Rs. 7000.

Q.20. Ramkali saved Rs 5 in the first week of a year and then increased her weekly saving by Rs 1.75. If in the nth week, her weekly saving becomes Rs 20.75, find n.

Solution. Let Amount saved in first week a = Rs. 5= Increment in saving every week d = Rs. 1.75= Rs. 20.75na = 1 5, 1.75a d= = ∴ [5 ( 1)1.75 20.75n+ − = ( 1) ]na a n d= + −∵

( 1)1.75 20.75 5 ( 1)1.75 15.75

1575 100 ( 1)100 175

1 9 9 1 10

nn

n

nn

⇒ − = −⇒ − =

⇒ − = ×

⇒ − =⇒ = + =

Hence, in 10th week, Ramkali’s saving becomes Rs. 20.75.

Exercise 5.3 Q.1. Find the sum of the following APs:

(i) to 10 terms. 2,7,12,…(ii) to 12 terms –37, –33, –29,…(iii) to 100 terms 0.6,1.7, 2.8,…

(iv) 1 1 1, , ,…15 12 10

to 100 terms



22

Solution. (i) Given AP is 2, 7,12,…Here, 2, 7 2 5, 10a d n= = − = =

[2 ( 1) ]2nnS a n= + − d ...

⇒ 1010 [2 2 (10 1)5] 5[4 45] 2452

S = × + − = + =

(ii) Given AP is 37, 33, 29,− − − … Here, 37, 33 37 4, 12a d n= − = − + = =

[2 ( 1) ]2nnS a n= + − d ...

⇒ 1212 [2( 37) (12 1)4] 6[ 74 44] 1802

S = − + − = − + = −

(iii) Given AP is 0.6,1.7, 2.8,… Here, 0.6, 1.7 0.6 1.1, 100a d n= = − = =

[2 ( 1) ]2nnS a n= + − d ...

⇒ 100100 [2(0.6) (100 1)1.1] 50[1.2 108.9] 5505

2S = + − = + =

(iv) Given AP is 1 1 1, , ,15 12 10

…

Here, 115

a = , 1 1 5 4 112 15 60 60

d −= − = =

11n =

[2 ( 1) ]2nnS a n= + − d ...

1111 1 12 (11 1)2 15 60

11 2 10 11 2 12 15 60 2 15 6

11 4 5 332 30 20

S ⎡ ⎤⎛ ⎞= + −⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦⎡ ⎤ ⎡= + = +⎢ ⎥ ⎢⎣ ⎦ ⎣

+⎡ ⎤= =⎢ ⎥⎣ ⎦

⎤⎥⎦

Q.2. Find the sums given below:

(i) 17 +10 +14 +…+ 842

(ii) 34 + 32 + 30 +…+10(iii) –5 + (–8) + (–11) + …+ (–230)

Solution. (i) Given AP is

17 10 14 842

+ + + +…



23

Here, 1 21 21 147, 10 7 72 2 2

a d 72

−= = − = − = =

84nl a= =

( 1) 847 7 ( 1) 8427 ( 1) 84 7 772

2 1 77 227

22 1 23

a n d

n

n

n

n

⇒ + − =

⇒ + − =

⇒ − = − =

⇒ − = ×

⇒ = +

=

=

Now, 2323[7 84]2

S = + [ (2nnS a 1)= +∵ , where l is last term]

23 2093912 2

= × =

(ii) Given AP is 34 32 30 10+ + + +…Here, 34, 32 34 2a d= = − = − 10nl a= =

( 1) 10 34 ( 1)( 2) 10 2( 1) 10 34 24

24 1 122

12 1 13

a n dn

n

n

n

⇒ + − =⇒ + − − =⇒ − − = − =

⇒ − = =

⇒ = +

−

=

Now, 1313[34 10]2

S = + [ (2nnS a )l= +∵ , where l is last term]

13 44 286= × =(iii) Given AP is 5 ( 8) ( 11) ( 230)− + − + − + + −…Here, 5, 8 5 3a d= − = − + = −

2 2230 ( 1) 230 5 ( 1)( 3) 230 3( 1) 230 5 225

225 1 753

75 1 76

nl a a ba n d

nn

n

n

= = − +

⇒ + − = −⇒ − + − − = −⇒ − − = − + = −

⇒ − = =

⇒ = + =

7676 [ 5 ( 230)]2

S = − + − [ (2nnS a )l= +∵ , where l is last term]

38( 235) 8930= − = −



24

Q.3. In an AP: (i) Given a = 5, d = an = 50. Find n and Sn. (ii) Given a = 7, a13 = 35, Find d and S13

(iii) Given a12 = 37, d = 3. Find a and S12

(iv) Given a3 = 15, S10 = 125. Find d and a10. (v) Given d = 5, S9 = 75. Find a and a9. (vi) Given a = 2, d = 8, Sn = 90. Find n and an. (vii) Given a = 8, an = 62, Sn = 210. Find n and d. (viii)Given an = 4, d = 2, Sn = –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 5, 3, 50na d a= = =∵ 50na =

( 1) 50 5 ( 1)3 50 3( 1) 50 5 45

45 1 153

15 1 16

a n dnn

n

n

⇒ + − =⇒ + − =⇒ − = −

⇒ − = =

⇒ = +

=

=

⇒ ( )2nnS a= + l

⇒ 1616 [5 50] 8 55 4402

S = + = × =

(ii) Given 137, 35a a= =∵ 13 35a =

7 (13 1) 35 12 35 7 28

28 7 12 3

dd

d

⇒ + − =⇒ = −

⇒ =

=

=

1313[7 35] [ ( )]2 2n

nS S= + = +∵ a l

13 42 13 21 273.2

= × = × =

(iii) Given 12 37, 3a d= =∵ 12 37a =

(12 1)3 37 37 33 4

aa

⇒ + − =⇒ = − =



25

1212 [4 37] [ ( )]2 2

6 41 246.

nnS S= + = +

= × =

∵ a l

(iv) Given 3 1015, 125a S= =∵ 3 15a =

(3 1) 15 2 15 = 15 – 2 ... (1)

a da d a d

⇒ + − =⇒ + = ⇒

∵ 10 125S =

10 [2 (10 1) ] 125 [ [2 ( 1) ]]2 2

125 2 9 255

2 9 25 ... (2)

nna d S a n

a d

a d

⇒ + − = = + −

⇒ + = =

⇒ + =

∵ d

From (1) and (2) we have

2(15 2 ) 9 25 30 4 9 25 5 25 30

5 15

d dd d

d

d

− + =⇒ − + =⇒ =

−⇒ =

−

= −

a n d

Putting the value of d in (1), we have

10

15 2( 1) 15 2 17

Now, 17 (10 1)( 1) [ ( 1) ]17 9 8.

n

aa

a a

= − −⇒ = + =

= + − − = + −

= − =

∵

(v) Given 95, 75d S= =∵ 9 75S =

9 [2 (9 1)5] 75 [ [2 ( 1) ]]2 2

9 [2 40] 752

75 2 50 2 409 3

50 50 120 2 403 370 1 3 2

nna S a

a

a

a

a

⇒ + − = = + −

⇒ + =

×⇒ + = =

−⇒ = − =

⇒ = − ×

⇒

∵ n d

35 3

a = −



26

9Now, (9 1)35 (9 1)5335 35 120 8540 .3 3

a a d= + −

−= + −

− − += + = =

3

(vi) Given 2, 8, 90na d S= = =∵ 90nS =

2

2

2

[2 ( 1) ] 902

[2 2 ( 1)8] 902

[2 4 4] 90

4 2 90 0

2 45 0

2 10 9 45 0 2 [ 5] 9( 5) 0 (2 9)( 5) 0

n a n d

n n

n n

n n

n n

n n nn n n

n n

⇒ + −

⇒ × + − =

⇒ + −

⇒ − − =

⇒ − −

⇒ − + − =⇒ − + − =⇒ + −

=

=

=

=

0

If , then 2 9n + = 92

n = −

If , then 5 0n − = 5n =

∵ n cannot be negative so we reject 92

n −=

∴ 5n =

5Now, ( 1)

2 (5 1)82 32 34

na a a n d= = + −

= + −= + =

(vii) Given 8, 62, 210n na a S= = =∵ 210nS =

⇒ [ ] 22 nn a a+ = 10

[8 62] 2102

70 2102

210 635

n

n

n

⇒ + =

⇒ × =

⇒ = =

1) ]

62na = ⇒ 8 (6 1) 62d+ − = [ (na a n d= + −∵



27

5 62 8 54

54 5

d

d

⇒ = − =

⇒ =

(viii) Given 4, 2, 14n na d S= = = −∵ 4 na =

( 1) 4 ( 1)2 4

2 2 4

a n da na n

⇒ + − =⇒ + − =

⇒ + − =n ________ (1) ⇒ 6 2a = −

[ ]

14

[ ] 14 From (1)2

n

n

Sn a a

= −

⇒ + = −

⇒ [6 2 4] 142n n− + = −

2

2

2

[10 2 ] 142

5 14 0

5 14 0

7 2 14 0 ( 7) 2( 7) 0 ( 7)( 2) 0

n n

n n

n n

n n nn n n

n n

⇒ −

⇒ − + =

⇒ − − =

⇒ − + − =⇒ − + − =⇒ − + =

= −

⇒ n = 7 or = n = – 2 ⇒ n = 7 [Rejecting n = – 2]

∴ 7n =Putting the value of n in eq. (1), we have

6 2 7

6 14 8aa

= − ×⇒ = − = −

(ix) Given 3, 8, 192a n S= = =∵ 192S =⇒ 8 192S = [ 8n ]=∵

8 [2 3 (8 1) 192 [ [2 ( 1) ]]2 2

4[6 7 ] 192192 6 7 48

4 7 48 6 42

42

nnd S a n

d

d

d

d

× + − = = + −

⇒ + =

⇒ + = =

⇒ = − =

⇒ =

∵ d

67

=

(x) Given 9 99, 28, 144n l a S= = = =



28

d∵ 9 28a =

9

(9 1) 28 [ ( 1) ] 8 28

144

na d a a na d

S

⇒ + − = = + −⇒ + =

=

∵

( )9 [ 28] 144 [2 2

144 2 28 329

32 28 4

n nna S

a

a

⇒ + = = +

×⇒ + = =

⇒ = − =

∵ a a

Q.4. How many terms of the AP: 9, must be taken to give a sum of 636? 17, 25,…Solution. Given AP is 9,17, 25,…Here, , 9, 17 9 8a d= = − = 636nS =

2

Sn = [2 ( 1) ] 6362

[2(9) ( 1)8] 6362

[18 8 8] 6362

[4 5] 636

4 5 636 0

n a n d

n n

n n

n n

n n

+ − =

⇒ + − =

⇒ + − =

⇒ +

⇒ + − =

=

– 5 25 + 4 × 4 × 636 = 2 × 4

25 + 10176 = – 58

n ±⇒

±

∴

5 101 8

106 96 53 or or 12.8 8 2

n

n n

− ±⇒ =

−⇒ = ⇒ −

∵ n cannot be negative so we reject 534

n = −

∴ 12n =Hence, sum of 12 terms of given AP has sum 636.

Q.5. 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. Given that and 1 5, 45 na l a= = = 400nS = ∵ 45na =



29

( 1) 45 [ ( 1) ] 5 ( 1) 45 ( 1) 45 5 40

na n d a a n dn dn d

⇒ + − = = + −

⇒ + − =⇒ − = − =

...(i) ( 1) 40n d− = Now, 400nS =

⇒ [ ] 42 nn a a+ = 00

[5 45] 4002

25 400400 1625

n

n

n

⇒ + =

⇒ =

⇒ = =

Putting the value of n in eq. (i), we have

(16 1) 40

15 4040 8 15 3

dd

d

− =⇒ =

⇒ = =

Hence, and 16n = 8 .3

d =

Q.6. The first and 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. Given that 1 17, 350, 9na a l a d= = = = =∵ 350nl a= =

( 1) 350 17 ( 1)9 350 9( 1) 350 17 333

333 1 379

37 1 38

a n dnn

n

n

⇒ + − =⇒ + − =⇒ − = −

⇒ − = =

⇒ = +

=

=

Now, ( )2nnS a= + l

⇒ 3838 38[ ] [17 350] 19 367 69732 2

S a l= + = + = × =

Hence, sum of 38 terms of given AP is 6973. Q.7. Find the sum of first 22 terms of an AP in which and 22nd term is 149. = 7dSolution. Given that 22= 7, = 149, = 22d a n

∵ 22 149a =



30

=

( 1) 149 (22 1)7 149 147 149 149 147 2

a n da

aa

⇒ + − =⇒ + − =⇒ + =⇒ = −

Now, 22 2222[ ] [2 149] 11 151 1661

2 2nS a a= + = + = × =

Hence, sum of first 22 terms of given AP is 1661. Q.8. Find the sum of first 51 terms of an AP whose second and third terms are 14

and 18 respectively. Solution. Let a and d be first term and common difference of the given AP.

Given that 2 314, 18, 51a a n= = = ∵ d = a3 – a2 = 18 – 14 = 4 Now, a = a2 – d = 14 – 4 = 10

51 [2 ( 1) ]251[2 10 (51 1)4]251[20 200]251 220 51 110 56102

nS a n d= + −

= × + −

= +

= × = × =

Hence, sum of first 51 terms of given AP is 5610. Q.9. If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find the

sum of first n terms. Solution. Let a be the first term and d be the common difference of given AP.

7 497 [2 (7 1) ] 492

7 [2 6 ] 492

3 7 7 3 ...(i)

S

a d

a d

a da d

=

⇒ + − =

⇒ + =

⇒ + =⇒ = −

17Also, 28917 [2 (17 1)] 2892

289 8 1717

S

a

a d

=

⇒ + − =

⇒ + = =



31

7 3 8 17 [From (i)] 5 17 7 10

10 25

d dd

d

⇒ − + =⇒ = − =

⇒ = =


7 3 2

7 6 1aa

= − ×⇒ = − =

Now, [2 ( 1) ]2nnS a n= + − d

2

[2 1 ( 1)2]2[1 1]

n n

n n n n n

= × + −

= + − = × =

Hence, sum of first n terms of given AP is 2.nQ.10. Show that form an AP, where a1 2 n, ,…, ,…a a a n is defined as below.

(i) (ii) n = 3 + 4a n

n

n = 9 – 5a n

Also find the sum of the first 15 terms in each case. Solution. (i) Given that ...(1) 3 4na = +

So, , 1 3 4(1) 7a = + = , 2 3 4(2) 11a = + = 3 3 4(3) 15a = + =

1 2 1

2 3 2

11 7 415 11 4

d a ad a a

= − = − == − = − =

(say) 1 2 4d d d= = =∴ Given sequence forms an AP Here, and 7, 4a d= = 15n =

∴ [2 ( 1) ]2nnS a n= + − d

1515 S [2(7) (15 12

15 15 [14 56] 702 2

15 35 525

⇒ = + −

= + = ×

= × =

)4]

n

(ii) Given that 9 5na = −

1

2

3

So, 9 5(1) 4 9 5(2) 1 9 5(3) 6

aaa

= − == − = −= − = −

1 2 1

2 3 2

1 4 56 1 5

d a ad a a

= − = − − = −= − = − + = −



32

∵ 1 2 5d d d= = − =∴ Given sequence forms an AP Here, and 4, 5a d= = − 15n =

∵ [2 ( 1) ]2nnS a n= + − d

15

15 S [2(4) (15 1)( 5)]215 15 [8 70] ( 62) 4652 2

⇒ = + − −

= − = − = −

Q.11. If the sum of the first n terms of an AP is what is the first term (that is S

24 – ,n n1? What is the sum of two terms ? What is the second term ? Similarly, find

the 3rd, the 10th and the nth terms. Solution.

...(1) 24nS n n= −Putting in eq. (1), we have 1n = 2

1 4(1) (1) 4 1 3S = − = − =∴ 1 1 3a a S= = =Putting , in eq. (1), we have 2n = 2

2 4(2) (2) 8 4S = − = −

2

1 2

2

2

4 4 3 4 4 3 1

Sa a

aa

⇒ =⇒ + =⇒ + =⇒ = − =

2 3 3S a+ = 34 3a+ = 3 3 4 1a = − = −

Putting in eq. (1), we have 3n = 2

3 4(3) (3) 12 9 3S = − = − = ⇒ ⇒ ⇒ 2 1 1 3 2d a a= − = − = −∴ 10 ( 1) 3 (10 1)( 2a a n d= + − = + − − )

10 3 18 15a = − = −

na n= −

⇒ ( 1) 3 ( 1)( 2) 3 2 2na a n d n n= + − = + − − = − + 5 2 ⇒

Q.12. Find the sum of the first 40 positive integers divisible by 6. Solution. Positive integers divisible by 6 are 6,12,18, 24,30,36, 42,…

Here, , d = 6 1 2 3 46, 12, 18, 24a a a a a= = = = =



33

Using formula, [2 ( 1) ]2nnS a n= + − d

⇒ 4040 [2(6) (40 1)6]2

S = + −

20[12 234]20(246) 4920

= += =

Hence, sum of first 40 positive integers divisible by 6 is 4920. Q.13. Find the sum of first 15 multiples of 8. Solution. Multiples of 8 are 8 ,16, 24,32, 40, 48,…

Here, , d = 8 1 2 3 48, 16, 24, 32a a a a a= = = = =


1515 [2(8) (15 1)8]2

15[16 112]2

15 128 9602

S⇒ = + −

= +

= × =

Hence, sum of first 15 multiples of 8 is 960. Q.14. Find the sum of the odd numbers between 0 and 50. Solution. Odd numbers between 0 and 50 are 1, 3,5,7,9, , 49…

Here, 1 2 3 41, 3, 5, 7a a a a a= = = = =and , d = 2 49nl a= =Also, 49nl a= =

( 1) 49 1 ( 1)2 49

( 1) 49 1 4848 1 242

24 1 25

a n dnn

n

n

⇒ + − =⇒ + − =

⇒ − = − =

⇒ − = =

⇒ = + =

Now, Sn = 2n (a + l)

⇒ S25 = 252

(1 + 49)

= 252

× 50 = 625

Hence, sum of the odd numbers between 0 and 50 is 625. Q.15. A contract on 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,



34

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 the contractor has to pay as penalty, if he has delayed the work by 30 days?

Solution. Penalty for delay of one, two, three days are Rs. 200, Rs. 250, Rs. 300, Rs. 350, ... Here, 1 200, Rs. 50 and 30a a d n= = = =Amount of penalty given for a delay of 30 days 30S=

[2 ( 1) ]2

30Rs [2(200) (30 1)50]2

Rs 15[400 1450]Rs 15(1850) Rs 27750

n a n d= + − ⊕

= +

= += =

−

Q.16. 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. Let amount of prize given to 1st student Rs. x= Amount of prize given to 2nd student Rs. ( 20)x= − Amount of prize given to 3rd student Rs. [ 20 20] Rs. ( 40)x x= − − = − and so on. So, required sequence is Rs. x, Rs. Rs. ( 4( 20),x − 0),x − … which forms an AP with Rs. , 20 and 7a x d n= = − =Using formula,

[2 ( 1) ]2nnS a n= + − d

⇒ 77 [2( ) (7 1)( 20)]2

S x= + − −

⇒ 77 [2 120] 7( 60)2

S x x= − = −

But, 7( 60) 700700 60 1007

100 60 160

x

x

xx

− =

⇒ − = =

⇒ =⇒ =

+

Hence, values of the prizes are Rs. 160, Rs. 140, Rs. 120, Rs. 100, Rs. 80, Rs. 60, Rs. 40. Q.17. In a school, students thought of planting trees in and around the school to

reduce air pollution. It was decided that 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



35

and so on till Class XII. There are three sections of each class. How many trees will be planted by the students?

Solution. Number of trees planted by three sections of class I 3 1 3= × = Number of trees planted by three sections of class II 3 2 6= × = Number of trees planted by three sections of class III 3 3 9= × = - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - Number of trees planted by three sections of class XII 3 12 36= × = ∴ Required AP is 3,6,9, ,36…Here, and 1 2 33, 6, 9a a a a= = = = 2 136, 12, 6 3 3nl a n d a a= = = = − = − = Total number of trees planted by students

1212[ ] [3 36] 6 39 234

2 2nS a l= = + = + = × =

Q.18. A spiral is made up of successive semicircles, with centres alternately at A and B, starting with centre at A, of radii 0.5 cm, 1.0 cm, 1.5 cm, 2.0 cm, ... as shown in the figure. What is the total length of such a spiral made up of

thirteen consecutive semicircles? (Take 22π = )7

Solution. Let length of first semicircle 1l = 1 (0.5)2

r ππ π= = =

length of second semi circle 2l = 2 (1)rπ π π= = =

length of third semi circle 3l = 33(1.5)2

r ππ π= = =

and length of fourth semi circle 4l = 4 (2) 2rπ π π= = = and so on.

Here, 1 2 3 43, , , 2 ... and 13

2 2a a a a a nπ ππ π= = = = = =

3, , , 2 , - - - is an AP.2 2π ππ π

2 12

2 2 2d a a π π π ππ −

= − = − = =



36

Length of whole spiral 13S=

[2 ( 1) ]213 2 (13 1)2 2 2

13[ 6 ]2

13 13 227 72 2 7

n a n d

π π

π π

π

= + −

143

⎡ ⎤⎛ ⎞ ⎛ ⎞= + −⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦

= +

= × = × × =

Hence, total length of the spiral made up of thirteen consecutive semicircles is 143 cm. Q.19. 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 (see fig below). In how many rows are the 200 logs placed and how many logs are in the top row?

Solution. Number of logs in the bottom (1st row) 20= Number of logs in the 2nd row 19= Number of log in the 3rd row 18= and so on. Here, 1 2 3 2 120, 19, 18, 19 20 1a a a a d a a= = = = = − = − = −Let S denote the number logs. n


[2(20) ( 1)( 1)]2

[40 1]2

[41 ]2

nnS n

n n

n n

= + −

= − +

= −

−



37

2

2

2

2

[41 ] 2002

41 400

41 400 0

41 400 0

16 25 400 0 ( 16) 25( 16) 0

( 16)( 25) 0

n n

n n

n n

n n

n n nn n n

n n

⇒ − =

⇒ − =

− + + =

⇒ − + =

⇒ − − + =⇒ − − − =

⇒ − − =

⇒ n = 16 or 25 Case I: When n = 2 5 , 25 ( 1) 20 (25 1)( 1) 20 24 4a a n d= + − = + − − = − = −Which is impossible ∴ is rejected 25n =Case II: When n = 16 16 ( 1) 20 (16 1)( 1) 20 15a a n d= + − = + − − = − = 5Hence, there are 16 rows and 5 logs are in the top row.

Q.20. In a potato race a bucket is placed at the starting point, which is 5 cm from the first potato, and the other potatoes are placed 3 m apart in a straight line. There are ten potatoes in the line (see fig. below)

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 she continues in the same way until all the potatoes are in the bucket. What is the total distance the competitor has to run?

Solution. Distance covered to pick up the 1st potato 2(5) m 10 m= = Distance between successive potatoes 3 m= ∴ Distance covered to pick up the 2nd potato 2(5 3) m 16 m= + = Distance covered to pick up the 3rd potato 2(5 3 3) m 22 m= + + = Here, 1 2 310, 16, 22, ,a a a a= = = = … and 2 1 16 10 6d a a= − = − = 10n = ∴ Total distance the competitor has to run 10S=

[2 ( 1) ]2n a n d= + −



38

10 [2(10) (10 1)6]2

5[20 54]5 74 370

= + −

= += × =

Hence, the competitor has to run 370 m in all.

Exercise 5.4 (Optional) Q.1. Which term of the AP 1 is its first negative term? 21,117,113,…Solution. Given AP is 121 ,117,113,…

Here, 1 2 3 2 1121, 117, 113, 117 121 4a a a a d a a= = = = = − = − = −Using formula, ( 1)na a n d= + −

121 ( 1)( 4)121 4 4125 4

nnn

= + − −= − += −

As per condition, 0na <

125 4 0 125 4

4 125125

4131 or 324

nn

n

n

n n

⇒ − <⇒ <

⇒ <

⇒ >

> =

∴ 32n =Hence, 32nd term is the first negative term of given AP.

Q.2. The sum of the third and the seventh term of an AP is 6 and their product is 8. Find the sum of first sixteen terms of the AP.

Solution. Let a and d be the first term and common difference of given AP.

Case I:

3 7 6 [ (3 1) ] [ (7 1) ] 6 [ ( 1) ]

2 6 6 2 8 6

4 3 ...(i)

n

a aa d a d a a n

a d a da da d

+ =⇒ + − + + − = = + −

⇒ + + + =⇒ + =

⇒ + =

∵ d

Case II:



39

3 7

2

2

2

( ) 8 [ (3 1) ][ (7 1) ] 8 [ ( 1) ] ( 2 )( 6 ) 8 [3 4 2 ][3 4 6 ] 8 [From (i), 3 4 ]

(3 2 )(3 2 ) 8

9 4 8

4 9 81 4

1 2

n

a aa d a d a a n d

a d a dd d d d a d

d d

d

d

d

d

=

⇒ + − + − = = + −

⇒ + + =⇒ − + − + = = −

⇒ − + =

⇒ − =

⇒ = −

⇒ =

⇒ = ±

∵

When 1=2

d

Putting 12

d = in eq. (i), we have

14 32

2 3 3 2 1

a

aa

⎛ ⎞+ =⎜ ⎟⎝ ⎠

⇒ + =⇒ = − =

16

16

Now, [2 ( 1) ]216 1 [2(1) (16 1) ]2 2

158 22

4 15 198 8 × 2 2

76

nnS a n d

S

S

= + −

⇒ = + −

⎡ ⎤= +⎢ ⎥⎣ ⎦+⎡ ⎤= =⎢ ⎥⎣ ⎦

=

Hence, the sum of first 16 terms of the AP is 76

When 1= – .2

d

Putting 12

d = − in eq. (i), we have

14 32

2 3 3 2 5

a

a a

⎛ ⎞+ − =⎜ ⎟⎝ ⎠

⇒ − = ⇒ = + =



40

16

16

Now, [2 ( 1) ]216 12(5) (16 1)2 2

15 20 15 58 10 8 8 ×2 2

20

nnS a n d

S

S

= + −

⎡ ⎤⎛ ⎞= + − −⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦−⎡ ⎤ ⎡ ⎤= − = =⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

=

2

Hence, the sum of first sixteen terms of the AP is 20. Q.3. A ladder has rungs 25 cm apart (see Fig. below). The rungs decrease uniformly

in length from 45 cm at the bottom to 25 cm at the top. If the top and bottom

rungs are 2 12

m apart, what is the length of the wood required for the rungs?

Solution. The gap between two consecutive rungs is 25 cm and the top and bottom rungs are 2.5 m i.e. 250 cm apart.

∴ Number of rungs = 25025

= 10



41

The rungs are decreasing uniformly in length from 45 cm at the bottom to 25 cm at the top. Therefore, lengths of the rungs form an AP with first term a = 45 cm and 10th term = 25 cm

∴ Length of the wood required for rungs = sum of 10 terms of an AP with first term 45 cm and last term = 25 cm

= 102

[45 + 25] cm = 350 cm.

Q.4. The houses of a row are numbered consecutively from 1 to 49. Show that there is a value of x such that the sum of the numbers of the houses preceding the house numbered x is equal to the sum of the numbers of the houses following it. Find this value of x.

Solution. Let x denote the number of any house. Here, 1 1, 1a a d= = =As per condition,

1 49

49 [1 49] [2(1) ( 1)(1)]2 2

Using [2 ( 1) and ( ), where is last term2 2

1 49 [2 2] (50) [2 1]2 2 2

( 1) ( 1) 49(25)2 2

[ 1 1] 12252

2

x x

n n

S S Sx x

n nS a n d S a l l

x xx x

x x x x

x x x

x

− = −

= + − + −

⎡ ⎤= + − = +⎢ ⎥⎣ ⎦−

⇒ + − = − + −

− +⇒ = −

⇒ − + + =

⇒

2

2 1225

1225 35

x

xx

× =

⇒ =⇒ =

Q.5. A small terrace at a football ground comprises of 15 steps each of which is 50 m

long and built of solid concrete.

Each step has a rise of 14

m and a tread of 12

m (see Fig. below). Calculate the

total volume of concrete required to build the terrace.



42

Solution. Volume of concrete required to build the first step

3 31 1 2550 m m4 2 4

⎛ ⎞ ⎛ ⎞= × × =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

Volume of concrete required to build the second step

3 32 1 2550 m m4 2 2

⎛ ⎞ ⎛ ⎞= × × =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

Volume of concrete required to build the third step

3 33 1 7550 m m4 2 4

⎛ ⎞ ⎛ ⎞= × × =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

and so on up to 15 steps.

Here, 1 2 325 25 75, , , 15.4 2 4

a a a a n= = = = =

2 125 25 50 25 252 4 4 4

d a a −= − = − = =

Total volume of concrete required to build the terrace

15 [2 ( 1) ]215 25 252 (15 1)2 4 4

15 25 14 252 2 4

15 25 1752 2 2

15 200 750.2 2

nS a n d= = + −

⎡ ⎤⎛ ⎞= + −⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦×⎡ ⎤= +⎢ ⎥⎣ ⎦

⎡ ⎤= +⎢ ⎥⎣ ⎦

= × =

Hence, total volume of concrete required to build the terrace is 750 3m .

Documents

5 ARITHMETIC PROGRESSIONS - testlabz.com · 5 ARITHMETIC PROGRESSIONS Exercise 5.1 Q.1. In which of the following situations, does the list of numbers involved make an arithmetic