# NCERT Solutions For Class 10 Maths Chapter 5 Arithmetic Progressions

## NCERT Solutions for Class 10 Maths Chapter 5 â€“ CBSE Term II Free PDF Download

NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progression is presented here for the benefit of the students preparing for the CBSE Term II Class 10 Maths examination. It is very important for the students to get well versed with these solutions of NCERT to get a good score in Class 10 examination. These NCERT problems are solved by the experts at BYJUâ€™S. These solutions will help you understand and master different types of questions on arithmetic progressions. NCERT Solutions help you to attain perfection in solving different kinds of questions.

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## Exercise 5.1 Page: 99

1. In which of the following situations, does the list of numbers involved make as 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.

Solution:

We can write the given condition as;

Taxi fare for 1 km = 15

Taxi fare for first 2 kms = 15+8 = 23

Taxi fare for first 3 kms = 23+8 = 31

Taxi fare for first 4 kms = 31+8 = 39

And so onâ€¦â€¦

Thus, 15, 23, 31, 39 â€¦ forms an A.P. because every next term is 8 more than the preceding term.

(ii) The amount of air present in a cylinder when a vacuum pump removes 1/4Â of the air remaining in the cylinder at a time.

Solution:

Let the volume of air in a cylinder, initially, beÂ VÂ litres.

In each stroke, the vacuum pump removesÂ 1/4thÂ of airÂ remaining in the cylinder at a time. Or we can say, after every stroke, 1-1/4 = 3/4th part of air will remain.

Therefore, volumes will beÂ V, 3V/4 , (3V/4)2Â , (3V/4)3â€¦and so on

Clearly, we can see here, the adjacent terms of this series do not have the common difference between them. Therefore, this series is not an A.P.

(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.

Solution:

We can write the given condition as;

Cost of digging a well for first metre = Rs.150

Cost of digging a well for first 2 metres = Rs.150+50 = Rs.200

Cost of digging a well for first 3 metres = Rs.200+50 = Rs.250

Cost of digging a well for first 4 metres =Rs.250+50 = Rs.300

And so on..

Clearly, 150, 200, 250, 300 â€¦ forms an A.P. with a common difference of 50 between each term.

(iv) The amount of money in the account every year, when Rs 10000 is deposited at compound interest at 8% per annum.

Solution:

We know that if Rs. PÂ is deposited atÂ r% compound interest per annum for n years, the amount of money will be:

P(1+r/100)n

Therefore, after each year, the amount of money will be;

10000(1+8/100), 10000(1+8/100)2, 10000(1+8/100)3â€¦â€¦

Clearly, the terms of this series do not have the common difference between them. Therefore, this isÂ not an A.P.

2. Write first four terms of the A.P. when the first term a and the common difference are given as follows:

(i)Â aÂ = 10,Â dÂ = 10
(ii)Â aÂ = -2,Â dÂ = 0
(iii)Â aÂ = 4,Â dÂ = â€“ 3
(iv)Â aÂ = -1Â dÂ = 1/2
(v)Â aÂ = â€“ 1.25,Â dÂ = â€“ 0.25

Solutions:

(i)Â aÂ = 10,Â dÂ = 10

Let us consider, the Arithmetic Progression series beÂ a1,Â a2,Â a3,Â a4,Â a5Â â€¦

a1Â =Â aÂ = 10

a2Â =Â a1+dÂ = 10+10 = 20

a3Â =Â a2+dÂ = 20+10 = 30

a4Â =Â a3+dÂ = 30+10 = 40

a5Â =Â a4+dÂ = 40+10 = 50

And so onâ€¦

Therefore, the A.P. series will be 10, 20, 30, 40, 50 â€¦

And First four terms of this A.P. will be 10, 20, 30, and 40.

(ii)Â aÂ = â€“ 2,Â dÂ = 0

Let us consider, the Arithmetic Progression series beÂ a1,Â a2,Â a3,Â a4,Â a5Â â€¦

a1Â =Â aÂ = -2

a2Â =Â a1+dÂ = â€“ 2+0 = â€“ 2

a3Â =Â a2+d = â€“ 2+0 = â€“ 2

a4Â =Â a3+dÂ = â€“ 2+0 = â€“ 2

Therefore, the A.P. series will be â€“ 2, â€“ 2, â€“ 2, â€“ 2 â€¦

And, First four terms of this A.P. will be â€“ 2, â€“ 2, â€“ 2 and â€“ 2.

(iii)Â aÂ = 4,Â dÂ = â€“ 3

Let us consider, the Arithmetic Progression series beÂ a1,Â a2,Â a3,Â a4,Â a5Â â€¦

a1Â =Â aÂ = 4

a2Â =Â a1+dÂ = 4-3 = 1

a3Â =Â a2+dÂ = 1-3 = â€“ 2

a4Â =Â a3+dÂ = -2-3 = â€“ 5

Therefore, the A.P. series will be 4, 1, â€“ 2 â€“ 5 â€¦

And, first four terms of this A.P. will be 4, 1, â€“ 2 and â€“ 5.

(iv)Â aÂ = â€“ 1,Â dÂ = 1/2

Let us consider, the Arithmetic Progression series beÂ a1,Â a2,Â a3,Â a4,Â a5Â â€¦

a2Â =Â a1+dÂ = -1+1/2 = -1/2

a3Â =Â a2+dÂ = -1/2+1/2 = 0

a4Â =Â a3+dÂ = 0+1/2 = 1/2

Thus, the A.P. series will be-1, -1/2, 0, 1/2

And First four terms of this A.P. will be -1, -1/2, 0 and 1/2.

(v)Â aÂ = â€“ 1.25,Â dÂ = â€“ 0.25

Let us consider, the Arithmetic Progression series beÂ a1,Â a2,Â a3,Â a4,Â a5Â â€¦

a1Â =Â aÂ = â€“ 1.25

a2Â =Â a1Â +Â dÂ = â€“ 1.25-0.25 = â€“ 1.50

a3Â =Â a2Â +Â dÂ = â€“ 1.50-0.25 = â€“ 1.75

a4Â =Â a3Â +Â dÂ = â€“ 1.75-0.25 = â€“ 2.00

Therefore, the A.P series will be 1.25, â€“ 1.50, â€“ 1.75, â€“ 2.00 â€¦â€¦..

And first four terms of this A.P. will be â€“ 1.25, â€“ 1.50, â€“ 1.75 and â€“ 2.00.

3. For the following A.P.s, write the first term and the common difference.
(i) 3, 1, â€“ 1, â€“ 3 â€¦
(ii) -5, â€“ 1, 3, 7 â€¦
(iii) 1/3, 5/3, 9/3, 13/3 â€¦.
(iv) 0.6, 1.7, 2.8, 3.9 â€¦

Solutions

(i) Given series,

3, 1, â€“ 1, â€“ 3 â€¦

First term,Â aÂ = 3

Common difference,Â dÂ = Second term â€“ First term

â‡’Â  1 â€“ 3 = -2

â‡’Â  d = -2

(ii) Given series, â€“ 5, â€“ 1, 3, 7 â€¦

First term,Â aÂ = -5

Common difference,Â dÂ = Second term â€“ First term

â‡’Â ( â€“ 1)-( â€“ 5) = â€“ 1+5 = 4

(iii) Given series, 1/3, 5/3, 9/3, 13/3 â€¦.

First term,Â aÂ = 1/3

Common difference,Â dÂ = Second term â€“ First term

â‡’Â 5/3 â€“ 1/3 = 4/3

(iv) Given series, 0.6, 1.7, 2.8, 3.9 â€¦

First term,Â aÂ = 0.6

Common difference,Â dÂ = Second term â€“ First term

â‡’Â 1.7 â€“ 0.6

â‡’Â 1.1

4. Which of the following are APs? If they form an A.P. find the common differenceÂ dÂ and write three more terms.

(i) 2, 4, 8, 16 â€¦
(ii) 2, 5/2, 3, 7/2 â€¦.
(iii) -1.2, -3.2, -5.2, -7.2 â€¦
(iv) -10, â€“ 6, â€“ 2, 2 â€¦
(v) 3, 3 +Â âˆš2, 3Â + 2âˆš2, 3Â + 3âˆš2
(vi) 0.2, 0.22, 0.222, 0.2222 â€¦.
(vii) 0, â€“ 4, â€“ 8, â€“ 12 â€¦
(viii) -1/2, -1/2, -1/2, -1/2 â€¦.
(ix) 1, 3, 9, 27 â€¦
(x)Â a, 2a, 3a, 4aÂ â€¦
(xi)Â a,Â a2,Â a3,Â a4Â â€¦
(xii) âˆš2, âˆš8, âˆš18, âˆš32Â â€¦
(xiii) âˆš3, âˆš6, âˆš9, âˆš12Â â€¦
(xiv) 12, 32, 52, 72Â â€¦
(xv) 12, 52, 72, 73Â â€¦

Solution

(i) Given to us,

2, 4, 8, 16 â€¦

Here, the common difference is;

a2Â â€“Â a1Â = 4 â€“ 2 = 2

a3Â â€“Â a2Â = 8 â€“ 4 = 4

a4Â â€“Â a3Â = 16 â€“ 8 = 8

Since,Â an+1Â â€“Â an or the common difference is not the same every time.

Therefore, the given series are not forming an A.P.

(ii) Given, 2, 5/2, 3, 7/2 â€¦.

Here,

a2Â â€“Â a1Â =Â 5/2-2 = 1/2

a3Â â€“Â a2Â =Â 3-5/2 = 1/2

a4Â â€“Â a3Â =Â 7/2-3 = 1/2

Since, an+1Â â€“Â anÂ or the common difference is same every time.

Therefore,Â dÂ = 1/2Â and the given series are in A.P.

The next three terms are;

a5Â = 7/2+1/2 = 4

a6Â = 4Â +1/2 = 9/2

a7Â = 9/2Â +1/2 = 5

(iii)Â Given, -1.2, â€“ 3.2, -5.2, -7.2 â€¦

Here,

a2Â â€“Â a1Â = (-3.2)-(-1.2) = -2

a3Â â€“Â a2Â = (-5.2)-(-3.2) = -2

a4Â â€“Â a3Â = (-7.2)-(-5.2) = -2

Since, an+1Â â€“Â anÂ or common difference is same every time.

Therefore,Â dÂ = -2 and the given series are in A.P.

Hence, next three terms are;

a5Â = â€“ 7.2-2 = -9.2

a6Â = â€“ 9.2-2 = â€“ 11.2

a7Â = â€“ 11.2-2 = â€“ 13.2

(iv) Given, -10, â€“ 6, â€“ 2, 2 â€¦

Here, the terms and their difference are;

a2Â â€“Â a1Â = (-6)-(-10) = 4

a3Â â€“Â a2Â = (-2)-(-6) = 4

a4Â â€“Â a3Â = (2 -(-2) = 4

Since,Â an+1Â â€“Â anÂ or the common difference is same every time.

Therefore,Â dÂ = 4 and the given numbers are in A.P.

Hence, next three terms are;

a5Â = 2+4 = 6

a6Â = 6+4 = 10

a7Â = 10+4 = 14

(v) Given, 3, 3+âˆš2, 3+2âˆš2, 3+3âˆš2

Here,

a2Â â€“Â a1Â = 3+âˆš2-3 = âˆš2

a3Â â€“Â a2Â = (3+2âˆš2)-(3+âˆš2) = âˆš2

a4Â â€“Â a3Â = (3+3âˆš2) â€“ (3+2âˆš2) = âˆš2

Since,Â an+1Â â€“Â anÂ or the common difference is same every time.

Therefore,Â dÂ =Â âˆš2Â and the given series forms a A.P.

Hence, next three terms are;

a5Â = (3+âˆš2)Â +âˆš2Â = 3+4âˆš2

a6Â = (3+4âˆš2)+âˆš2Â = 3+5âˆš2

a7Â = (3+5âˆš2)+âˆš2Â = 3+6âˆš2

(vi)Â 0.2, 0.22, 0.222, 0.2222 â€¦.

Here,

a2Â â€“Â a1Â =Â 0.22-0.2 = 0.02

a3Â â€“Â a2Â =Â 0.222-0.22 = 0.002

a4Â â€“Â a3Â =Â 0.2222-0.222 = 0.0002

Since,Â an+1Â â€“Â anÂ or the common difference is not same every time.

Therefore, and the given series doesnâ€™t forms a A.P.

(vii)Â 0, -4, -8, -12 â€¦

Here,

a2Â â€“Â a1Â =Â (-4)-0 = -4

a3Â â€“Â a2Â =Â (-8)-(-4) = -4

a4Â â€“Â a3Â =Â (-12)-(-8) = -4

Since,Â an+1Â â€“Â anÂ or the common difference is same every time.

Therefore,Â dÂ =Â -4Â and the given series forms a A.P.

Hence, next three terms are;

a5Â =Â -12-4 = -16

a6Â =Â -16-4 = -20

a7Â =Â -20-4 = -24

(viii) -1/2, -1/2, -1/2, -1/2 â€¦.

Here,

a2Â â€“Â a1Â = (-1/2) â€“ (-1/2) = 0

a3Â â€“Â a2Â = (-1/2) â€“ (-1/2) = 0

a4Â â€“Â a3Â = (-1/2) â€“ (-1/2) = 0

Since,Â an+1Â â€“Â anÂ or the common difference is same every time.

Therefore,Â dÂ = 0 and the given series forms a A.P.

Hence, next three terms are;

a5Â = (-1/2)-0 = -1/2

a6Â =Â (-1/2)-0 = -1/2

a7Â =Â (-1/2)-0 = -1/2

(ix) 1, 3, 9, 27 â€¦

Here,

a2Â â€“Â a1Â =Â 3-1 = 2

a3Â â€“Â a2Â =Â 9-3 = 6

a4Â â€“Â a3Â =Â 27-9 = 18

Since,Â an+1Â â€“Â anÂ or the common difference is not same every time.

Therefore, and the given series doesnâ€™t form a A.P.

(x)Â a, 2a, 3a, 4aÂ â€¦

Here,

a2Â â€“Â a1Â =Â 2aâ€“aÂ =Â a

a3Â â€“Â a2Â =Â 3a-2aÂ =Â a

a4Â â€“Â a3Â =Â 4a-3aÂ =Â a

Since,Â an+1Â â€“Â anÂ or the common difference is same every time.

Therefore,Â dÂ =Â aÂ and the given series forms a A.P.

Hence, next three terms are;

a5Â =Â 4a+aÂ = 5a

a6Â = 5a+aÂ = 6a

a7Â =Â 6a+aÂ = 7a

(xi)Â a,Â a2,Â a3,Â a4Â â€¦

Here,

a2Â â€“Â a1Â =Â a2â€“aÂ = a(a-1)

a3Â â€“Â a2Â =Â a3Â â€“Â a2Â =Â a2(a-1)

a4Â â€“Â a3Â =Â a4Â â€“Â a3Â =Â a3(a-1)

Since,Â an+1Â â€“Â anÂ or the common difference is not same every time.

Therefore, the given series doesnâ€™t forms a A.P.

(xii) âˆš2, âˆš8, âˆš18, âˆš32Â â€¦

Here,

a2Â â€“Â a1Â = âˆš8-âˆš2Â Â = 2âˆš2-âˆš2Â = âˆš2

a3Â â€“Â a2Â = âˆš18-âˆš8Â = 3âˆš2-2âˆš2Â = âˆš2

a4Â â€“Â a3Â = 4âˆš2-3âˆš2Â = âˆš2

Since,Â an+1Â â€“Â anÂ or the common difference is same every time.

Therefore,Â dÂ =Â âˆš2Â and the given series forms a A.P.

Hence, next three terms are;

a5Â = âˆš32+âˆš2Â = 4âˆš2+âˆš2Â = 5âˆš2Â = âˆš50

a6Â = 5âˆš2+âˆš2Â = 6âˆš2Â = âˆš72

a7Â = 6âˆš2+âˆš2Â = 7âˆš2Â = âˆš98

(xiii)Â âˆš3, âˆš6, âˆš9, âˆš12Â â€¦

Here,

a2Â â€“Â a1Â = âˆš6-âˆš3Â = âˆš3Ã—âˆš2-âˆš3Â = âˆš3(âˆš2-1)

a3Â â€“Â a2Â = âˆš9-âˆš6Â = 3-âˆš6Â = âˆš3(âˆš3-âˆš2)

a4Â â€“Â a3Â = âˆš12Â â€“ âˆš9Â = 2âˆš3Â â€“ âˆš3Ã—âˆš3 = âˆš3(2-âˆš3)

Since,Â an+1Â â€“Â anÂ or the common difference is not same every time.

Therefore, the given series doesnâ€™t form a A.P.

(xiv) 12, 32, 52, 72Â â€¦

Or, 1, 9, 25, 49 â€¦..

Here,

a2Â âˆ’Â a1Â = 9âˆ’1 = 8

a3Â âˆ’Â a2Â = 25âˆ’9 = 16

a4Â âˆ’Â a3Â = 49âˆ’25 = 24

Since,Â an+1Â â€“Â anÂ or the common difference is not same every time.

Therefore, the given series doesnâ€™t form a A.P.

(xv) 12, 52, 72, 73 â€¦

Or 1, 25, 49, 73 â€¦

Here,

a2Â âˆ’Â a1Â = 25âˆ’1 = 24

a3Â âˆ’Â a2Â = 49âˆ’25 = 24

a4Â âˆ’Â a3Â = 73âˆ’49 = 24

Since,Â an+1Â â€“Â anÂ or the common difference is same every time.

Therefore,Â dÂ = 24Â and the given series forms a A.P.

Hence, next three terms are;

a5Â = 73+24 = 97

a6Â = 97+24 = 121

a7Â = 121+24 = 145

## Exercise 5.2 Page: 105

1. Fill in the blanks in the following table, given thatÂ aÂ is the first term,Â dÂ the common difference andÂ anÂ theÂ nthÂ term of the A.P.

Solutions:

(i)Â Given,

First term, aÂ = 7

Common difference,Â dÂ = 3

Number of terms,Â nÂ = 8,

We have to find the nth term, anÂ = ?

As we know, for an A.P.,

anÂ =Â a+(nâˆ’1)d

Putting the values,

=> 7+(8 âˆ’1) 3

=> 7+(7) 3

=> 7+21 = 28

Hence,Â anÂ = 28

(ii) Given,

First term, aÂ = -18

Common difference,Â dÂ = ?

Number of terms,Â nÂ = 10

Nth term, anÂ = 0

As we know, for an A.P.,

anÂ =Â a+(nâˆ’1)d

Putting the values,

0 = âˆ’ 18 +(10âˆ’1)d

18 = 9d

dÂ = 18/9 = 2

Hence, common difference,Â dÂ = 2

(iii) Given,

First term, aÂ = ?

Common difference,Â dÂ = -3

Number of terms,Â nÂ = 18

Nth term, anÂ = -5

As we know, for an A.P.,

anÂ =Â a+(nâˆ’1)d

Putting the values,

âˆ’5 =Â a+(18âˆ’1) (âˆ’3)

âˆ’5 =Â a+(17) (âˆ’3)

âˆ’5 =Â aâˆ’51

aÂ = 51âˆ’5 = 46

Hence,Â aÂ = 46

(iv)Â Given,

First term, aÂ = -18.9

Common difference,Â dÂ = 2.5

Number of terms,Â nÂ = ?

Nth term, anÂ = 3.6

As we know, for an A.P.,

anÂ =Â aÂ +(nÂ âˆ’1)d

Putting the values,

3.6 = âˆ’ 18.9+(nÂ âˆ’1)2.5

3.6 + 18.9 = (nâˆ’1)2.5

22.5 = (nâˆ’1)2.5

(nÂ â€“ 1) = 22.5/2.5

nÂ â€“ 1 = 9

nÂ = 10

Hence,Â nÂ = 10

(v)Â Given,

First term, aÂ = 3.5

Common difference,Â dÂ = 0

Number of terms,Â nÂ = 105

Nth term, anÂ = ?

As we know, for an A.P.,

anÂ =Â a+(nÂ âˆ’1)d

Putting the values,

anÂ = 3.5+(105âˆ’1) 0

anÂ = 3.5+104Ã—0

anÂ = 3.5

Hence,Â anÂ = 3.5

2. Choose the correct choice in the following and justify:
(i) 30thÂ term of the A.P: 10,7, 4, â€¦, is
(A)Â 97 (B)Â 77 (C)Â âˆ’77 (D) âˆ’87

(ii) 11thÂ term of the A.P. -3, -1/2, ,2 â€¦. is
(A) 28 (B) 22 (C) â€“ 38 (D)

Â

Solutions:

(i) Given here,

A.P. = 10, 7, 4, â€¦

Therefore, we can find,

First term,Â aÂ = 10

Common difference,Â dÂ =Â a2Â âˆ’Â a1Â = 7âˆ’10 = âˆ’3

As we know, for an A.P.,

anÂ =Â aÂ +(nâˆ’1)d

Putting the values;

a30Â = 10+(30âˆ’1)(âˆ’3)

a30Â = 10+(29)(âˆ’3)

a30Â = 10âˆ’87 = âˆ’77

Hence, the correct answer is optionÂ C.

(ii) Given here,

A.P. =Â -3, -1/2, ,2 â€¦

Therefore, we can find,

First termÂ aÂ = â€“ 3

Common difference,Â dÂ =Â a2Â âˆ’Â a1Â = (-1/2) -(-3)

â‡’(-1/2)Â + 3 = 5/2

As we know, for an A.P.,

anÂ =Â a+(nâˆ’1)d

Putting the values;

a11Â = -3+(11-1)(5/2)

a11Â = -3+(10)(5/2)

a11Â = -3+25

a11Â = 22

Hence, the answer is option B.

3. In the following APs find the missing term in the boxes.

Solutions:

(i)Â For the given A.P., 2,2 , 26

The first and third term are;

aÂ = 2

a3Â = 26

As we know, for an A.P.,

anÂ =Â a+(nÂ âˆ’1)d

Therefore, putting the values here,

a3Â = 2+(3-1)d

26 = 2+2d

24 = 2d

dÂ = 12

a2Â = 2+(2-1)12

= 14

Therefore, 14 is the missing term.

(ii)Â For the given A.P., , 13, ,3

a2Â = 13 and

a4Â = 3

As we know, for an A.P.,

anÂ =Â a+(nâˆ’1)Â d

Therefore, putting the values here,

a2Â =Â aÂ +(2-1)d

13 =Â a+dÂ â€¦â€¦â€¦â€¦â€¦â€¦.Â (i)

a4Â =Â a+(4-1)d

3 =Â a+3dÂ â€¦â€¦â€¦â€¦..Â (ii)

On subtracting equationÂ (i)Â fromÂ (ii), we get,

â€“ 10 = 2d

dÂ = â€“ 5

From equationÂ (i), putting the value of d,we get

13 =Â a+(-5)

aÂ = 18

a3Â = 18+(3-1)(-5)

= 18+2(-5) = 18-10 = 8

Therefore, the missing terms are 18 and 8 respectively.

(iii) For the given A.P.,

aÂ = 5 and

a4Â = 19/2

As we know, for an A.P.,

anÂ =Â a+(nâˆ’1)d

Therefore, putting the values here,

a4Â =Â a+(4-1)d

19/2 =Â 5+3d

(19/2) â€“ 5 = 3d

3d =Â 9/2

d = 3/2

a2Â =Â a+(2-1)d

a2Â =Â 5+3/2

a2Â =Â 13/2

a3Â =Â a+(3-1)d

a3Â =Â 5+2Ã—3/2

a3Â =Â 8

Therefore, the missing terms are 13/2 and 8 respectively.

(iv) For the given A.P.,

aÂ = âˆ’4 and

a6Â = 6

As we know, for an A.P.,

anÂ =Â aÂ +(nâˆ’1)Â d

Therefore, putting the values here,

a6Â = a+(6âˆ’1)d

6 = âˆ’ 4+5d

10 = 5d

dÂ = 2

a2Â =Â a+dÂ = âˆ’ 4+2 = âˆ’2

a3Â =Â a+2dÂ = âˆ’ 4+2(2) = 0

a4Â =Â a+3dÂ = âˆ’ 4+ 3(2) = 2

a5Â =Â a+4dÂ = âˆ’ 4+4(2) = 4

Therefore, the missing terms are âˆ’2, 0, 2, and 4 respectively.

(v) For the given A.P.,

a2Â = 38

a6Â = âˆ’22

As we know, for an A.P.,

anÂ =Â a+(nÂ âˆ’1)d

Therefore, putting the values here,

a2Â =Â a+(2âˆ’1)d

38 =Â a+dÂ â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦.Â (i)

a6Â =Â a+(6âˆ’1)d

âˆ’22 =Â a+5dÂ â€¦â€¦â€¦â€¦â€¦â€¦â€¦.Â (ii)

On subtracting equationÂ (i)Â fromÂ (ii), we get

âˆ’ 22 âˆ’ 38 = 4d

âˆ’60 = 4d

dÂ = âˆ’15

aÂ =Â a2Â âˆ’Â dÂ = 38 âˆ’ (âˆ’15) = 53

a3Â =Â aÂ + 2dÂ = 53 + 2 (âˆ’15) = 23

a4Â =Â aÂ + 3dÂ = 53 + 3 (âˆ’15) = 8

a5Â =Â aÂ + 4dÂ = 53 + 4 (âˆ’15) = âˆ’7

Therefore, the missing terms are 53, 23, 8, and âˆ’7 respectively.

4. Which term of the A.P. 3, 8, 13, 18, â€¦ is 78?

Solutions:

Given the A.P. series as3, 8, 13, 18, â€¦

First term, aÂ = 3

Common difference, dÂ =Â a2Â âˆ’Â a1Â = 8 âˆ’ 3 = 5

LetÂ the nthÂ term of given A.P. be 78. Now as we know,

anÂ =Â a+(nâˆ’1)d

Therefore,

78 = 3+(nÂ âˆ’1)5

75 = (nâˆ’1)5

(nâˆ’1) = 15

nÂ = 16

Hence, 16thÂ term of this A.P. is 78.

5. Find the number of terms in each of the following A.P.

(i) 7, 13, 19, â€¦, 205

Solutions:

(i) Given, 7, 13, 19, â€¦, 205 is the A.P

Therefore

First term, aÂ = 7

Common difference, dÂ =Â a2Â âˆ’Â a1Â = 13 âˆ’ 7 = 6

Let there areÂ nÂ terms in this A.P.

anÂ = 205

As we know, for an A.P.,

anÂ =Â aÂ + (nÂ âˆ’ 1)Â d

Therefore, 205 = 7 + (nÂ âˆ’ 1) 6

198 = (nÂ âˆ’ 1) 6

33 = (nÂ âˆ’ 1)

nÂ = 34

Therefore, this given series has 34 terms in it.

First term, aÂ = 18

Common difference, d = a2-a1 =

d = (31-36)/2 = -5/2

Let there are n terms in this A.P.

anÂ = -47

As we know, for an A.P.,

anÂ =Â a+(nâˆ’1)d

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

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

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

(n-1) = -130/-5

(n-1) =Â 26

nÂ = 27

Therefore, this given A.P. has 27 terms in it.

6.Â Check whether -150 is a term of the A.P. 11, 8, 5, 2, â€¦

Solution:

For the given series, A.P. 11, 8, 5, 2..

First term, aÂ = 11

Common difference, dÂ =Â a2âˆ’a1Â = 8âˆ’11 = âˆ’3

Let âˆ’150 be theÂ nthÂ term of this A.P.

As we know, for an A.P.,

anÂ =Â a+(nâˆ’1)d

-150 = 11+(nÂ -1)(-3)

-150 = 11-3nÂ +3

-164 = -3n

nÂ = 164/3

Clearly,Â nÂ is not an integer but a fraction.

Therefore, â€“ 150 is not a term of this A.P.

7. Find the 31stÂ term of an A.P. whose 11thÂ term is 38 and the 16thÂ term is 73.

Solution:

Given that,

11th term, a11Â = 38

and 16th term, a16Â = 73

We know that,

anÂ =Â a+(nâˆ’1)d

a11Â =Â a+(11âˆ’1)d

38 =Â a+10dÂ â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦.Â (i)

In the same way,

a16Â =Â aÂ +(16âˆ’1)d

73 =Â a+15dÂ â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦Â (ii)

On subtracting equationÂ (i)Â fromÂ (ii), we get

35 = 5d

dÂ = 7

From equationÂ (i), we can write,

38 =Â a+10Ã—(7)

38 âˆ’ 70 =Â a

aÂ = âˆ’32

a31Â =Â aÂ +(31âˆ’1)Â d

= âˆ’ 32 + 30 (7)

= âˆ’ 32 + 210

= 178

Hence, 31stÂ term is 178.

8. An A.P. consists of 50 terms of which 3rdÂ term is 12 and the last term is 106. Find the 29thÂ term.

Solution: Given that,

3rd term, a3Â = 12

50th term, a50Â = 106

We know that,

anÂ =Â a+(nâˆ’1)d

a3Â =Â a+(3âˆ’1)d

12 =Â a+2dÂ â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦.Â (i)

In the same way,

a50Â =Â a+(50âˆ’1)d

106 =Â a+49dÂ â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦.Â (ii)

On subtracting equationÂ (i)Â fromÂ (ii), we get

94 = 47d

dÂ = 2 = common difference

From equationÂ (i), we can write now,

12 =Â a+2(2)

aÂ = 12âˆ’4 = 8

a29Â =Â a+(29âˆ’1)Â d

a29Â = 8+(28)2

a29Â = 8+56 = 64

Therefore, 29thÂ term is 64.

9. If the 3rdÂ and the 9thÂ terms of an A.P. are 4 and âˆ’ 8 respectively. Which term of this A.P. is zero.
Solution:

Given that,

3rd term, a3Â = 4

and 9th term, a9Â = âˆ’8

We know that,

anÂ =Â a+(nâˆ’1)d

Therefore,

a3Â =Â a+(3âˆ’1)d

4 =Â a+2dÂ â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦Â (i)

a9Â =Â a+(9âˆ’1)d

âˆ’8 =Â a+8dÂ â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦Â (ii)

On subtracting equationÂ (i)Â fromÂ (ii), we will get here,

âˆ’12 = 6d

dÂ = âˆ’2

From equationÂ (i), we can write,

4 =Â a+2(âˆ’2)

4 =Â aâˆ’4

aÂ = 8

LetÂ nthÂ term of this A.P. be zero.

anÂ =Â a+(nâˆ’1)d

0 = 8+(nâˆ’1)(âˆ’2)

0 = 8âˆ’2n+2

2nÂ = 10

nÂ = 5

Hence, 5thÂ term of this A.P. is 0.

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

Solution:

We know that, for an A.P series;

anÂ =Â a+(nâˆ’1)d

a17Â =Â a+(17âˆ’1)d

a17Â =Â aÂ +16d

In the same way,

a10Â =Â a+9d

As it is given in the question,

a17Â âˆ’Â a10Â = 7

Therefore,

(aÂ +16d)âˆ’(a+9d) = 7

7dÂ = 7

dÂ = 1

Therefore, the common difference is 1.

11.Â Which term of the A.P. 3, 15, 27, 39,.. will be 132 more than its 54thÂ term?

Solution:

Given A.P. is 3, 15, 27, 39, â€¦

first term, aÂ = 3

common difference, dÂ =Â a2Â âˆ’Â a1Â = 15 âˆ’ 3 = 12

We know that,

anÂ =Â a+(nâˆ’1)d

Therefore,

a54Â =Â a+(54âˆ’1)d

â‡’3+(53)(12)

â‡’3+636 = 639

a54 = 639+132=771

We have to find the term of this A.P. which is 132 more than a54, i.e.771.

LetÂ nthÂ term be 771.

anÂ =Â a+(nâˆ’1)d

771 = 3+(nÂ âˆ’1)12

768 = (nâˆ’1)12

(nÂ âˆ’1) = 64

nÂ = 65

Therefore, 65thÂ term was 132 more than 54thÂ term.

Or another method is;

LetÂ nthÂ term be 132 more than 54thÂ term.

nÂ = 54Â + 132/2

= 54Â + 11 =Â 65thÂ term

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

Solution:

Let, the first term of two APs beÂ a1Â andÂ a2Â respectively

And the common difference of these APs beÂ d.

For the first A.P.,we know,

anÂ =Â a+(nâˆ’1)d

Therefore,

a100Â =Â a1+(100âˆ’1)d

=Â a1Â + 99d

a1000Â =Â a1+(1000âˆ’1)d

a1000Â =Â a1+999d

For second A.P., we know,

anÂ =Â a+(nâˆ’1)d

Therefore,

a100Â =Â a2+(100âˆ’1)d

=Â a2+99d

a1000Â =Â a2+(1000âˆ’1)d

=Â a2+999d

Given that, difference between 100thÂ term of the two APs = 100

Therefore, (a1+99d) âˆ’ (a2+99d) = 100

a1âˆ’a2Â = 100â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦..Â (i)

Difference between 1000thÂ terms of the two APs

(a1+999d) âˆ’ (a2+999d) =Â a1âˆ’a2

From equationÂ (i),

This difference,Â a1âˆ’a2Â = 100

Hence, the difference between 1000thÂ terms of the two A.P. will be 100.

13. How many three digit numbers are divisible by 7?

Solution:

First three-digit number that is divisible by 7 are;

First number = 105

Second number = 105+7 = 112

Third number = 112+7 =119

Therefore, 105, 112, 119, â€¦

All are three digit numbers are divisible by 7 and thus, all these are terms of an A.P. having first term as 105 and common difference as 7.

As we know, the largest possible three-digit number is 999.

When we divide 999 by 7, the remainder will be 5.

Therefore, 999-5 = 994 is the maximum possible three-digit number that is divisible by 7.

Now the series is as follows.

105, 112, 119, â€¦, 994

Let 994 be theÂ nth term of this A.P.

first term, aÂ = 105

common difference, dÂ = 7

anÂ = 994

nÂ = ?

As we know,

anÂ =Â a+(nâˆ’1)d

994 = 105+(nâˆ’1)7

889 = (nâˆ’1)7

(nâˆ’1) = 127

nÂ = 128

Therefore, 128 three-digit numbers are divisible by 7.

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

Solution:

The first multiple of 4 that is greater than 10 is 12.

Next multiple will be 16.

Therefore, the series formed as;

12, 16, 20, 24, â€¦

All these are divisible by 4 and thus, all these are terms of an A.P. with first term as 12 and common difference as 4.

When we divide 250 by 4, the remainder will be 2. Therefore, 250 âˆ’ 2 = 248 is divisible by 4.

The series is as follows, now;

12, 16, 20, 24, â€¦, 248

Let 248 be theÂ nthÂ term of this A.P.

first term, aÂ = 12

common difference, dÂ = 4

anÂ =Â 248

As we know,

anÂ =Â a+(nâˆ’1)d

248 = 12+(n-1)Ã—4

236/4 =Â n-1

59 Â =Â n-1

nÂ = 60

Therefore, there are 60 multiples of 4 between 10 and 250.

15. For what value ofÂ n, are theÂ nthÂ terms of two APs 63, 65, 67, and 3, 10, 17, â€¦ equal?

Solution:

Given two APs as; 63, 65, 67,â€¦ and 3, 10, 17,â€¦.

Taking first AP,

63, 65, 67, â€¦

First term, aÂ = 63

Common difference, dÂ =Â a2âˆ’a1Â = 65âˆ’63 = 2

We know, nthÂ term of this A.P. =Â anÂ =Â a+(nâˆ’1)d

an= 63+(nâˆ’1)2 = 63+2nâˆ’2

anÂ = 61+2nÂ â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦.Â (i)

Taking second AP,

3, 10, 17, â€¦

First term, aÂ = 3

Common difference, dÂ =Â a2Â âˆ’Â a1Â = 10 âˆ’ 3 = 7

We know that,

nthÂ term of this A.P. = 3+(nâˆ’1)7

anÂ = 3+7nâˆ’7

anÂ = 7nâˆ’4 â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦..Â (ii)

Given,Â nthÂ term of these A.P.s are equal to each other.

Equating both these equations, we get,

61+2nÂ = 7nâˆ’4

61+4 = 5n

5nÂ = 65

nÂ = 13

Therefore, 13thÂ terms of both these A.P.s are equal to each other.

Â

16. Determine the A.P. whose third term is 16 and the 7thÂ term exceeds the 5thÂ term by 12.

Solutions:

Given,

Third term, a3Â = 16

As we know,

aÂ +(3âˆ’1)dÂ = 16

a+2dÂ = 16 â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦.Â (i)

It is given that, 7thÂ term exceeds the 5thÂ term by 12.

a7Â âˆ’Â a5Â = 12

[a+(7âˆ’1)d]âˆ’[aÂ +(5âˆ’1)d]= 12

(a+6d)âˆ’(a+4d) = 12

2dÂ = 12

dÂ = 6

From equationÂ (i), we get,

a+2(6) = 16

a+12 = 16

aÂ = 4

Therefore, A.P. will be4, 10, 16, 22, â€¦

17. Find the 20thÂ term from the last term of the A.P. 3, 8, 13, â€¦, 253.

Solution:

Given A.P. is3, 8, 13, â€¦, 253

Common difference, d= 5.

Therefore, we can write the given AP in reverse order as;

253, 248, 243, â€¦, 13, 8, 5

Now for the new AP,

first term, aÂ = 253

and common difference, dÂ = 248 âˆ’ 253 = âˆ’5

nÂ = 20

Therefore, using nth term formula, we get,

a20Â =Â a+(20âˆ’1)d

a20Â = 253+(19)(âˆ’5)

a20Â = 253âˆ’95

aÂ = 158

Therefore, 20thÂ term from the last term of the AP 3, 8, 13, â€¦, 253.is 158.

18. The sum of 4thÂ and 8thÂ terms of an A.P. is 24 and the sum of the 6thÂ and 10thÂ terms is 44. Find the first three terms of the A.P.

Solution:

We know that, the nth term of the AP is;

anÂ =Â a+(nâˆ’1)d

a4Â =Â a+(4âˆ’1)d

a4Â =Â a+3d

In the same way, we can write,

a8Â =Â a+7d

a6Â =Â a+5d

a10Â =Â a+9d

Given that,

a4+a8Â = 24

a+3d+a+7dÂ = 24

2a+10dÂ = 24

a+5dÂ = 12 â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦Â (i)

a6+a10Â = 44

aÂ +5d+a+9dÂ = 44

2a+14dÂ = 44

a+7dÂ = 22 â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦..Â (ii)

On subtracting equationÂ (i)Â fromÂ (ii), we get,

2dÂ = 22 âˆ’ 12

2dÂ = 10

dÂ = 5

From equationÂ (i), we get,

a+5dÂ = 12

a+5(5) = 12

a+25 = 12

aÂ = âˆ’13

a2Â =Â a+dÂ = âˆ’ 13+5 = âˆ’8

a3Â =Â a2+dÂ = âˆ’ 8+5 = âˆ’3

Therefore, the first three terms of this A.P. are âˆ’13, âˆ’8, and âˆ’3.

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:

It can be seen from the given question, that the incomes of Subba Rao increases every year by Rs.200 and hence, forms an AP.

Therefore, after 1995, the salaries of each year are;

5000, 5200, 5400, â€¦

Here,Â first term, aÂ = 5000

and common difference, dÂ = 200

Let afterÂ nthÂ year, his salary be Rs 7000.

Therefore, by the nth term formula of AP,

anÂ =Â a+(nâˆ’1)Â d

7000 = 5000+(nâˆ’1)200

200(nâˆ’1)= 2000

(nâˆ’1) = 10

nÂ = 11

Therefore, in 11th year, his salary will be Rs 7000.

20. Ramkali saved Rs 5 in the first week of a year and then increased her weekly saving by Rs 1.75. If in theÂ nthÂ week, her weekly savings become Rs 20.75, findÂ n.

Solution:

Given that, Ramkali saved Rs.5 in first week and then started saving each week by Rs.1.75.

Hence,

First term, aÂ = 5

and common difference, dÂ = 1.75

Also given,

anÂ = 20.75

Find, nÂ = ?

As we know, by the nth term formula,

anÂ =Â a+(nâˆ’1)d

Therefore,

20.75 = 5+(nÂ -1)Ã—1.75

15.75 = (nÂ -1)Ã—1.75

(nÂ -1) = 15.75/1.75 = 1575/175

= 63/7 = 9

nÂ -1 = 9

nÂ = 10

Hence,Â nÂ is 10.

## Exercise 5.3 Page: 112

1. Find the sum of the following APs.

(i) 2, 7, 12 ,â€¦., to 10 terms.
(ii) âˆ’ 37, âˆ’ 33, âˆ’ 29 ,â€¦, to 12 terms
(iii) 0.6, 1.7, 2.8 ,â€¦â€¦.., to 100 terms
(iv) 1/15, 1/12, 1/10, â€¦â€¦ , to 11 terms

Solutions:

(i) Given, 2, 7, 12 ,â€¦, to 10 terms

For this A.P.,

first term, aÂ = 2

And common difference, dÂ =Â a2Â âˆ’Â a1Â = 7âˆ’2 = 5

nÂ = 10

We know that, the formula for sum of nth term in AP series is,

SnÂ =Â n/2Â [2aÂ +(n-1)d]

S10Â = 10/2Â [2(2)+(10Â -1)Ã—5]

= 5[4+(9)Ã—(5)]

= 5 Ã— 49 = 245

(ii) Given, âˆ’37, âˆ’33, âˆ’29 ,â€¦, to 12 terms

For this A.P.,

first term, aÂ = âˆ’37

And common difference, dÂ =Â a2âˆ’Â a1

d= (âˆ’33)âˆ’(âˆ’37)

= âˆ’ 33 + 37 = 4

nÂ = 12

We know that, the formula for sum of nth term in AP series is,

SnÂ =Â n/2Â [2a+(n-1)d]

S12Â = 12/2Â [2(-37)+(12-1)Ã—4]

= 6[-74+11Ã—4]

= 6[-74+44]

= 6(-30) = -180

(iii) Given, 0.6, 1.7, 2.8 ,â€¦, to 100 terms

For this A.P.,

first term, aÂ = 0.6

Common difference, dÂ =Â a2Â âˆ’Â a1Â = 1.7 âˆ’ 0.6 = 1.1

nÂ = 100

We know that, the formula for sum of nth term in AP series is,

SnÂ =Â n/2[2aÂ +(n-1)d]

S12Â = 50/2Â [1.2+(99)Ã—1.1]

= 50[1.2+108.9]

= 50[110.1]

= 5505

(iv) Given, 1/15, 1/12, 1/10, â€¦â€¦ , to 11 terms

For this A.P.,

First term, a = 1/5

Common difference, d = a2 â€“a1 = (1/12)-(1/5) = 1/60

And number of terms n = 11

We know that, the formula for sum of nth term in AP series is,

SnÂ =Â n/2Â [2aÂ +Â (nÂ â€“ 1)Â d]

= 11/2(2/15 + 10/60)

= 11/2 (9/30)

= 33/20

2. Find the sums given below:

(ii) 34 + 32 + 30 + â€¦â€¦â€¦.. + 10
(iii) âˆ’ 5 + (âˆ’ 8) + (âˆ’ 11) + â€¦â€¦â€¦â€¦ + (âˆ’ 230)

Solutions:

(i)

First term, aÂ = 7

nth term, an = 84

Let 84 be theÂ nthÂ term of this A.P., then as per the nth term formula,

an =Â a(n-1)d

84 = 7+(nÂ â€“ 1)Ã—7/2

77 = (n-1)Ã—7/2

22 =Â nâˆ’1

nÂ = 23

We know that, sum of n term is;

SnÂ =Â n/2 (aÂ +Â l) , l = 84
Sn
Â = 23/2Â (7+84)

SnÂ  = (23Ã—91/2) = 2093/2

(ii) Given, 34 + 32 + 30 + â€¦â€¦â€¦.. + 10

For this A.P.,

first term, aÂ = 34

common difference, dÂ =Â a2âˆ’a1Â = 32âˆ’34 = âˆ’2

nth term, an= 10

Let 10 be theÂ nthÂ term of this A.P., therefore,

an=Â aÂ +(nâˆ’1)d

10 = 34+(nâˆ’1)(âˆ’2)

âˆ’24 = (nÂ âˆ’1)(âˆ’2)

12 =Â nÂ âˆ’1

nÂ = 13

We know that, sum of n terms is;

SnÂ =Â n/2 (aÂ +l) , l = 10

= 13/2 (34Â + 10)

= (13Ã—44/2) = 13 Ã— 22

= 286

(iii)Â Given, (âˆ’5) + (âˆ’8) + (âˆ’11) + â€¦â€¦â€¦â€¦ + (âˆ’230)

For this A.P.,

First term, aÂ = âˆ’5

nth term,Â an= âˆ’230

Common difference, dÂ =Â a2âˆ’a1Â = (âˆ’8)âˆ’(âˆ’5)

â‡’d = âˆ’ 8+5 = âˆ’3

Let âˆ’230 be theÂ nthÂ term of this A.P., and by the nth term formula we know,

an=Â a+(nâˆ’1)d

âˆ’230 = âˆ’ 5+(nâˆ’1)(âˆ’3)

âˆ’225 = (nâˆ’1)(âˆ’3)

(nâˆ’1) = 75

nÂ = 76

And, Sum of n term,

SnÂ =Â n/2 (aÂ +Â l)

= 76/2Â [(-5) +Â (-230)]

= 38(-235)

= -8930

3. In an AP
(i) GivenÂ aÂ = 5,Â dÂ = 3,Â anÂ = 50, findÂ nÂ andÂ Sn.
(ii) GivenÂ aÂ = 7,Â a13Â = 35, findÂ dÂ andÂ S13.
(iii) GivenÂ a12Â = 37,Â dÂ = 3, findÂ aÂ andÂ S12.
(iv) GivenÂ a3Â = 15,Â S10Â = 125, findÂ dÂ andÂ a10.
(v) GivenÂ dÂ = 5,Â S9Â = 75, findÂ aÂ andÂ a9.
(vi) GivenÂ aÂ = 2,Â dÂ = 8,Â SnÂ = 90, findÂ nÂ andÂ an.
(vii) GivenÂ aÂ = 8,Â anÂ = 62,Â SnÂ = 210, findÂ nÂ andÂ d.
(viii) GivenÂ anÂ = 4,Â dÂ = 2,Â SnÂ = âˆ’ 14, findÂ nÂ andÂ a.
(ix) GivenÂ aÂ = 3,Â nÂ = 8,Â SÂ = 192, findÂ d.
(x) GivenÂ lÂ = 28,Â SÂ = 144 and there are total 9 terms. FindÂ a.

Solutions:

(i) Given that,Â aÂ = 5,Â dÂ = 3,Â anÂ = 50

AsÂ we know, from the formula of the nth term in an AP,

anÂ =Â aÂ +(nÂ âˆ’1)d,

Therefore, putting the given values, we get,

â‡’ 50 = 5+(nÂ -1)Ã—3

â‡’ 3(nÂ -1) = 45

â‡’Â nÂ -1 = 15

â‡’Â nÂ = 16

Now, sum of n terms,

SnÂ =Â n/2 (aÂ +an)

SnÂ = 16/2 (5Â + 50) = 440

(ii)Â Given that,Â aÂ = 7,Â a13Â = 35

AsÂ we know, from the formula of the nth term in an AP,

anÂ =Â a+(nâˆ’1)d,

Therefore, putting the given values, we get,

â‡’ 35 = 7+(13-1)d

â‡’ 12dÂ = 28

â‡’Â dÂ = 28/12 = 2.33

Now,Â SnÂ =Â n/2 (a+an)

S13Â =Â 13/2 (7+35) = 273

(iii) Given that,Â a12Â = 37,Â dÂ = 3

AsÂ we know, from the formula of the nth term in an AP,

anÂ =Â a+(nÂ âˆ’1)d,

Therefore, putting the given values, we get,

â‡’Â a12Â =Â a+(12âˆ’1)3

â‡’ 37 =Â a+33

â‡’Â aÂ = 4

Now, sum of nth term,

SnÂ =Â n/2 (a+an)

SnÂ =Â 12/2 (4+37)

= 246

(iv) Given that,Â a3Â = 15,Â S10Â = 125

AsÂ we know, from the formula of the nth term in an AP,

anÂ =Â aÂ +(nâˆ’1)d,

Therefore, putting the given values, we get,

a3Â =Â a+(3âˆ’1)d

15 =Â a+2dÂ â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦..Â (i)

Sum of the nth term,

SnÂ =Â n/2Â [2a+(n-1)d]

S10Â =Â 10/2Â [2a+(10-1)d]

125 = 5(2a+9d)

25 = 2a+9dÂ â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦..Â (ii)

On multiplying equationÂ (i)Â byÂ (ii), we will get;

30 = 2a+4dÂ â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦.Â (iii)

By subtracting equationÂ (iii)Â fromÂ (ii), we get,

âˆ’5 = 5d

dÂ = âˆ’1

From equationÂ (i),

15 =Â a+2(âˆ’1)

15 =Â aâˆ’2

aÂ = 17 = First term

a10Â =Â a+(10âˆ’1)d

a10Â = 17+(9)(âˆ’1)

a10Â = 17âˆ’9 = 8

(v) Given that,Â dÂ = 5,Â S9Â = 75

As, sum of n terms in AP is,

SnÂ =Â n/2Â [2aÂ +(nÂ -1)d]

Therefore, the sum of first nine terms are;

S9Â = 9/2Â [2aÂ +(9-1)5]

25 = 3(a+20)

25 = 3a+60

3aÂ = 25âˆ’60

aÂ = -35/3

As we know, the nth term can be written as;

anÂ =Â a+(nâˆ’1)d

a9Â =Â a+(9âˆ’1)(5)

= -35/3+8(5)

= -35/3+40

= (35+120/3) = 85/3

(vi) Given that,Â aÂ = 2,Â dÂ = 8,Â SnÂ = 90

As, sum of n terms in an AP is,

SnÂ =Â n/2Â [2aÂ +(nÂ -1)d]

90 =Â n/2Â [2aÂ +(nÂ -1)d]

â‡’ 180 =Â n(4+8nÂ -8) =Â n(8n-4) = 8n2-4n

â‡’ 8n2-4n â€“180 = 0

â‡’ 2n2â€“n-45 = 0

â‡’ 2n2-10n+9n-45 = 0

â‡’ 2n(nÂ -5)+9(nÂ -5) = 0

â‡’ (n-5)(2n+9) = 0

So,Â nÂ = 5 (as n only be a positive integer)

âˆ´Â a5Â = 8+5Ã—4 = 34

(vii) Given that,Â aÂ = 8,Â anÂ = 62,Â SnÂ = 210

As, sum of n terms in an AP is,

SnÂ =Â n/2 (aÂ +Â an)

210 =Â n/2 (8 +62)

â‡’ 35nÂ = 210

â‡’Â nÂ = 210/35 = 6

Now, 62 = 8+5d

â‡’ 5dÂ = 62-8 = 54

â‡’Â dÂ = 54/5 = 10.8

(viii) Given that,Â nth term, anÂ = 4,Â common difference, dÂ = 2,Â sum of n terms, SnÂ = âˆ’14.

AsÂ we know, from the formula of the nth term in an AP,

anÂ =Â a+(nÂ âˆ’1)d,

Therefore, putting the given values, we get,

4 =Â a+(nÂ âˆ’1)2

4 =Â a+2nâˆ’2

a+2nÂ = 6

aÂ = 6 âˆ’ 2nÂ â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦.Â (i)

As we know, the sum of n terms is;

SnÂ =Â n/2 (a+an)

-14 =Â n/2 (a+4)

âˆ’28 =Â nÂ (a+4)

âˆ’28 =Â nÂ (6 âˆ’2nÂ +4) {From equationÂ (i)}

âˆ’28 =Â nÂ (âˆ’ 2nÂ +10)

âˆ’28 = âˆ’ 2n2+10n

2n2Â âˆ’10nÂ âˆ’ 28 = 0

n2Â âˆ’5nÂ âˆ’14 = 0

n2Â âˆ’7n+2nÂ âˆ’14 = 0

nÂ (nâˆ’7)+2(nÂ âˆ’7) = 0

(nÂ âˆ’7)(nÂ +2) = 0

EitherÂ nÂ âˆ’ 7 = 0 orÂ nÂ + 2 = 0

nÂ = 7 orÂ nÂ = âˆ’2

However,Â nÂ can neither be negative nor fractional.

Therefore,Â nÂ = 7

From equationÂ (i), we get

aÂ = 6âˆ’2n

aÂ = 6âˆ’2(7)

= 6âˆ’14

= âˆ’8

(ix) Given that,Â first term, aÂ = 3,

Number of terms,Â nÂ = 8

AndÂ sum of n terms, SÂ = 192

As we know,

SnÂ =Â n/2Â [2a+(nÂ -1)d]

192 = 8/2Â [2Ã—3+(8Â -1)d]

192 = 4[6 +7d]

48 = 6+7d

42 = 7d

d =Â 6

(x) Given that,Â lÂ = 28,SÂ = 144 and there are total of 9 terms.

Sum of n terms formula,

SnÂ =Â n/2 (aÂ +Â l)

144 = 9/2(a+28)

(16)Ã—(2) =Â a+28

32 =Â a+28

aÂ = 4

4. How many terms of the AP. 9, 17, 25 â€¦ must be taken to give a sum of 636?

Solutions:

Let there beÂ nÂ terms of the AP. 9, 17, 25 â€¦

For this A.P.,

First term, aÂ = 9

Common difference, dÂ =Â a2âˆ’a1Â = 17âˆ’9 = 8

As, the sum of n terms, is;

SnÂ =Â n/2Â [2a+(nÂ -1)d]

636 =Â n/2Â [2Ã—a+(8-1)Ã—8]

636 =Â n/2Â [18+(n-1)Ã—8]

636 =Â nÂ [9 +4nÂ âˆ’4]

636 =Â nÂ (4nÂ +5)

4n2Â +5nÂ âˆ’636 = 0

4n2Â +53nÂ âˆ’48nÂ âˆ’636 = 0

nÂ (4nÂ + 53)âˆ’12 (4nÂ + 53) = 0

(4nÂ +53)(nÂ âˆ’12) = 0

Either 4n+53 = 0 orÂ nâˆ’12 = 0

nÂ = (-53/4) orÂ nÂ = 12

nÂ cannot be negative or fraction, therefore,Â nÂ = 12 only.

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,

first term, aÂ = 5

last term, lÂ = 45

Sum of the AP, SnÂ = 400

As we know, the sum of AP formula is;

SnÂ =Â n/2 (a+l)

400 =Â n/2(5+45)

400 =Â n/2(50)

Number of terms, nÂ =16

As we know, the last term of AP series can be written as;

l = a+(nÂ âˆ’1)d

45 = 5 +(16 âˆ’1)d

40 = 15d

Common difference, dÂ = 40/15 = 8/3

6. The first and the last term 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,

First term, aÂ = 17

Last term, lÂ = 350

Common difference, dÂ = 9

Let there beÂ nÂ terms in the A.P., thus the formula for last term can be written as;

l = a+(nÂ âˆ’1)d

350 = 17+(nÂ âˆ’1)9

333 = (nâˆ’1)9

(nâˆ’1) = 37

nÂ = 38

SnÂ =Â n/2 (a+l)

S38Â = 38/2 (17+350)

= 19Ã—367

= 6973

Thus, this A.P. contains 38 terms and the sum of the terms of this A.P. is 6973.

7. Find the sum of first 22 terms of an AP in whichÂ dÂ = 7 and 22ndÂ term is 149.
Solution:

Given,

Common difference, dÂ = 7

22nd term, a22Â = 149

Sum of first 22 term, S22Â = ?

By the formula of nth term,

anÂ =Â a+(nâˆ’1)d

a22Â =Â a+(22âˆ’1)d

149 =Â a+21Ã—7

149 =Â a+147

aÂ = 2 = First term

Sum of n terms,

SnÂ =Â n/2(a+an)

S22 = 22/2 (2+149)

= 11Ã—151

= 1661

8. Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.
Solution:

Given that,

Second term, a2Â = 14

Third term, a3Â = 18

Common difference, dÂ =Â a3âˆ’a2Â = 18âˆ’14 = 4

a2Â =Â a+d

14 =Â a+4

aÂ = 10 = First term

Sum of n terms;

SnÂ =Â n/2Â [2aÂ + (nÂ â€“ 1)d]

S51Â = 51/2Â [2Ã—10Â (51-1) 4]

= 51/2 [20+(50)Ã—4]

= 51 Ã— 220/2

= 51 Ã— 110

= 5610

9. If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of firstÂ nÂ terms.

Solution:

Given that,

S7Â = 49

S17Â = 289

We know, Sum of n terms;

SnÂ =Â n/2Â [2aÂ + (nÂ â€“ 1)d]

Therefore,

S7=Â 7/2Â [2aÂ +(nÂ -1)d]

S7Â = 7/2Â [2aÂ + (7 -1)d]

49 = 7/2Â [2aÂ +6d]

7 = (a+3d)

aÂ + 3dÂ = 7 â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦.Â (i)

In the same way,

S17Â = 17/2Â [2a+(17-1)d]

289 = 17/2 (2aÂ +16d)

17 = (a+8d)

aÂ +8dÂ = 17 â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦.Â (ii)

Subtracting equationÂ (i)Â from equationÂ (ii),

5dÂ = 10

dÂ = 2

From equationÂ (i), we can write it as;

a+3(2) = 7

a+Â 6 = 7

a =Â 1

Hence,

SnÂ =Â n/2[2a+(n-1)d]

=Â n/2[2(1)+(nÂ â€“ 1)Ã—2]

=Â n/2(2+2n-2)

=Â n/2(2n)

=Â n2

10. Show thatÂ a1,Â a2Â â€¦ ,Â anÂ , â€¦ form an AP whereÂ anÂ is defined as below

(i)Â anÂ = 3+4n
(ii)Â anÂ = 9âˆ’5n
Also find the sum of the first 15 terms in each case.

Solutions:

(i)Â anÂ = 3+4n

a1Â = 3+4(1) = 7

a2Â = 3+4(2) = 3+8 = 11

a3Â = 3+4(3) = 3+12 = 15

a4Â = 3+4(4) = 3+16 = 19

We can see here, the common difference between the terms are;

a2Â âˆ’Â a1Â = 11âˆ’7 = 4

a3Â âˆ’Â a2Â = 15âˆ’11 = 4

a4Â âˆ’Â a3Â = 19âˆ’15 = 4

Hence,Â akÂ + 1 âˆ’ akÂ is the same value every time. Therefore, this is an AP with common difference as 4 and first term as 7.

Now, we know, the sum of nth term is;

SnÂ =Â n/2[2a+(nÂ -1)d]

S15Â = 15/2[2(7)+(15-1)Ã—4]

= 15/2[(14)+56]

= 15/2(70)

= 15Ã—35

= 525

(ii)Â anÂ = 9âˆ’5n

a1Â = 9âˆ’5Ã—1 = 9âˆ’5 = 4

a2Â = 9âˆ’5Ã—2 = 9âˆ’10 = âˆ’1

a3Â = 9âˆ’5Ã—3 = 9âˆ’15 = âˆ’6

a4Â = 9âˆ’5Ã—4 = 9âˆ’20 = âˆ’11

We can see here, the common difference between the terms are;

a2Â âˆ’Â a1Â = âˆ’1âˆ’4 = âˆ’5

a3Â âˆ’Â a2Â = âˆ’6âˆ’(âˆ’1) = âˆ’5

a4Â âˆ’Â a3Â = âˆ’11âˆ’(âˆ’6) = âˆ’5

Hence, akÂ + 1Â âˆ’Â akÂ is same every time. Therefore, this is an A.P. with common difference as âˆ’5 and first term as 4.

Now, we know, the sum of nth term is;

SnÂ =Â n/2Â [2aÂ +(n-1)d]

S15Â = 15/2[2(4)Â +(15Â -1)(-5)]

= 15/2[8 +14(-5)]

= 15/2(8-70)

= 15/2(-62)

= 15(-31)

= -465

11. If the sum of the firstÂ nÂ terms of an AP is 4nÂ âˆ’Â n2, what is the first term (that isÂ S1)? What is the sum of first two terms? What is the second term? Similarly find the 3rd, the10thÂ and theÂ nthÂ terms.

Solution:

Given that,

SnÂ = 4nâˆ’n2

First term,Â aÂ =Â S1Â = 4(1) âˆ’ (1)2Â = 4âˆ’1 = 3

Sum of first two terms =Â S2= 4(2)âˆ’(2)2Â = 8âˆ’4 = 4

Second term,Â a2Â =Â S2Â âˆ’Â S1Â = 4âˆ’3 = 1

Common difference, dÂ =Â a2âˆ’aÂ = 1âˆ’3 = âˆ’2

Nth term, anÂ =Â a+(nâˆ’1)dÂ

= 3+(nÂ âˆ’1)(âˆ’2)

= 3âˆ’2nÂ +2

= 5âˆ’2n

Therefore,Â a3Â = 5âˆ’2(3) = 5-6 = âˆ’1

a10Â = 5âˆ’2(10) = 5âˆ’20 = âˆ’15

Hence, the sum of first two terms is 4. The second term is 1.

The 3rd, the 10th, andÂ the nthÂ terms are âˆ’1, âˆ’15, and 5 âˆ’ 2nÂ respectively.

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

Solution:

The positive integers that are divisible by 6 are 6, 12, 18, 24 â€¦.

We can see here, that this series forms an A.P. whose first term is 6 and common difference is 6.

aÂ = 6

dÂ = 6

S40Â =Â ?

By the formula of sum of n terms, we know,

SnÂ =Â n/2Â [2aÂ +(nÂ â€“ 1)d]

Therefore, putting n = 40, we get,

S40Â = 40/2Â [2(6)+(40-1)6]

= 20[12+(39)(6)]

= 20(12+234)

= 20Ã—246

= 4920

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

Solution:

The multiples of 8 are 8, 16, 24, 32â€¦

The series is in the form of AP, having first term as 8 and common difference as 8.

Therefore,Â aÂ = 8

dÂ = 8

S15Â = ?

By the formula of sum of nth term, we know,

SnÂ =Â n/2Â [2a+(n-1)d]

S15Â = 15/2Â [2(8)Â + (15-1)8]

=Â 15/2[16 +(14)(8)]

=Â 15/2[16 +112]

= 15(128)/2

= 15 Ã— 64

= 960

14. Find the sum of the odd numbers between 0 and 50.

Solution:

The odd numbers between 0 and 50 are 1, 3, 5, 7, 9 â€¦ 49.

Therefore, we can see that these odd numbers are in the form of A.P.

Hence,

First term, aÂ = 1

Common difference, dÂ = 2

Last term, l = 49

By the formula of last term, we know,

lÂ =Â a+(nâˆ’1)Â d

49 = 1+(nâˆ’1)2

48 = 2(nÂ âˆ’ 1)

nÂ âˆ’ 1 = 24

nÂ = 25 = Number of terms

By the formula of sum of nth term, we know,

SnÂ =Â n/2(aÂ +l)

S25Â = 25/2 (1+49)

= 25(50)/2

=(25)(25)

= 625

15.Â 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, 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:

We can see, that the given penalties are in the form of A.P. having first term as 200 and common difference as 50.

Therefore, aÂ = 200 and dÂ = 50

Penalty that has to be paid if contractor has delayed the work by 30 days =Â S30

By the formula of sum of nth term, we know,

SnÂ =Â n/2[2a+(nÂ -1)d]

Therefore,

S30= 30/2[2(200)+(30 â€“ 1)50]

= 15[400+1450]

= 15(1850)

= 27750

Therefore, the contractor has to pay Rs 27750 as penalty.

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 the cost of 1stÂ prize beÂ Rs. P.

Cost of 2ndÂ prize =Â Rs. PÂ âˆ’ 20

And cost of 3rdÂ prize =Â Rs. PÂ âˆ’ 40

We can see that the cost of these prizes are in the form of A.P., having common difference as âˆ’20 and first term asÂ P.

Thus, aÂ =Â P and dÂ = âˆ’20

Given that,Â S7Â = 700

By the formula of sum of nth term, we know,

SnÂ =Â n/2Â [2aÂ + (nÂ â€“ 1)d]

7/2Â [2aÂ + (7 â€“ 1)d]Â = 700

aÂ + 3(âˆ’20) = 100

aÂ âˆ’60 = 100

aÂ = 160

Therefore, the value of each of the prizes was Rs 160, Rs 140, Rs 120, Rs 100, Rs 80, Rs 60, and Rs 40.

17. 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:

It can be observed that the number of trees planted by the students is in an AP.

1, 2, 3, 4, 5â€¦â€¦â€¦â€¦â€¦â€¦..12

First term,Â aÂ = 1

Common difference,Â dÂ = 2âˆ’1 = 1

SnÂ =Â n/2Â [2aÂ +(n-1)d]

S12Â = 12/2Â [2(1)+(12-1)(1)]

= 6(2+11)

= 6(13)

= 78

Therefore, number of trees planted by 1 section of the classes = 78

Number of trees planted by 3 sections of the classes = 3Ã—78 = 234

Therefore, 234 trees will be planted by the students.

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, 1.0 cm, 1.5 cm, 2.0 cm, â€¦â€¦â€¦ as shown in figure. What is the total length of such a spiral made up of thirteen consecutive semicircles? (TakeÂ Ï€ = 22/7)

Solution:

We know,

Perimeter of a semi-circle = Ï€r

Therefore,

P1Â = Ï€(0.5) = Ï€/2 cm

P2Â = Ï€(1) = Ï€ cm

P3Â = Ï€(1.5) = 3Ï€/2 cm

Where, P1,Â P2,Â P3Â are the lengths of the semi-circles.

Hence we got a series here, as,

Ï€/2, Ï€, 3Ï€/2, 2Ï€, â€¦.

P1Â = Ï€/2 cm

P2Â = Ï€ cm

Common difference, dÂ =Â P2 â€“ P1 =Â Ï€ â€“ Ï€/2 = Ï€/2

First term =Â P1=Â aÂ =Â Ï€/2 cm

By the sum of n term formula, we know,

SnÂ =Â n/2Â [2aÂ + (nÂ â€“ 1)d]

Therefor, Sum of the length of 13 consecutive circles is;

S13Â =Â 13/2Â [2(Ï€/2)Â + (13Â â€“ 1)Ï€/2]

= Â 13/2Â [Ï€Â + 6Ï€]

=13/2Â (7Ï€)

=Â 13/2 Ã— 7 Ã—Â 22/7

= 143 cm

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. In how many rows are the 200 logs placed and how many logs are in the top row?

Solution:Â

We can see that the numbers of logs in rows are in the form of an A.P.20, 19, 18â€¦

For the given A.P.,

First term, aÂ = 20 and common difference, dÂ =Â a2âˆ’a1Â = 19âˆ’20 = âˆ’1

Let a total of 200 logs be placed inÂ nÂ rows.

Thus, SnÂ = 200

By the sum of nth term formula,

SnÂ =Â n/2Â [2aÂ +(nÂ -1)d]

S12Â = 12/2Â [2(20)+(nÂ -1)(-1)]

400 =Â nÂ (40âˆ’n+1)

400 =Â nÂ (41-n)

400 = 41nâˆ’n2

n2âˆ’41nÂ + 400 = 0

n2âˆ’16nâˆ’25n+400 = 0

n(nÂ âˆ’16)âˆ’25(nÂ âˆ’16) = 0

(nÂ âˆ’16)(nÂ âˆ’25) = 0

Either (nÂ âˆ’16) = 0 orÂ nâˆ’25 = 0

nÂ = 16 orÂ nÂ = 25

By the nth term formula,

anÂ =Â a+(nâˆ’1)d

a16Â = 20+(16âˆ’1)(âˆ’1)

a16Â = 20âˆ’15

a16Â = 5

Similarly, the 25th term could be written as;

a25Â = 20+(25âˆ’1)(âˆ’1)

a25Â = 20âˆ’24

= âˆ’4

It can be seen, the number of logs in 16thÂ row is 5 as the numbers cannot be negative.

Therefore, 200 logs can be placed in 16 rows and the number of logs in the 16thÂ row is 5.

Â

20. In a potato race, a bucket is placed at the starting point, which is 5 m from the first potato and 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 she 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 metres) run by a competitor is 2Ã—5+2Ã—(5+3)]

Solution:

The distances of potatoes from the bucket are 5, 8, 11, 14â€¦, which is in the form of AP.

Given, the distance run by the competitor for collecting these potatoes are two times of the distance at which the potatoes have been kept.

Therefore, distances to be run w.r.t distances of potatoes, could be written as;

10, 16, 22, 28, 34,â€¦â€¦â€¦.

Hence, the first term, aÂ = 10 and dÂ = 16âˆ’10 = 6

S10Â =?

By the formula of sum of n terms, we know,

S10Â = 10/2Â [2(10)+(10Â -1)(6)]

= 5[20+54]

= 5(74)

= 370

Therefore, the competitor will run a total distance of 370 m.

## Exercise 5.4 Page: 115

1. Which term of the AP: 121, 117, 113, . . ., is its first negative term? [Hint: Find n for an < 0]

Solution:

Given the AP series is 121, 117, 113, . . .,

Thus, first term, a = 121

Common difference, d = 117-121= -4

By the nth term formula,

anÂ =Â a+(nÂ âˆ’1)d

Therefore,

anÂ =Â 121+(nâˆ’1)(-4)

= 121-4n+4

=125-4n

To find the first negative term of the series, an < 0

Therefore,

125-4n < 0

125 < 4n

n>125/4

n>31.25

Therefore, the first negative term of the series is 32nd term.

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

Solution:

From the given statements, we can write,

a3 + a7 = 6 â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦.(i)

And

a3 Ã—a7 = 8 â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦..(ii)

By the nth term formula,

anÂ =Â a+(nâˆ’1)d

Third term, a3 = a+(3 -1)d

a3 = a + 2dâ€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦(iii)

And Seventh term, a7= a+(7-1)d

a7 = a + 6d â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦..(iv)

From equation (iii) and (iv), putting in equation(i), we get,

a+2d +a+6d = 6

2a+8d = 6

a+4d=3

or

a = 3â€“4d â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦(v)

Again putting the eq.(iii) and (iv), in eq. (ii), we get,

(a+2d)Ã—(a+6d) = 8

Putting the value of a from equation (v), we get,

(3â€“4d +2d)Ã—(3â€“4d+6d) = 8

(3 â€“2d)Ã—(3+2d) = 8

32 â€“ 2d2 = 8

9 â€“ 4d2 = 8

4d2 = 1

d = 1/2 or -1/2

Now, by putting both the values of d, we get,

a = 3 â€“ 4d = 3 â€“ 4(1/2) = 3 â€“ 2 = 1, when d = 1/2

a = 3 â€“ 4d = 3 â€“ 4(-1/2) = 3+2 = 5, when d = -1/2

We know, the sum of nth term of AP is;

SnÂ =Â n/2Â [2aÂ +(nÂ â€“ 1)d]

So, when a = 1 and d=1/2

Then, the sum of first 16 terms are;

S16Â =Â 16/2Â [2Â +(16-1)1/2] = 8(2+15/2) = 76

And when a = 5 and d= -1/2

Then, the sum of first 16 terms are;

S16Â =Â 16/2Â [2(5)+(16-1)(-1/2)] = 8(5/2)=20

3. A ladder has rungs 25 cm apart. (see Fig. 5.7). The rungs decrease uniformly in length from 45 cm at the bottom to 25 cm at the top. If the top and the bottom rungs are
apart, what is the length of the wood required for the rungs? [Hint: Number of rungs = -250/25 ].

Solution:

Given,

Distance between the rungs of the ladder is 25cm.

Distance between the top rung and bottom rung of the ladder is =
= 5/2 Ã—100cm

= 250cm

Therefore, total number of rungs = 250/25 + 1 = 11

As we can see from the figure, the ladder has rungs in decreasing order from top to bottom. Thus, we can conclude now, that the rungs are decreasing in an order of AP.

And the length of the wood required for the rungs will be equal to the sum of the terms of AP series formed.

So,

First term, a = 45

Last term, l = 25

Number of terms, n = 11

Now, as we know, sum of nth terms is equal to,

Sn= n/2(a+ l)

Sn= 11/2(45+25) = 11/2(70) = 385 cm

Hence, the length of the wood required for the rungs is 385cm.

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. [Hint :Sx â€“ 1 = S49 â€“ Sx ]

Solution:

Given,

Row houses are numbers from 1,2,3,4,5â€¦â€¦.49.

Thus we can see the houses numbered in a row are in the form of AP.

So,

First term, a = 1

Common difference, d=1

Let us say the number of xth houses can be represented as;

Sum of nth term of AP = n/2[2a+(n-1)d]

Sum of number of houses beyond x house = Sx-1

= (x-1)/2[2(1)+(x-1-1)1]

= (x-1)/2 [2+x-2]

= x(x-1)/2 â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦(i)

By the given condition, we can write,

S49 â€“ Sx = {49/2[2(1)+(49-1)1]}â€“{x/2[2(1)+(x-1)1]}

= 25(49) â€“ x(x + 1)/2 â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦.(ii)

As per the given condition, eq.(i) and eq(ii) are equal to each other;

Therefore,

x(x-1)/2 = 25(49) â€“ x(x+1)/2

x = Â±35

As we know, the number of houses cannot be a negative number. Hence, the value of x is 35.

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 1 4 m and a tread of 1 2 m. (see Fig. 5.8). Calculate the total volume of concrete required to build the terrace. [Hint : Volume of concrete required to build the first step = Â¼ Ã—1/2 Ã—50 m3.]

Solution:

As we can see from the given figure, the first step is Â½ m wide, 2nd step is 1m wide and 3rd step is 3/2m wide. Thus we can understand that the width of step by Â½ m each time when height is Â¼ m. And also, given length of the steps is 50m all the time. So, the width of steps forms a series AP in such a way that;

Â½ , 1, 3/2, 2, â€¦â€¦..

Volume of steps = Volume of Cuboid

Now,

Volume of concrete required to build the first step = Â¼ Ã—1/2 Ã—50 = 25/4

Volume of concrete required to build the second step =Â¼ Ã—1Ã—50 = 25/2

Volume of concrete required to build the second step = Â¼ Ã—3/2 Ã—50 = 75/4

Now, we can see the volumes of concrete required to build the steps, are in AP series;

25/4 , 25/2 , 75/4 â€¦..

Thus, applying the AP series concept,

First term, a = 25/4

Common difference, d = 25/2 â€“ 25/4 = 25/4

As we know, the sum of n terms is;

Sn = n/2[2a+(n-1)d] = 15/2(2Ã—(25/4 )+(15/2 -1)25/4)

Upon solving, we get,

Sn = 15/2 (100)

Sn=750

Hence, the total volume of concrete required to build the terrace is 750 m3.

This chapter comes under unit 3 algebra and this unit has 20 marks allotted in the examination. Students can expect an average of 3 questions from arithmetic progressions. Along with Class 10 examinations, this topic is very important from the point of competitive exams.

### Sub-topics of Class 10 Chapter 5 Arithmetic Progression

5.1 Introduction

In this chapter, we shall discuss patterns which we come across in our day-to-day life in which succeeding terms are obtained by adding a fixed number to the preceding terms. We shall also see how to find their nth terms and the sum of n consecutive terms, and use this knowledge in solving some daily life problems.

5.2 Arithmetic Progressions

The topic describes Arithmetic Progressions, its definition and relatable terms along with fine examples. You will also learn about Finite Arithmetic Progressions and Infinite Arithmetic Progressions. The general form of AP is a, a + d, a + 2d, a + 3d,â€¦

5.3 nth Term of an AP
This topic discusses various methods to determine the nth term of an AP. The concepts are explained with different types of problems solving techniques and finding the nth term of an AP. The examples mentioned in the chapter will help you while solving the exercise problems.

5.4 Sum of First n Terms of an AP
The topics discuss different techniques to find the sum of the first n terms of an AP. It also provides suitable examples which show different techniques to find the sum of the first n terms of AP.

5.5 Summary
It gives an overview of the entire chapter and the important topics explained in the entire chapter. By going through the summary part you can cover the entire chapter in a few points which help in memorizing the essential concepts.

### List of Exercise from Class 10 Maths Chapter 5 Arithmetic progression

Exercise 5.1â€“ 4 questions 1 MCQ and 3 descriptive type questions
Exercise 5.2â€“ 20 questions, 1 fill in the blanks, 2 MCQâ€™s, 7 Short answer questions and 10 Long answer questions
Exercise 5.3â€“ 20 Questions 3 fill in the blanks, 4 daily life examples, and 13 descriptive type questions
Exercise 5.4 5 Questions- 5 Long answer questions

This NCERT Solutions for Class 10 Maths is a perfect study material that will help you solve different kinds of problems. Solving these NCERT Solutions will help you understand the topic completely and help you lay a greater foundation for future studies.

In this chapter, students will discuss patterns in succeeding terms obtained by adding a fixed number to the preceding terms. They also, see how to find nth terms and the sum of n consecutive terms. Students will learn arithmetic progression effectively when they solve daily life problems.

This chapter has Arithmetic Progression Derivation of the nthÂ term and sum of the first n terms of an A.P. and their application in solving daily life problems. This is one of the important chapters from the point of the Class 10 examination. An arithmetic progression is a very basic and important topic to study as almost all the competitive exams will ask questions on arithmetic progression.

### Key Features of NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic progressions

• Has answers to different types of questions such as MCQs and long answer questions.
• Solving this NCERT Solutions will make you get well versed with important formulas.
• Act as a basis to solve arithmetic progression problems asked in competitive examination.
• Has answers to all the exercise questions provided in the NCERT textbook
• Provide you with the necessary practise of solving questions
• You can solve different types of questions with varying difficulties.
• Different examples taken from day to day life will help you understand the topic thoroughly.

Keep visiting BYJUâ€™S to get complete assistance for CBSE Class 10Â first and second term exams. At BYJUâ€™S, students can get several sample papers, question papers, notes, textbooks, videos, animations and effective preparation tips which can help you to score well in the class 10 CBSE Term II exams.

For more information on the study resources we provide, register with BYJUâ€™S website or download BYJUâ€™S Learning App for a customised learning experience.

Students can also make use of the other solutions of CBSE syllabus to understand the types of questions that would arise in the CBSE Term II exams.

 Also AccessÂ NCERT Exemplar for Class 10 Maths Chapter 5 Arithmetic ProgressionsÂ CBSE Notes for Class 10 Maths Chapter 5 Arithmetic ProgressionsÂ

## Frequently Asked Questions on NCERT Solutions for Class 10 Maths Chapter 5

### What kind of questions are there in NCERT Solutions for Class 10 Maths Chapter 5?

Chapter 5 of NCERT Solutions for Class 10 Maths has multiple choice questions, descriptive type of questions, long answer type questions, short answer type questions, fill in the blanks and daily life examples. By the end of this chapter, students can increase their problem-solving skills and time management skills. This helps in procuring high marks in their CBSE Term II exams.

### Is it necessary to learn all the topics provided in NCERT Solutions for Class 10 Maths Chapter 5?

Yes, it is compulsory to learn all the topics provided in NCERT Solutions for Class 10 Maths Chapter 5 to score high marks in Class 10 second term exams. These solutions are designed by subject matter experts who have assembled model questions covering all the exercise questions from the textbook. They also focus on cracking the Solutions of Maths in such a way that it is easy for the students to understand.

### Are NCERT Solutions for Class 10 Maths Chapter 5 enough to attend all the questions that come in the CBSE term II exams?

Yes, it is enough to attend all the questions that come in the CBSE Term II exams by utilizing the NCERT Solutions for Class 10 Maths Chapter 5. Practising this chapter can make them learn the concepts flawlessly. These questions have been devised, as per the latest term-wise CBSE syllabus and the guidelines. This makes the students score good marks in the CBSE final second term examination.