ncert solutions class 10 maths chapter 5 A.P.

Chapter 5: Arithmetic Progression

NCERT Solutions for Class 10 Maths Chapter 5 A.P. is represented here the solution of Arithmetic Progression of class 10 Maths. It is very important for the students because every question solve in simple language to understand the student. This chapter is solved by the the expert of maths.

Exercise 5.1

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: According to the question

Taxi fare for 1 km =  Rs 15

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

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

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

And so on……

Thus, 15, 23, 31, 39 … is forms an A.P. because every preceding term is 8 less than the next term .

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

Solution:  Let the volume of air in a cylinder be V litres.

In each stroke, the vacuum pump removes 1/4th of air remaining in the cylinder at a time.

Remains air in after first stroke = (1-1/4) V = 3/4 V

Remains air in after second stroke = (3/4 V)²

Remains air in after third stroke =  (3/4 V)³

Therefore, volumes will be V, 3V/4 , (3V/4)² , (3V/4)^{³   .   .   .} 

Clearly, we can see here, this series do not have same 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: According to the question

Cost of digging a well for first metre = 150

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

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

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

And so on..

Clearly, 150, 200, 250, 300 … 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. P is deposited at r% compound interest per annum for n years,

the amount of money will be  =  P(1+r/100)ⁿ

The amount of money after 1 year = 10000(1+8/100) = Rs 10800

The amount of money after 2 year = 10000(1+8/100)²                                                               

=  Rs 11,664

The amount of money after 3 year =  10000(1+8/100)³

= Rs 12,597.12

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

10,000, 10,800, 11,664, 12,597.12  .   .    .  . .

Clearly, the terms of this series do not have the common difference between them.

Therefore, this is 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) a = 10, d = 10
(ii) a = -2, d = 0
(iii) a = 4, d = – 3
(iv) a = -1 d = 1/2
(v) a = – 1.25, d = – 0.25

Solutions: (i) a = 10, d = 10

Let, the Arithmetic Progression series be a₁, a₂, a₃, a₄, a₅ …

a₁ = a = 10

a₂ = a+d = 10+10 = 20

a₃ = a + 2d = 10+ 2×10 = 30

a₄ = a + 3d = 10 + 3×10 = 40

a₅ = a + 4d = 10 + 4×10 = 50

And so on…

Therefore, the A.P. will be 10, 20, 30, 40, 50 …

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

(ii) a = – 2, d = 0

Let , the A.P. series be a₁, a₂, a₃, a₄, a₅ …

a₁ = a = -2

a₂ = a+ d = – 2+0 = – 2

a₃ = a + 2d = – 2+0 = – 2

a₄ = a + 3d = – 2+0 = – 2

Therefore, the A.P. series will be – 2, – 2, – 2, – 2 …

And, First four terms of this A.P. will be – 2, – 2, – 2 and – 2.

(iii) a = 4, d = – 3

Let , the A.P. series be a₁, a₂, a₃, a₄, a₅ …

a₁ = a = 4

a₂ = a₁ + d = 4 – 3 = 1

a₃ = a₂+d = 1 – 3 = – 2

a₄ = a₃+d = -2  – 3 = – 5

Therefore, the A.P. series will be 4, 1, – 2 – 5 …

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

(iv) a = – 1, d = 1/2

Let, the A.P. series be a₁, a₂, a₃, a₄, a₅ …

a₂ = a₁+d = -1+1/2 = -1/2

a₃ = a₂+d = -1/2+1/2 = 0

a₄ = a₃+d = 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) a = – 1.25, d = – 0.25

Let, the A.P. series be a₁, a₂, a₃, a₄, a₅ …

a₁ = a = – 1.25

a₂ = a₁ + d = – 1.25-0.25 = – 1.50

a₃ = a₂ + d = – 1.50-0.25 = – 1.75

a₄ = a₃ + d = – 1.75-0.25 = – 2.00

Therefore, the A.P series will be  1.25, – 1.50, – 1.75, – 2.00 ……..

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

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

Solutions (i) Given series,

3, 1, – 1, – 3 …

a₁  = 3, a₂ = 1

Common difference, d = a₂  –  a₁

⇒  1 – 3 = -2

⇒  d = -2

(ii) Given series, – 5, – 1, 3, 7 …

a₁  = -5,  a₂  =  -1

Common difference, d = a₂  –  a₁

⇒ ( – 1)-( – 5) = – 1+5 = 4

(iii) Given series, 1/3, 5/3, 9/3, 13/3 ….

a₁  = 1/3, a₂ = 5/3

Common difference, d = a₂  –  a₁

⇒ 5/3 – 1/3 = 4/3

(iv) Given series, 0.6, 1.7, 2.8, 3.9 …

a₁  = 0.6, a₂ = 1.7

Common difference, d = a₂  –  a₁

⇒ 1.7 – 0.6

⇒ 1.1

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

(i) 2, 4, 8, 16 …
(ii) 2, 5/2, 3, 7/2 ….
(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) -1/2, -1/2, -1/2, -1/2 ….
(ix) 1, 3, 9, 27 …
(x) a, 2a, 3a, 4a …
(xi) a, a², a³, a⁴ …
(xii) √2, √8, √18, √32 …
(xiii) √3, √6, √9, √12 …
(xiv) 1², 3², 5², 7² …
(xv) 1², 5², 7², 7³ …

Solution (i) Given,

2, 4, 8, 16 …

Here, the common difference is;

a₂ – a₁ = 4 – 2 = 2

a₃ – a₂ = 8 – 4 = 4

a₄ – a₃ = 16 – 8 = 8

Since, the common difference is not the same every time.

Therefore, the given series not an A.P.

(ii) Given, 2, 5/2, 3, 7/2 ….

Here,

a₂ – a₁ = 5/2-2 = 1/2

a₃ – a₂ = 3-5/2 = 1/2

a₄ – a₃ = 7/2-3 = 1/2

Since, the common difference is same every time.

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

The next three terms are;

a₅ = 7/2 +1/2 = 4

a₆ = 4 +1/2 = 9/2

a₇ = 9/2 +1/2 = 5

(iii) Given, -1.2, – 3.2, -5.2, -7.2 …

Here,

a₂ – a₁ = (-3.2) – (-1.2) = -2

a₃ – a₂ = (-5.2) – (-3.2) = -2

a₄ – a₃ = (-7.2) – (-5.2) = -2

Since,  common difference is same every time.

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

Hence, next three terms are;

a₅ = – 7.2 – 2 = -9.2

a₆ = – 9.2 – 2 = – 11.2

a₇ = – 11.2 – 2 = – 13.2

(iv) Given, -10, – 6, – 2, 2 …

Here, the terms and their difference are;

a₂ – a₁ = (-6) – (-10) = 4

a₃ – a₂ = (-2) – (-6) = 4

a₄ – a₃ = (2)  – (-2) = 4

Since, the common difference is same every time.

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

Hence, next three terms are;

a₅ = 2+4 = 6

a₆ = 6+4 = 10

a₇ = 10+4 = 14

(v) Given, 3, 3+√2, 3+2√2, 3+3√2

Here,

a₂ – a₁ = 3+√2-3 = √2

a₃ – a₂ = (3+2√2)-(3+√2) = √2

a₄ – a₃ = (3+3√2) – (3+2√2) = √2

Since, the common difference is same every time.

Therefore, d = √2 and the given series forms a A.P.

Hence, next three terms are;

a₅ = (3+√2) +√2 = 3+4√2

a₆ = (3+4√2)+√2 = 3+5√2

a₇ = (3+5√2)+√2 = 3+6√2

(vi) 0.2, 0.22, 0.222, 0.2222 ….

Here,

a₂ – a₁ = 0.22  – 0.2 = 0.02

a₃ – a₂ = 0.222 – 0.22 = 0.002

a₄ – a₃ = 0.2222 – 0.222 = 0.0002

Since, the common difference is not same every time.

Therefore, and the given series doesn’t forms a A.P.

(vii) 0, -4, -8, -12 …

Here,

a₂ – a₁ = (-4)-0 = -4

a₃ – a₂ = (-8)-(-4) = -4

a₄ – a₃ = (-12)-(-8) = -4

Since, the common difference is same every time.

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

Hence, next three terms are;

a₅ = -12-4 = -16

a₆ = -16-4 = -20

a₇ = -20-4 = -24

(viii) -1/2, -1/2, -1/2, -1/2 ….

Here,

a₂ – a₁ = (-1/2) – (-1/2) = 0

a₃ – a₂ = (-1/2) – (-1/2) = 0

a₄ – a₃ = (-1/2) – (-1/2) = 0

Since, the common difference is same every time.

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

Hence, next three terms are;

a₅ = (-1/2)-0 = -1/2

a₆ = (-1/2)-0 = -1/2

a₇ = (-1/2)-0 = -1/2

(ix) 1, 3, 9, 27 …

Here,

a₂ – a₁ = 3-1 = 2

a₃ – a₂ = 9-3 = 6

a₄ – a₃ = 27-9 = 18

Since,the common difference is not same every time.

Therefore, and the given series doesn’t form a A.P.

(x) a, 2a, 3a, 4a …

Here,

a₂ – a₁ = 2a–a = a

a₃ – a₂ = 3a-2a = a

a₄ – a₃ = 4a-3a = a

Since,  the common difference is same every time.

Therefore, d = a and the given series forms a A.P.

Hence, next three terms are;

a₅ = 4a+a = 5a

a₆ = 5a+a = 6a

a₇ = 6a+a = 7a

(xi) a, a², a³, a⁴ …

Here,

a₂ – a₁ = a²–a = a(a-1)

a₃ – a₂ = a³ – a² = a²(a-1)

a₄ – a₃ = a⁴ – a³ = a³(a-1)

Since,the common difference is not same every time.

Therefore, the given series doesn’t forms a A.P.

(xii) √2, √8, √18, √32 …

Here,

a₂ – a₁ = √8-√2  = 2√2-√2 = √2

a₃ – a₂ = √18-√8 = 3√2-2√2 = √2

a₄ – a₃ = 4√2-3√2 = √2

Since, the common difference is same every time.

Therefore, d = √2 and the given series forms a A.P.

Hence, next three terms are;

a₅ = √32+√2 = 4√2+√2 = 5√2 = √50

a₆ = 5√2+√2 = 6√2 = √72

a₇ = 6√2+√2 = 7√2 = √98

(xiii) √3, √6, √9, √12 …

Here,

a₂ – a₁ = √6-√3 = √3×√2-√3 = √3(√2-1)

a₃ – a₂ = √9-√6 = 3-√6 = √3(√3-√2)

a₄ – a₃ = √12 – √9 = 2√3 – √3×√3 = √3(2-√3)

Since, the common difference is not same every time.

Therefore, the given series doesn’t form a A.P.

(xiv) 1², 3², 5², 7² …

Or, 1, 9, 25, 49 …..

Here,

a₂ − a₁ = 9−1 = 8

a₃ − a₂ = 25−9 = 16

a₄ − a₃ = 49−25 = 24

Since, the common difference is not same every time.

Therefore, the given series doesn’t form a A.P.

(xv) 1², 5², 7², 73 …

Or 1, 25, 49, 73 …

Here,

a₂ − a₁ = 25−1 = 24

a₃ − a₂ = 49−25 = 24

a₄ − a₃ = 73−49 = 24

Since, the common difference is same every time.

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

Hence, next three terms are;

a₅ = 73+24 = 97

a₆ = 97+24 = 121

a₇ = 121+24 = 145

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Exercise 5.2

1. Fill in the blanks in the following table, given that a is the first term, d the common difference and a_n the n^th term of the A.P.

 

Solutions:

(i) Given, First term, a = 7

Common difference, d = 3

Number of terms, n = 8,

We have to find the nth term, a_n = ?

From formula

a_n = a+(n−1)d

⇒  7+(8 −1) 3

⇒  7+(7) 3

⇒   7+21 = 28

Hence, a_n = 28

(ii) Given, First term, a = -18

Common difference, d = ?

Number of terms, n = 10

nth term, a_n = 0

We know that

a_n = a+(n−1)d

0 = − 18 +(10−1)d

⇒  18 = 9d

d = 18/9 = 2

Hence, common difference, d = 2

(iii) Given, First term, a = ?

Common difference, d = -3

Number of terms, n = 18

Nth term, a_n = -5

We know that

a_n = a+(n−1)d

⇒   −5 = a+(18−1) (−3)

⇒   −5 = a+(17) (−3)

⇒   −5 = a−51

a = 51−5 = 46

Hence, a = 46

(iv) Given, First term, a = -18.9

Common difference, d = 2.5

Number of terms, n = ?

nth term, a_n = 3.6

We know that

a_n = a +(n −1)d

3.6 = − 18.9+(n −1)2.5

⇒ 3.6 + 18.9 = (n−1)2.5

⇒ 22.5 = (n−1)2.5

⇒ (n – 1) = 22.5/2.5

⇒ n – 1 = 9

⇒ n = 10

Hence, n = 10

(v) Given, First term, a = 3.5

Common difference, d = 0

Number of terms, n = 105

nth term, a_n = ?

We know that,

a_n = a+(n −1)d

⇒ a_n = 3.5+(105−1) 0

⇒ a_n = 3.5+104×0

⇒ a_n = 3.5

Hence, a_n = 3.5

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

(i) 30^th term of the A.P: 10,7, 4, …, is

(A) 97 (B) 77 (C) −77 (D) −87

(ii) 11^th term of the A.P. -3, -1/2, ,2 …. is

(A) 28   (B) 22    (C) – 38   (D) -48(½)

Solutions:  (i) Given here,

A.P. = 10, 7, 4, …

Therefore, we can find,

First term, a = 10

Common difference, d = a₂ − a₁ = 7−10 = −3

We know that,

a_n = a +(n−1)d

⇒  a₃₀ = 10+(30−1)(−3)

⇒  a₃₀ = 10+(29)(−3)

⇒   a₃₀ = 10−87 = −77

Hence, the correct answer is option C.

(ii) Given here,

A.P. = -3, -1/2, ,2 …

Therefore, we can find,

First term a = – 3

Common difference, d = a₂ − a₁ = (-1/2) -(-3)

⇒(-1/2) + 3 = 5/2

We know that,

a_n = a+(n−1)d

⇒ a₁₁ = -3+(11-1)(5/2)

⇒  a₁₁ = -3+(10)(5/2)

⇒ a₁₁ = -3+25

a₁₁ = 22

Hence, the answer is option B.

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

Solutions: (i) The first and third term are;

a = 2,  a₃ = 26

We know that,

a_n = a+(n −1)d

⇒ a₃ = 2+(3-1)d

⇒ 26 = 2+2d

⇒ 24 = 2d

⇒ d = 12

Hence, a₂ = 2+(2-1)12

= 14

Therefore, 14 is the missing term.

(ii) For the given A.P., , 13, ,3

a₂ = 13,  a₄ = 3

We know that,

a_n = a+(n−1) d

a₂ = a +(2-1)d

13 = a+d ………………. (i)

a₄ = a+(4-1)d

3 = a+3d ………….. (ii)

On subtracting equation (i) from (ii), we get,

– 10 = 2d

d = – 5

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

13 = a+(-5)

a = 18

a₃ = 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.,

a = 5, a₄ = 19/2

We know that,

a_n = a+(n−1)d

a₄ = a+(4-1)d

⇒ 19/2 = 5+3d

⇒ (19/2) – 5 = 3d

⇒ 3d = 9/2

⇒d = 3/2

a₂ = a+(2-1)d

⇒ a₂ = 5+3/2

⇒ a₂ = 13/2

a₃ = a+(3-1)d

⇒ a₃ = 5+2×3/2

a₃ = 8

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

(iv) For the given A.P.,

a = −4, a₆ = 6

We know that,

a_n = a +(n−1) d

Therefore, putting the values here,

a₆ = a+(6−1)d

⇒ 6 = − 4+5d

⇒ 10 = 5d

⇒ d = 2

a₂ = a+d = − 4+2 = −2

a₃ = a+2d = − 4+2(2) = 0

a₄ = a+3d = − 4+ 3(2) = 2

a₅ = a+4d = − 4+4(2) = 4

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

(v) For the given A.P.,

a₂ = 38, a₆ = −22

We know that,

a_n = a+(n −1)d

Therefore, putting the values here,

a₂ = a+(2−1)d

38 = a+d ……………………. (i)

a₆ = a+(6−1)d

−22 = a+5d …………………. (ii)

On subtracting equation (i) from (ii), we get

− 22 − 38 = 4d

⇒ −60 = 4d

⇒ d = −15

a = a₂ − d = 38 − (−15) = 53

a₃ = a + 2d = 53 + 2 (−15) = 23

a₄ = a + 3d = 53 + 3 (−15) = 8

a₅ = a + 4d = 53 + 4 (−15) = −7

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

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

Solutions: Given the A.P. series as 3, 8, 13, 18, …

First term, a = 3

Common difference, d = a₂ − a₁ = 8 − 3 = 5

Let the n^th term of given A.P. be 78. we know,

a_n = a+(n−1)d

⇒  78 = 3+(n −1)5

⇒ 75 = (n−1)5

⇒ (n−1) = 15

n = 16

Hence, 16^th term of this A.P. is 78.

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

(i) 7, 13, 19, …, 205

(ii) 18, 15½, 13 . . . 47

Solutions:

(i) Given, 7, 13, 19, …, 205 is the A.P

First term, a = 7

Common difference, d = a₂ − a₁ = 13 − 7 = 6

Let there are n terms in this A.P.

a_n = 205

We know that,

a_n = a + (n − 1) d

Therefore, 205 = 7 + (n − 1) 6

198 = (n − 1) 6

33 = (n − 1)

n = 34

Therefore, this given series has 34 terms in it.

(ii) 18, 15½, 13 . . . 47

First term, a = 18,  a₂ = 15½ = 31/2

Common difference, d = a₂-a₁ =

=  31/2 –  18

d = (31-36)/2 = – 5/2

Let there are n terms in this A.P.

a_n = -47

We know that,

a_n = a+(n−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) = 26

n = 27

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

6. Check whether -150 is a term of the A.P. 11, 8, 5, 2, …

Solution:

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

First term, a = 11

Common difference, d = a₂−a₁ = 8−11 = −3

Let  a_n = -150

We know that,

a_n = a+(n−1)d

⇒ -150 = 11+(n -1)(-3)

⇒ -150 = 11-3n +3

⇒ -164 = -3n

⇒  n = 164/3

Clearly, n is not an integer but a fraction.

Therefore, – 150 is not a term of this A.P.

7. Find the 31^st term of an A.P. whose 11^th term is 38 and the 16^th term is 73.

Solution: Given that,

11^th term, a₁₁ = 38

and 16^th term, a₁₆ = 73

We know that,

a_n = a+(n−1)d

⇒ a₁₁ = a+(11−1)d

38 = a+10d ………………………………. (i)

Similarly,

a₁₆ = a +(16−1)d

73 = a+15d ………………………………………… (ii)

On subtracting equation (i) from (ii), we get

35 = 5d

d = 7

From equation (i), we can write,

38 = a+10×(7)

⇒ 38 − 70 = a

⇒ a = −32

a₃₁ = a +(31−1) d

= − 32 + 30 (7)

= − 32 + 210

= 178

Hence, 31^st term is 178.

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

Solution: Given that,

3^rd term, a₃ = 12

50^th term, a₅₀ = 106

We know that,

a_n = a+(n−1)d

a₃ = a+(3−1)d

12 = a+2d ……………………………. (i)

Similarly,

a₅₀ = a+(50−1)d

106 = a+49d …………………………. (ii)

On subtracting equation (i) from (ii), we get

94 = 47d

d = 2 = common difference

From equation (i), we can write now,

12 = a+2(2)

⇒ a = 12−4 = 8

a₂₉ = a+(29−1) d

⇒ a₂₉ = 8+(28)2

a₂₉ = 8+56 = 64

Therefore, 29^th term is 64.

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

Solution:

Given that,

3^rd term, a₃ = 4

and 9^th term, a₉ = −8

We know that,

a_n = a+(n−1)d

Therefore,

a₃ = a+(3−1)d

4 = a+2d ……………………………………… (i)

a₉ = a+(9−1)d

−8 = a+8d ………………………………………………… (ii)

On subtracting equation (i) from (ii), we will get here,

−12 = 6d

d = −2

From equation (i), we can write,

4 = a+2(−2)

⇒ 4 = a−4

a = 8

Let n^th term of this A.P. be zero.

a_n = a+(n−1)d

⇒ 0 = 8+(n−1)(−2)

⇒ 0 = 8−2n+2

⇒ 2n = 10

n = 5

Hence, 5^th term of this A.P. is 0.

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

Solution: Given that,

a_n = a+(n−1)d

⇒ a₁₇ = a+(17−1)d

⇒ a₁₇ = a +16d

Similarly

a₁₀ = a+9d

According to the question,

a₁₇ − a₁₀ = 7

Therefore,

(a +16d) − (a+9d) = 7

7d = 7

d = 1

Therefore, the common difference is 1.

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

Solution:  Given A.P. is  3, 15, 27, 39, …

first term, a = 3

common difference, d = a₂ − a₁ = 15 − 3 = 12

We know that,

a_n = a+(n−1)d

a₅₄ = a+(54−1)d

⇒ 3 + (53)(12)

⇒3 + 636 = 639

Given that,

     a_n   =  132 + a₅₄ 

⇒ a+(n−1)d = 132 + 639

771 = 3+(n −1)12

768 = (n−1)12

(n −1) = 64

n = 65

Therefore, 65^th term was 132 more than 54^th term.

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

Solution: Let, the first term of two APs be a₁ and a₂ respectively

And the common difference o = d.

For the first A.P.,we know,

a_n = a+(n−1)d

a₁₀₀ = a₁+(100−1)d

= a₁ + 99d

a₁₀₀₀ = a₁+(1000−1)d

a₁₀₀₀ = a₁+999d

For second A.P., we know,

a_n = a+(n−1)d

a₁₀₀ = a₂+(100−1)d

= a₂+99d

a₁₀₀₀ = a₂+(1000−1)d

= a₂+999d

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

Therefore, (a₁+99d) − (a₂+99d) = 100

a₁−a₂ = 100……………………………………………………………….. (i)

Difference between 1000^th terms of the two APs

(a₁+999d) − (a₂+999d) = a₁−a₂

From equation (i),

This difference, a₁−a₂ = 100

Hence, the difference between 1000^th 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;

Therefore, 105, 112, 119, …

a = 105, d = 112 – 105 = 7

The maximum possible three-digit number that is divisible by 7 = 994

Now the series is as follows.

105, 112, 119,  …    994

Let 994 be the nth term of this A.P.

first term, a = 105

common difference, d = 7

a_n = 994

We  know that.

a_n = a+(n−1)d

994 = 105+(n−1)7

889 = (n−1)7

(n−1) = 127

n = 128

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

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

Solution: Multiples of 4 lies between 10 and 250

12, 16, 20, 24, … 248

Let 248 be the n^th term of this A.P.

first term, a = 12

common difference, d = 4

a_n = 248

We know that,

a_n = a+(n−1)d

⇒ 248 = 12+(n-1)×4

⇒ 236/4 = n-1

⇒ 59  = n-1

n = 60

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

15. For what value of n, are the n^th terms of two APs 63, 65, 67, and 3, 10, 17, … equal?

Solution:

Given two APs as; 63, 65, 67,… and 3, 10, 17,….

Taking first AP,

63, 65, 67, …

First term, a = 63

Common difference, d = a₂−a₁ = 65−63 = 2

We know, n^th term of this A.P. = a_n = a+(n−1)d

a_n= 63+(n−1)2 = 63+2n−2

a_n = 61+2n ………………………………………. (i)

Taking second AP,

3, 10, 17, …

First term, a = 3

Common difference, d = a₂ − a₁ = 10 − 3 = 7

We know that,

n^th term of this A.P. = 3+(n−1)7

a_n = 3+7n−7

a_n = 7n−4 ……………………………………………………….. (ii)

Given, n^th term of these A.P.s are equal to each other.

Equating both these equations, we get,

61+2n = 7n−4

61+4 = 5n

5n = 65

n = 13

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

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

Solutions: Given,

Third term, a₃ = 16

We know that,

a +(3−1)d = 16

a+2d = 16 ………………………………………. (i)

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

a₇ − a₅ = 12

⇒ [a+(7−1)d] − [a +(5−1)d]= 12

⇒ (a+6d) − (a+4d) = 12

⇒ 2d = 12

⇒ d = 6

From equation (i), we get,

a+2(6) = 16

⇒ a+12 = 16

⇒ a = 4

Therefore, A.P. will be 4, 10, 16, 22, …

17. Find the 20^th term from the last term of the A.P. 3, 8, 13, …, 253.

Solution: Given A.P. is  3, 8, 13, …, 253

Common difference, d= 5.

a = 3, l = 253

Therefore the formula of nth term from last

a_n  = l – (n-1)d

a₂₀   = 253 – (20 – 1)×5

Therefore, using nth term formula, we get,

a₂₀ = a+(20−1)d

a₂₀ = 253 – (19)(5)

a₂₀ = 253−95

a = 158

Therefore, 20^th term from the last term of the AP 3, 8, 13, …, 253.is 158.

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

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

a_n = a+(n−1)d

a₄ = a+(4−1)d

a₄ = a+3d

And

a₈ = a+7d

a₆ = a+5d

a₁₀ = a+9d

Given that,

a₄+ a₈ = 24

a + 3d + a + 7d = 24

2a + 10d = 24

a+5d = 12 …………………………………………………… (i)

a₆ + a₁₀ = 44

a + 5d + a + 9d = 44

2a +14d = 44

a + 7d = 22 …………………………………….. (ii)

On subtracting equation (i) from (ii), we get,

2d = 22 − 12

2d = 10

d = 5

From equation (i), we get,

a + 5d = 12

a + 5(5) = 12

a + 25 = 12

a = −13

a₂ = a+d = − 13+5 = −8

a₃ = a₂+d = − 8+5 = −3

Therefore, the first three terms of this A.P. are −13, −8, and −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: 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, …

Here, first term, a = 5000

and  d = 200

Let after n^th year, his salary be Rs 7000.

Therefore, by the n^th term formula of AP,

a_n = a+(n−1) d

7000 = 5000+(n−1)200

200(n−1)= 2000

(n−1) = 10

n = 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 the n^th week, her weekly savings become Rs 20.75, find n.

Solution: Given that,

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

Hence,

First term, a = 5

and common difference, d = 1.75

Also given,

a_n = 20.75

Find, n = ?

As we know, by the n^th term formula,

a_n = a+(n−1)d

Therefore,

20.75 = 5 + (n -1)×1.75

⇒ 15.75 = (n -1)×1.75

⇒ (n -1) = 15.75/1.75 = 1575/175

⇒ 63/7 = 9

⇒ n -1 = 9

n = 10

Hence, n is 10.

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Exercis 5.3

1. Find the sum of the following APs.

(i) 2, 7, 12 ,…., to 10 terms.
(ii) − 37, − 33, − 29 ,…, to 12 terms
(iii) 0.6, 1.7, 2.8 ,…….., to 100 terms
(iv) 1/15, 1/12, 1/10, …… , to 11 terms

Solutions: (i) Given, 2, 7, 12 ,…, to 10 terms

first term, a = 2

And common difference, d = a₂ − a₁ = 7−2 = 5

n = 10

We know that,

= 5[4+(9)×(5)]

= 5 × 49 = 245

(ii) Given, −37, −33, −29 ,…, to 12 terms

For this A.P.,

first term, a = −37

And common difference, d = a₂− a₁

d= (−33)−(−37)

= − 33 + 37 = 4

n = 12

We know that,

= 6[-74 + 11×4]

= 6[-74 + 44]

= 6(-30) = -180

(iii)  Given, 0.6, 1.7, 2.8 ,…, to 100 terms

first term, a = 0.6

Common difference, d = a₂ − a₁ = 1.7 − 0.6 = 1.1

n = 100

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

= 50[1.2+108.9]

= 50[110.1]

= 5505

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

First term, a = 1/5

Common difference, d = a₂ –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,

= 33/20

2. Find the sums given below:

(i) 7 + 10½ + 14 + . . . . . . . + 84

(ii) 34 + 32 + 30 + ……….. + 10

(iii) − 5 + (− 8) + (− 11) + ………… + (− 230)

Solutions:

(i) Given A.P., 7 + 10½ + 14 + . . . . . . . + 84

First term, a = 7

n^th term, a_n = 84

Common diffrence, d =  10½ – 7

We know that,

a_n = a + (n-1)d

⇒ 84 = 7+(n – 1)×7/2

⇒ 77 = (n-1)×7/2

⇒ 22 = n−1

n = 23

We know that, sum of n term is;

⇒ S_n  = (23×91/2) = 2093/2

⇒ S_n    =   1046½

 

(ii)  Given, 34 + 32 + 30 + ……….. + 10

For this A.P.,

first term, a = 34

common difference, d = a₂−a₁ = 32−34 = −2

n^th term, a_n= 10 = l

Let 10 be the n^th term of this A.P., therefore,

a_n= a +(n−1)d

⇒ 10 = 34+(n−1)(−2)

⇒ −24 = (n −1)(−2)

⇒ 12 = n −1

⇒  n = 13

We know that, sum of n terms is;

= (13×44/2) = 13 × 22

= 286

(iii) Given, (−5) + (−8) + (−11) + ………… + (−230)

For this A.P.,

First term, a = −5

nth term, a_n= −230

Common difference, d = a₂−a₁ = (−8)−(−5)

⇒d = − 8+5 = −3

Let −230 be the n^th term of this A.P. Then

a_n= a + (n−1)d

−230 = − 5+(n−1)(−3)

−225 = (n−1)(−3)

(n−1) = 75

n = 76

And, Sum of n term,

= 38(-235)

= -8930

3. In an AP
(i) Given a = 5, d = 3, a_n = 50, find n and S_n.
(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, S_n = 90, find n and a_n.
(vii) Given a = 8, a_n = 62, S_n = 210, find n and d.
(viii) Given a_n = 4, d = 2, S_n = − 14, find n and a.
(ix) Given a = 3, n = 8, S = 192, find d.
(x) Given l = 28, S = 144 and there are total 9 terms. Find a.

Solutions:

(i) Given that, a = 5, d = 3, a_n = 50

From the formula of the nth term in an AP,

a_n = a +(n −1)d,

Therefore, putting the given values, we get,

⇒ 50 = 5+(n -1)×3

⇒ 3(n -1) = 45

⇒ n -1 = 15

⇒ n = 16

Now, sum of n terms,

S_n = 16/2 (5 + 50) = 440

(ii) Given that, a = 7, a₁₃ = 35

From the formula of the nth term in an AP,

a_n = a+(n−1)d,

Therefore, putting the given values, we get,

⇒ 35 = 7+(13-1)d

⇒ 12d = 28

⇒ d = 28/12 = 2.33

Now,

(iii) Given that, a₁₂ = 37, d = 3

As we know, from the formula of the n^th term in an AP,

a_n = a+(n −1)d,

Therefore, putting the given values, we get,

⇒ a₁₂ = a+(12−1)3

⇒ 37 = a+33

⇒ a = 4

Now, sum of nth term,

= 246

(iv) Given that, a₃ = 15, S₁₀ = 125

From the formula of the nth term in an AP,

a_n = a +(n−1)d,

Therefore, putting the given values, we get,

a₃ = a+(3−1)d

15 = a+2d ………………………….. (i)

Sum of the nth term,

⇒ 125 = 5(2a+9d)

⇒ 25 = 2a+9d ……………………….. (ii)

On multiplying eq(i) by 2

30 = 2a+4d ………………………………. (iii)

By subtracting equation (iii) from (ii), we get,

−5 = 5d

d = −1

From equation (i),

15 = a+2(−1)

15 = a−2

a = 17 = First term

a₁₀ = a+(10−1)d

⇒  a₁₀ = 17+(9)(−1)

⇒ a₁₀ = 17−9 = 8

(v) Given that, d = 5, S₉ = 75

Since,

Therefore, the sum of first nine terms are;

⇒ 25 = 3(a+20)

⇒ 25 = 3a+60

⇒ 3a = 25−60

a = -35/3

The n^th term can be written as;

a_n = a+(n−1)d

a₉ = a+(9−1)(5)

= -35/3+8(5)

= -35/3+40

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

(vi) Given that, a = 2, d = 8, S_n = 90

Sum of n terms in an AP is,

90 = n/2 [2a +(n -1)d]

⇒ 180 = n(4+8n -8)

⇒ n(8n-4) = 8n²-4n

⇒ 8n²-4n –180 = 0

⇒ 2n²–n-45 = 0

⇒ 2n²-10n+9n-45 = 0

⇒ 2n(n -5)+9(n -5) = 0

⇒ (n-5)(2n+9) = 0

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

∴   a₅ = 8+5×4 = 34

(vii) Given that, a = 8, a_n = 62, S_n = 210

Sum of n terms in an AP is,

210 = n/2 (8 +62)

⇒ 35n = 210

⇒ n = 210/35 = 6

Now, 62 = 8+5d

⇒ 5d = 62-8 = 54

⇒ d = 54/5 = 10.8

(viii) Given that, a_n = 4,  d = 2, S_n = −14.

From the formula of the n^th term in an AP,

a_n = a+(n −1)d,

Therefore, putting the given values, we get,

4 = a+(n −1)2

4 = a+2n−2

a+2n = 6

a = 6 − 2n …………………………………………. (i)

The sum of n terms is;

-14 = n/2 (a+4)

⇒ −28 = n (a+4)

⇒ −28 = n (6 −2n +4) {From equation (i)}

⇒ −28 = n (− 2n +10)

⇒ −28 = − 2n²+10n

⇒ 2n² −10n − 28 = 0

⇒ n² −5n −14 = 0

⇒ n² −7n+2n −14 = 0

⇒ n (n−7)+2(n −7) = 0

⇒ (n −7)(n +2) = 0

Either n − 7 = 0 or n + 2 = 0

n = 7 or n = −2

However, n can neither be negative nor fractional.

Therefore, n = 7

From equation (i), we get

a = 6−2n

⇒ a = 6−2(7)

⇒  a= 6−14

⇒ a = −8

(ix) Given that,  a = 3,

Number of terms, n = 8

And S_n  = 192

As we know,

⇒ 192 = 4[6 +7d]

⇒ 48 = 6+7d

⇒ 42 = 7d

d = 6

(x) Given that, l = 28, S = 144, n = 9 terms.

Sum of n terms formula,

⇒ (16)×(2) = a+28

⇒ 32 = a+28

⇒ a = 4

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

Solutions: Let there be n terms of the AP. 9, 17, 25 …

a = 9

 d = a₂−a₁ = 17−9 = 8

The sum of n terms, is;

⇒ 636 = n/2 [18+(n-1)×8]

⇒ 636 = n [9 +4n −4]

⇒ 636 = n (4n +5)

⇒ 4n² +5n −636 = 0

⇒ 4n² +53n −48n −636 = 0

⇒ n (4n + 53)−12 (4n + 53) = 0

⇒ (4n +53)(n −12) = 0

Either 4n+53 = 0 or n−12 = 0

n = (-53/4) or n = 12

n cannot be negative or fraction, therefore, n = 12 .

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,

 a = 5,  l = 45,  S_n = 400

The sum of AP formula is;

⇒ 400 = n/2(5+45)

⇒ 400 = n/2(50)

⇒ n =16

The last term of AP series can be written as;

l = a+(n −1)d

⇒ 45 = 5 +(16 −1)d

⇒ 40 = 15d

Common difference, d = 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,

 a = 17, l = 350, d = 9

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

l = a+(n −1)d

⇒ 350 = 17+(n −1)9

⇒ 333 = (n−1)9

⇒ (n−1) = 37

⇒ n = 38

Sum of n term

= 19×367

= 6973

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

7. Find the sum of first 22 terms of an AP in which d = 7 and 22^nd term is 149.

Solution: Given,

d = 7, a₂₂ = 149, S₂₂ = ?

By the formula of nth term,

a_n = a+(n−1)d

⇒ a₂₂ = a+(22−1)d

⇒ 149 = a+21×7

⇒ 149 = a+147

a = 2 = First term

Sum of n terms,

= 11×151

= 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,

 a₂ = 14, a₃ = 18, d = a₃−a₂ = 18−14 = 4

a₂ = a+d

⇒ 14 = a+4

a = 10

Sum of n terms;

= 51/2 [20+(50)×4]

= 51 × 220/2

= 51 × 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 first n terms.

Solution: Given that,

S₇ = 49, S₁₇ = 289

We know, Sum of n terms;

⇒ S₇ = 7/2 [2a + (7 -1)d]

⇒ 49 = 7/2 [2a +6d]

⇒ 7 = (a+3d)

a + 3d = 7 …………………………………. (i)

Similarly,

⇒ 289 = 17/2 (2a +16d)

⇒ 17 = (a+8d)

a +8d = 17 ………………………………. (ii)

Subtracting equation (i) from equation (ii),

5d = 10

d = 2

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

a+3(2) = 7

a+ 6 = 7

a = 1

Hence,

= n²

10. Show that a₁, a₂ … , a_n , … form an AP where a_n is defined as below

(i) a_n = 3+4n
(ii) a_n = 9−5n
Also find the sum of the first 15 terms in each case.

Solutions: (i)Given,  a_n = 3+4n

a₁ = 3+4(1) = 7

a₂ = 3+4(2) = 3+8 = 11

a₃ = 3+4(3) = 3+12 = 15

a₄ = 3+4(4) = 3+16 = 19

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

a₂ − a₁ = 11−7 = 4

a₃ − a₂ = 15−11 = 4

a₄ − a₃ = 19−15 = 4

Therefore, this is an AP with common difference as 4 and first term as 7.

The sum of nth term is;

= 15/2[(14)+56]

= 15/2(70)

= 15×35

= 525

(ii) Given, a_n = 9−5n

a₁ = 9−5×1 = 9−5 = 4

a₂ = 9−5×2 = 9−10 = −1

a₃ = 9−5×3 = 9−15 = −6

a₄ = 9−5×4 = 9−20 = −11

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

a₂ − a₁ = −1−4 = −5

a₃ − a₂ = −6−(−1) = −5

a₄ − a₃ = −11−(−6) = −5

Therefore, this is an A.P. with common difference as −5 and first term as 4.

The sum of nth term is;

=  15/2(8-70)

=  15/2(-62)

=  15(-31)

=  -465

11. If the sum of the first n terms of an AP is 4n − n², what is the first term (that is S₁)? What is the sum of first two terms? What is the second term? Similarly find the 3^rd, the10^th and the n^th terms.

Solution: Given that,

S_n = 4n−n²

a = S₁ = 4(1) − (1)² = 4−1 = 3

Sum of first two terms = S₂= 4(2)−(2)² = 8−4 = 4

a₂ = S₂ − S₁ = 4−3 = 1

 d = a₂−a = 1−3 = −2

nth term, a_n = a+(n−1)d 

= 3+(n −1)(−2)

= 3−2n +2

= 5−2n

Therefore, a₃ = 5−2(3) = 5-6 = −1

a₁₀ = 5−2(10) = 5−20 = −15

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

The 3^rd, the 10^th, and the n^th terms are −1, −15, and 5 − 2n 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 …. .

a = 6, d = 6

S₄₀ = ?

By the formula of sum of n terms,

putting n = 40, we get,

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

= 20(12+234)

= 20×246

= 4920

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

Solution: The multiples of 8 are

8, 16, 24, 32…

The series is in the form of AP, a = 8

d = 8, S₁₅ = ?

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

= 15/2[16 +112]

= 15(128)/2

= 15 × 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  …   49

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

Hence,

a = 1, d = 2, l = 49

By the formula of last term, we know,

l = a+(n−1) d

49 = 1+(n−1)2

48 = 2(n − 1)

n − 1 = 24

n = 25 = Number of terms

By the formula of sum of nth term,

= 25(50)/2

=(25)(25)

= 625

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, 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: Penalities

200, 250, 300, 350   .  .  .

We can see, that the given penalties are in the form of A.P.

Therefore, a = 200 and d = 50

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

By the formula of sum of nth term,

= 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 1^st prize be Rs. a.

Cost of 2^nd prize =a − 20

And cost of 3^rd prize = a − 40

Hence, prizes

a, a – 20, a – 40, a – 60  .  .  .

We can see that the cost of these prizes are in the form of A.P.,

a = P and d = −20

Given that, S₇ = 700

By the formula of sum of nth term,

7/2 [2a + (7 – 1)d] = 700

⇒ 2a + 6d = 200

⇒ a + 3(−20) = 100

⇒ a −60 = 100

⇒ a = 160

The prizes are

Rs 160, Rs 140, Rs 120, Rs 100, Rs 80, Rs 60, 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: The number of trees planted by the students  in different section is an AP.

3, 6, 9, 12, 15………………..36

a = 3, d = 6 – 3 = 3

= 6(6+33)

= 6(39)

= 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, ……… as shown in figure. What is the total length of such a spiral made up of thirteen consecutive semicircles? (Take π = 22/7)

 

Solution: We know,

Perimeter of a semi-circle = πr

Therefore,

_l1 = π(0.5) = π/2 cm

_l2 = π(1) = π cm

_l3 = π(1.5) = 3π/2 cm

Where, l_1, l₂, P₃ are the lengths of the semi-circles.

Hence we got a series here,

π/2, π, 3π/2, 2π, ….

d = l₂ – l₁ = π – π/2 = π/2

First term = l₁= a = π/2 cm

By the sum of n term formula,

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

=13/2 (7π)

= 13/2 × 7 × 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:   The numbers of logs in rows are in the form of an A.P.

20, 19, 18…

 a = 20,  d = a₂−a₁ = 19−20 = −1

Let a total of 200 logs be placed in n rows.

 S_n = 200

By the sum of nth term formula,

⇒ 400 = n (40−n+1)

⇒ 400 = n (41-n)

⇒ 400 = 41n−n²

⇒ n²−41n + 400 = 0

⇒ n²−16n−25n+400 = 0

⇒ n(n −16)−25(n −16) = 0

⇒ (n −16)(n −25) = 0

n −16 = 0 or  n−25 = 0

n = 16 or n = 25

By the nth term formula,

a_n = a+(n−1)d

a₁₆ = 20+(16−1)(−1)

a₁₆ = 20−15

a₁₆ = 5

Similarly, the 25^th term could be written as;

a₂₅ = 20+(25−1)(−1)

a₂₅ = 20−24

= −4

It can be seen, the number of logs in 16^th row is 5

the numbers cannot be negative.

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

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 2×5+2×(5+3)]

Solution: 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 run by the competetor to collect the potatos

10, 16, 22, 28, 34,……….

a = 10,  d = 16−10 = 6

S₁₀ =?

By the formula of sum of n terms,

= 5[20+54]

= 5(74)

= 370

Therefore, the competitor will run a total distance = 370 m

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Exercise 5.4

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

Solution: Given the AP is 121, 117, 113, . . .,

a = 121, d = 117-121= -4

By the nth term formula,

= 121 – 4n+4

=125 – 4n

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

Therefore,

⇒ 125-4n < 0

⇒ 125 <  4n

⇒ n > 125/4

⇒ n > 31.25

Therefore, the first negative term of the series is 32^nd 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:  According to the question, we can write,

a_{3 +} a₇ = 6 …………………………….(i)

And

a₃ ×a₇ = 8 ……………………………..(ii)

By the nth term formula,

a_n = a+(n−1)d

 a₃ = a+(3 -1)d

a₃ = a + 2d………………………………(iii)

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

a₇ = a + 6d ………………………………..(iv)

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

a+2d +a+6d = 6

2a+8d = 6

a+4d=3

⇒ a = 3–4d …………………………………(v)

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

(a+2d)×(a+6d) = 8

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

(3–4d +2d)×(3–4d+6d) = 8

(3 –2d)×(3+2d) = 8

3² – 2d² = 8

9 – 4d² = 8

4d² = 1

d = 1/2 or -1/2

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

when d = 1/2

a = 3 – 4d = 3 – 4(1/2) = 3 – 2 = 1,

when d = -1/2

a = 3 – 4d = 3 – 4(-1/2) = 3+2 = 5,

The sum of nth term of AP is;

a = 1 and d=1/2, n = 16

= 8(2+15/2) = 76

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

Then, the sum of first 16 terms are;

= 8(5/2)=20

3. A ladder has rungs 25 cm apart. (see Fig.). 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 2½  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 =

2½ m = 5/2×100 cm = 250 cm

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, 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.

First term, a = 45

Last term, l = 25

Number of terms, n = 11

Now, sum of nth terms is equal to,

S_n= 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 – 1 = S49 – Sx ]

Solution: Given,

Row houses are numbers from 1,2,3,4,5…….49.

First term, a = 1

Common difference, d=1

Let us say the number of x^th houses can be represented as;

Sum of nth term of AP

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

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

= x(x-1)/2 ………………………………………(i)

By the given condition, we can write,

S₄₉ – S_x = {49/2[2(1)+(49-1)1]}–{x/2[2(1)+(x-1)1]}

= 25(49) – x(x + 1)/2 ………………………………….(ii)

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

Therefore,

x(x-1)/2 = 25(49) – x(x+1)/2

x = ±35

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 = ¼ ×1/2 ×50 m³.]

 

Solution: As we can see from the given figure, the first step is ½ m wide, 2^nd step is 1m wide and 3^rd step is 3/2m wide.

Thus we can understand that the width of step by ½ m each time when height is ¼ 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;

½ , 1, 3/2, 2, ……..

Volume of steps = Volume of Cuboid

= Length × Breadth Height

Now,

Volume of concrete required to build the first step = ¼ ×1/2 ×50 = 25/4

Volume of concrete required to build the second step =¼ ×1×50 = 25/2

Volume of concrete required to build the second step = ¼ ×3/2 ×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 …..

Thus, applying the AP series concept,

First term, a = 25/4

Common difference, d = 25/2 – 25/4 = 25/4

As we know, the sum of n terms is;

S_n = n/2[2a+(n-1)d] = 15/2(2×(25/4 )+(15/2 -1)25/4)

Upon solving, we get,

S_{n =} 15/2 (100)

S_n=750

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

Ncert solution Class 10

Chapter 1: Real Number

Ncert maths solution class 10 chapter 1.1

Ncert maths solution class 10 chapter 1.2

Chapter 2: Polynomials

Ncert maths solutions for class 10 chapter 2 Polynomials

Chapter 4:Quadratic Equations

NCERT Solutions Maths for Class 10 Chapter 4