NCERT Solutions Maths Class 10 Chapter 5

NCERT Solutions Maths Class 10 Chapter 5

 

NCERT Solutions Maths Class 10 Exercise 5.1

 

1. In which of the following situations, does the list of numbers involved make an arithmetic progression, and why?

(i) The taxi fare after each km when the fare is Rs 15 for the first km and Rs 8 for each additional km.

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

(iii) The cost of digging a well after every metre of digging, when it costs Rs 150 for the first metre and rises by Rs 50 for each subsequent metre.

(iv) The amount of money in the account every year, when Rs 10,000 is deposited at compound Interest at 8% per annum.

 

Solution: (i) Taxi fare for the first km = Rs 15

Taxi fare for 2 km = 15 + 8 = Rs 23

Taxi fare for 3 km = 23 + 8 = Rs 31

Taxi fare for 4 km = 31 + 8 = Rs 39

Therefore, the sequence is 15, 23, 31, 39…

The above sequence is an arithmetic progression because difference between any two consecutive terms is equal that is 8. (23 – 15 = 8, 31 – 23 = 8, 39 – 31 = 8, … )

 

(ii) Let the amount of air initially present in the cylinder = V

Amount of air left after pumping out air by vacuum pump

Amount of air left when vacuum pump again pumps out air

Therefore, the sequence we get is 

Let us find the difference between consecutive terms.

The difference between consecutive terms is not equal.

Therefore, the above sequence is not an arithmetic progression.

 

(iii) The cost of digging 1 metre of well = Rs 150

The cost of digging 2 metres of well = 150 + 50 = Rs 200

The cost of digging 3 metres of well = 200 + 50 = Rs 250

The cost of digging 4 metres of well = 250 + 50 = Rs 300

Therefore, we get a sequence of the form 150, 200, 250, 300, …

This sequence is an arithmetic progression because the difference between any two consecutive terms is equal that is 50. (200 – 150 = 250 – 200 = 300 – 250 = 50 … )


 (iv) Amount in the bank account after one year =  … (1)

Amount in the bank account after two years = … (2)

Amount in the bank account after three years = … (3)

Amount in the bank account after four years = … (4)

Here, we can find that (2) − (1) ≠ (3) − (2)

(The difference between any two consecutive terms is not equal.)

Therefore, the sequence is not an arithmetic progression.

 

2. Write first four terms of the AP, when the first term a and the common difference d 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

 

Solution: (i) a = 10, d = 10

Let the first four terms of the AP be a, a + d, a + 2d and a + 3d.

First term = a = 10

Second term = a + d = 10 + 10 = 20

Third term = a + 2d = 10 + 2 × 10 = 30

Fourth term = a + 3d = 10 + 3 × 10 = 40

Therefore, the first four terms are: 10, 20, 30, 40

 

(ii) a = 2, d = 0

Let the first four terms of the AP be a, a + d, a + 2d and a + 3d.

First term = a = –2

Second term = a + d = –2 + 0 = –2

Third term = a + 2d = –2 + 2 × 0 = –2

Fourth term = a + 3d = –2 + 3 × 0 = –2

Therefore, the first four terms are: –2, –2, –2, –2

 

(iii) a = 4, d = 3

Let the first four terms of the AP be a, a + d, a + 2d and a + 3d.

First term = a = 4

Second term = a + d = 4 – 3 = 1

Third term = a + 2d = 4 – 2 × 3 = –2

Fourth term = a + 3d = 4 – 3 × 3 = –5

Therefore, the first four terms are: 4, 1, –2, –5

 

(iv) a = 1, d = ½

Let the first four terms of the AP be a, a + d, a + 2d and a + 3d.

First term = a = –1

Second term = a + d = –1 + ½ = −½

Third term = a + 2d = −1 2 × ½ = 0

Fourth term = a + 3d = −1 3 × ½ = ½

Therefore, the first four terms are: –1, −½, 0, ½

 

(v) a = 1.25, d = 0.25

Let the first four terms of the AP be a, a + d, a + 2d and a + 3d.

First term = a = –1.25

Second term = a + d = –1.25 – 0.25 = –1.50

Third term = a + 2d = –1.25 – 2 × 0.25 = –1.75

Fourth term = a + 3d = –1.25 – 3 × 0.25 = –2.00

Therefore, the first four terms are: –1.25, –1.50, –1.75, –2.00

 

3. For the following APs, write the first term and the common difference:

(i) 3, 1, –1, –3 …

(ii) –5, –1, 3, 7 …

(iii) 

(iv) 0.6, 1.7, 2.8, 3.9 …

 

Solution:  (i) 3, 1, –1, –3 …

First term = a = 3

Common difference (d) = Second term – first term = Third term – second term, and so on

Therefore, the common difference (d) = 1 – 3 = –2

 

(ii) –5, –1, 3, 7…

First term = a = –5

Common difference (d) = Second term – first term = Third term – second term, and so on

Therefore, the common difference (d) = –1 – (–5) = –1 + 5 = 4

 

(iii) 

First term = a = 1/3

Common difference (d) = Second term – first term = Third term – second term, and so on

Therefore, the common difference (d) = 

 

(iv) 0.6, 1.7, 2.8, 3.9 …

First term = a = 0.6

Common difference (d) = Second term – first term = Third term – second term, and so on

Therefore, the common difference (d) = 1.7 − 0.6 = 1.1

 

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

(i) 2, 4, 8, 16 …

(ii) 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) 

(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) 11, 52, 72, 73 …

 

Solution: (i) 2, 4, 8, 16 …

Here, 4 – 2 ≠ 8 − 4

It is not an AP because the difference between consecutive terms is not equal.

 

(ii) 2, 5/2, 3, 7/2

Here, 5/2 – 2 = 3 – 5/2 = 1/2

It is an AP because the difference between consecutive terms is equal.

Common difference (d) = 1/2

Fifth term = 7/2 + ½ = 4 

Sixth term = 4 + 1/2  9/2

Seventh term = 9/2 + ½ = 5

Therefore, the next three terms are 4, 9/2 and 5.

 

(iii) −1.2, −3.2, −5.2, −7.2 …

Here, −3.2 − (−1.2) = −5.2 − (−3.2) = −7.2 − (−5.2) = −2

It is an AP because the difference between consecutive terms is equal.

Common difference (d) = −2

Fifth term = −7.2 – 2 = −9.2 

Sixth term = −9.2 – 2 = −11.2

Seventh term = −11.2 – 2 = −13.2

Therefore, the next three terms are −9.2, −11.2 and −13.2.

 

(iv) −10, −6, −2, 2 …

Here, −6 − (−10) = −2 − (−6) = 2 − (−2) = 4

It is an AP because the difference between consecutive terms is equal.

Common difference (d) = 4

Fifth term = 2 + 4 = 6

Sixth term = 6 + 4 = 10

Seventh term = 10 + 4 = 14

Therefore, the next three terms are 6, 10 and 14.

 

(v) 3, 3 + √2, 3 + 2√2, 3 + 3√2 …

Here, 3 + √2 3 = √2, 3 + 2√2 – (3 + √2) = √2 and 3 + 3√2 – (3 + 2√2) = √2

It is an AP because the difference between consecutive terms is equal.

Common difference (d) = √2

Fifth term = 3 + 3√2 + √2 = 3 + 4√2

Sixth term = 3 + 4√2 + √2 = 3 + 5√2

Seventh term = 3 + 5√2 + √2 = 3 + 6√2

Therefore, the next three terms are (3 + 4√2), (3 + 5√2) and (3 + 6√2).

 

(vi) 0.2, 0.22, 0.222, 0.2222 …

Here, 0.22 − 0.2 ≠ 0.222 − 0.22

It is not an AP because the difference between consecutive terms is not equal.

 

(vii) 0, −4, −8, −12 …

Here, −4 – 0 = −8 − (−4) = −12 − (−8) = −4

It is an AP because the difference between consecutive terms is equal.

Common difference (d) = −4

Fifth term = −12 – 4 = −16

Sixth term = −16 – 4 = −20

Seventh term = −20 – 4 = −24

Therefore, the next three terms are −16, −20 and −24.

 

(viii) 

Here,  

It is an AP because the difference between consecutive terms is equal.

Common difference (d) = 0

Fifth term = 

Sixth term = 

Seventh term = 

Therefore, the next three terms are 

 

(ix) 1, 3, 9, 27 …

Here, 3 – 1 ≠ 9 – 3

It is not an AP because the difference between consecutive terms is not equal.

 

(x) a, 2a, 3a, 4a

Here, 2– = 3− 2= 4− 3a

It is an AP because the difference between consecutive terms is equal.

Common difference (d) = a

Fifth term = 4= 5a  

Sixth term = 5= 6a

Seventh term = 6= 7a

Therefore, the next three terms are 5a, 6a and 7a.

 

(xi) aa2a3a4

Here, a2 – ≠ a3 − a2 

It is not an AP because the difference between consecutive terms is not equal.

 

(xii) √2, √8, √18, √32 … 

 √2, 2√2, 3√2, 4√2

Here, 2√2 √2 = 3√2 – 2√2 = 4√2 – 3√2 =√2  

It is an AP because the difference between consecutive terms is equal.

Common difference (d) = √2

Fifth term = 4√2+ √2 = 5√2 = √50 

Sixth term = 5√2+ √2 = 6√2 = √72

Seventh term = 6√2+ √2= 7√2 = √98

Therefore, next three terms are √50, √72, √98.


(xiii)  √3, √6, √9, √12 …

Here, √6 – √3 √9 – √6

It is not an AP because the difference between consecutive terms is not equal.

 

(xiv)  12, 32, 52, 72

Here, 32 – 12 ≠ 52 – 32

It is not an AP because the difference between consecutive terms is not equal.

 

(xv) 12, 52, 72, 73

 1, 25, 49, 73 …

Here, 52 – 12 = 72 – 52 = 73 – 72 = 24

It is an AP because the difference between consecutive terms is equal.

Common difference (d) = 24

Fifth term = 73 + 24 = 97

Sixth term = 97 + 24 = 121

Seventh term = 121 + 24 = 145

Therefore, the next three terms are 97, 121 and 145.

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