Class 10 Math Chapter 5 Exercise 5.1 Solution – Arithmetic Progressions NCERT Solution

NCERT Class 10th Math Chapter 5 Exercise 5.1 Question Answer and can take online test here.. CCL Chapter helps you in it..  NCERT Solution for Class 10 Maths Chapter 5 Arithmetic Progressions NCERT Solution for CBSE, HBSE, RBSE and Up Board and all other chapters of class 10 math.

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NCERT Class 10 Math Chapter 5 Arithmetic Progressions Exercise 5.1 Question Answer Solution.

Class 10 Math Chapter 5 Exercise 5.1 Solution

Question 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 ₹15 for the first km and ₹8 for each additional km.

Ans. Yes. 12, 23, 31, ……… forms an AP as each succeeding term is obtained by adding 8 in its preceding term.

(ii) The amount of air present in a cylinder when a vacuum pump removes \displaystyle \frac{1}{4} of the air remaining in the cylinder at a time.

Ans. No. Volumes are V, \displaystyle \left( {\frac{{3V}}{4}} \right),{{\left( {\frac{{3V}}{4}} \right)}^{2}},....

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

Ans. Yes. 150, 200, 250, …. form an AP.

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

Ans. No. Amounts are \displaystyle 10000\left( {1+\frac{8}{{100}}} \right),10000{{\left( {1+\frac{8}{{100}}} \right)}^{2}},10000{{\left( {1+\frac{8}{{100}}} \right)}^{3}},.......

Question 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

Ans. As we know, Ap is a, a + d, a + 2d, a + 3d, …..

Therefore, required first four terms of AP are as :

10, 10 + 10, 10 + 2(10), 10 + 3(10)

i.e. 10, 20, 30, 40.

(ii) a = –2,  d = 0

Ans. First four terms of AP are as :

-2, -2, -2, -2

(iii) a = 4,  d = – 3

Ans. First four terms of AP are as :

4, 4 +(-3), 4 + 2(-3), 4 + 3(-3)

i.e. 4, 1, -2, -5

(iv) a = – 1,  d = \displaystyle \frac{1}{2}

Ans. First four terms of AP are as :

\displaystyle -1,-1+\frac{1}{2},-1+2\left( {\frac{1}{2}} \right),-1+3\left( {\frac{1}{2}} \right)

i.e. \displaystyle -1,-\frac{1}{2},0,\frac{1}{2}

(v) a = – 1.25,  d = – 0.25

Ans. First four terms of AP are as :

-1.25, -1,25 +(-0.25), -1.25 + (-0.50), -1.25 + (-0.75)

i.e.  -1.25, -1.5, -1.75, -2.

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

(i) 3, 1, – 1, – 3, …….

Ans. First term (a) = 3,

Common difference (d) = a2 – a1 = 1 – 3 = -2

(ii) – 5, – 1, 3, 7, …….

Ans. First term = -5, Common Difference = -1 -(-5) = 4

(iii) \displaystyle \frac{1}{3},\frac{5}{3},\frac{9}{3},\frac{{13}}{3},.....

Ans. First term = \displaystyle \frac{1}{3},

Common Difference = \displaystyle \frac{5}{3}-\frac{1}{3}=\frac{4}{3}

(iv) 0.6, 1.7, 2.8, 3.9, ……

Ans. First term = 0.6, Common Difference = 1.7 -0.6 = 1.1

Question 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, …..

Ans. A series can be in AP if their common difference is same i.e. d = a2 – a1= a3 – a2 = a4 – a3
4 – 2 = 2,
8 – 4 = 4,

2≠4

Therefore, given series is not an AP.

(ii) \displaystyle 2,\frac{5}{2},3,\frac{7}{2},.......

Ans. Checking AP : \displaystyle \frac{5}{2}-2=3-\frac{5}{2}=\frac{1}{2}

Therefore, given series is in AP.

d = \displaystyle \frac{1}{2}

Next three terms : 4, \displaystyle \frac{9}{2}, 5

(iii) – 1.2, – 3.2, – 5.2, – 7.2…….

Ans. Checking AP : -3.2 -(-1.2) = -5.2 -(-3.2) = -2

Therefore, given series is in AP.

d = -2

Next three terms : -9.2, -11.2, -13.2

(iv) – 10,  – 6,  – 2,  2, …….

Ans. Checking AP : -6 -(-10) = -2 -(-6) = 4

Therefore, given series is in AP.

d = 4

Next three terms : 6, 10, 14

(v) \displaystyle 3,3+\sqrt{2},3+2\sqrt{2},3+3\sqrt{2},.......

Ans. Checking AP : \displaystyle (3+\sqrt{2})-3=(3+2\sqrt{2})-(3+\sqrt{2})=\sqrt{2}

Therefore, given series is in AP.

d = \displaystyle \sqrt{2}

Next three terms : \displaystyle 3+4\sqrt{2},3+6\sqrt{2},3+6\sqrt{2}

(vi) 0.2, 0.22, 0.222, 0.2222, …….

Ans. Checking AP : 0.22 – 0.2 ≠ 0.222 – 0.22

Therefore, given series is not in AP.

(vii) 0, – 4, – 8, –12, …….

Ans. Checking AP : -4 – 0 = -8 – (-4) = -4

Therefore, given series is in AP.

d = -4

Next three terms : -16, -20, -24

(viii) \displaystyle -\frac{1}{2},-\frac{1}{2},-\frac{1}{2},-\frac{1}{2},........

Ans. Checking AP : \displaystyle -\frac{1}{2}-\left( {-\frac{1}{2}} \right)=-\frac{1}{2}-\left( {-\frac{1}{2}} \right)=0

Therefore given series is in AP.

d = 0

Next three terms : \displaystyle -\frac{1}{2},-\frac{1}{2},-\frac{1}{2}

(ix) 1, 3, 9, 27,…….

Ans. Checking AP : 3 – 1 ≠ 9 – 3

Therefore, given series is not in AP.

(x) a, 2a, 3a, 4a, …….

Ans. Checking AP : 2a – a = 3a – 2a = a

Therefore, given series is in AP

d = a

Next three terms of AP  : 5a, 6a, 7a

(xi) a, a2, a3, a4, ………

Ans. Checking AP : a2 – a ≠ a3 – a2

Therefore, given series is not in AP

(xii) \displaystyle \sqrt{2},\sqrt{8},\sqrt{{18}},\sqrt{{32}},........

Ans. Checking AP : \displaystyle \sqrt{8}-\sqrt{2}=\sqrt{{18}}-\sqrt{8}=\sqrt{2}

Therefore, given series is in AP.

d = \displaystyle \sqrt{2}

Next three terms : \displaystyle \sqrt{{50}},\sqrt{{72}},\sqrt{{98}}

(xiii) \displaystyle \sqrt{3},\sqrt{6},\sqrt{8},\sqrt{{12}},........

Ans. Checking AP : \displaystyle \sqrt{6}-\sqrt{3}\ne \sqrt{9}-\sqrt{6}

Therefore, given series is not in AP.

(xiv) 12, 32, 52, 72, ………

Ans. Checking AP : 32 – 12 ≠ 52 – 32

Therefore, given series is not in AP.

(xv) 12, 52, 72, 73, ……… .

Ans. Checking AP : 52 – 1 = 72 – 52 = 24

Therefore it is in AP.

d = 24

Next three terms : 97, 121, 145

 

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