Careers360 Logo
School
Search Colleges, Exams, Schools & more
NCERT Solutions for Exercise 5.1 Class 10 Maths Chapter 5 - Arithmetic Progressions

NCERT Solutions for Exercise 5.1 Class 10 Maths Chapter 5 - Arithmetic Progressions

Edited By Ramraj Saini | Updated on Nov 17, 2023 04:51 PM IST | #CBSE Class 10th

NCERT Solutions For Class 10 Maths Chapter 5 Exercise 5.1

NCERT Solutions for Exercise 5.1 Class 10 Maths Chapter 5 Arithmetic Progressions are discussed here. These NCERT solutions are created by subject matter expert at Careers360 considering the latest syllabus and pattern of CBSE 2023-24. This class 10 ex 5.1 the focus is on the general equation of arithmetic progression along with the general equation it includes many problems based on common difference and first term. Class 10 maths ex 5.1 also gives a very small idea about finite and infinite A.P series.

This Story also Contains
  1. NCERT Solutions For Class 10 Maths Chapter 5 Exercise 5.1
  2. Download Free Pdf of NCERT Solutions for Class 10 Maths Chapter 5 Exercise 5.1
  3. Assess NCERT Solutions for Class 10 Maths Chapter 5 Exercise 5.1
  4. More About NCERT Solutions for Class 10 Maths Exercise 5.2
  5. Benefits of NCERT Solutions for Class 10 Maths Exercise 5.1
  6. NCERT Solutions of Class 10 Subject Wise
  7. Subject Wise NCERT Exemplar Solutions
NCERT Solutions for Exercise 5.1 Class 10 Maths Chapter 5 - Arithmetic Progressions
NCERT Solutions for Exercise 5.1 Class 10 Maths Chapter 5 - Arithmetic Progressions

NCERT solutions for 10th class Maths exercise 5.1 answers are designed as per the students demand covering comprehensive, step by step solutions of every problem. Practice these questions and answers to command the concepts, boost confidence and in depth understanding of concepts. Students can find all exercise together using the link provided below.

VMC VIQ Scholarship Test

Register for Vidyamandir Intellect Quest. Get Scholarship and Cash Rewards.

Apply
Pearson | PTE

Register now for PTE & Unlock 20% OFF : Use promo code: 'C360SPL20'. Valid till 15th NOV'24! Trusted by 3,500+ universities globally

Apply

Download Free Pdf of NCERT Solutions for Class 10 Maths Chapter 5 Exercise 5.1

Download PDF

Assess NCERT Solutions for Class 10 Maths Chapter 5 Exercise 5.1

Arithmetic Progressions Class 10 Chapter 5 Exercise: 5.1

Q1 (i) 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 \small Rs\hspace{1mm}15 for the first km and \small Rs\hspace{1mm}8 for each additional \small km .

Answer:

It is given that
Fare for 1^{st} \ km = Rs. 15
And after that \small Rs\hspace{1mm}8 for each additional \small km
Now,
Fare for 2^{nd} \ km = Fare of first km + Additional fare for 1 km
= Rs. 15 + 8 = Rs 23

Fare for 3^{rd} \ km = Fare of first km + Fare of additional second km + Fare of additional third km
= Rs. 23 + 8= Rs 31

Fare of n km = 15 + 8 \times (n - 1)
( We multiplied by n - 1 because the first km was fixed and for the rest, we are adding additional fare.

In this, each subsequent term is obtained by adding a fixed number (8) to the previous term.)

Now, we can clearly see that this is an A.P. with the first term (a) = 15 and common difference (d) = 8

Q1 (ii) In which of the following situations, does the list of numbers involved make an arithmetic progression, and why? (ii) The amount of air present in a cylinder when a vacuum pump removes \small \frac{1}{4} of the air remaining in the cylinder at a time.

Answer:

It is given that
vacum pump removes \small \frac{1}{4} of the air remaining in the cylinder at a time
Let us take initial quantity of air = 1

Now, the quantity of air removed in first step = 1/4

Remaining quantity after 1 st step

= 1-\frac{1}{4}= \frac{3}{4}

Similarly, Quantity removed after 2 nd step = Quantity removed in first step \times Remaining quantity after 1 st step

=\frac{3}{4}\times \frac{1}{4}= \frac{3}{16}
Now,

Remaining quantity after 2 nd step would be = Remaining quantity after 1 st step - Quantity removed after 2 nd step

=\frac{3}{4}- \frac{3}{16}= \frac{12-3}{16}= \frac{9}{16}
Now, we can clearly see that

After the second step the difference between second and first and first and initial step is not the same, hence

the common difference (d) is not the same after every step

Therefore, it is not an AP

Q1 (iii) In which of the following situations, does the list of numbers involved make an arithmetic progression, and why? (iii) The cost of digging a well after every meter of digging, when it costs \small Rs\hspace{1mm}150 for the first metre and rises by \small Rs\hspace{1mm}50 for each subsequent meter.

Answer:

It is given that
Cost of digging of 1st meter = Rs 150
and
rises by \small Rs\hspace{1mm}50 for each subsequent meter
Therefore,

Cost of digging of first 2 meters = cost of digging of first meter + cost of digging additional meter

Cost of digging of first 2 meters = 150 + 50

= Rs 200

Similarly,
Cost of digging of first 3 meters = cost of digging of first 2 meters + cost of digging of additional meter

Cost of digging of first 3 meters = 200 + 50

= Rs 250

We can clearly see that 150, 200,250, ... is in AP with each subsequent term is obtained by adding a fixed number (50) to the previous term.

Therefore, it is an AP with first term (a) = 150 and common difference (d) = 50

Q1 (iv) In which of the following situations, does the list of numbers involved make an arithmetic progression, and why? (iv) The amount of money in the account every year, when \small Rs\hspace{1mm}10000 is deposited at compound interest at \small 8\hspace{1mm}\% per annum .

Answer:

Amount in the beginning = Rs. 10000

Interest at the end of 1st year at the rate of \small 8\hspace{1mm}\%
is \small 8\hspace{1mm}\% of 10000 = \frac{8\times 10000}{100}= 800

Therefore, amount at the end of 1st year will be

= 10000 + 800

= 10800

Now,

Interest at the end of 2nd year at rate of \small 8\hspace{1mm}\%
is \small 8\hspace{1mm}\% of 10800 = \frac{8\times 10800}{100}= 864

Therefore,, amount at the end of 2 nd year

= 10800 + 864 = 11664

Since each subsequent term is not obtained by adding a unique number to the previous term; hence, it is not an AP

Q2 (i) Write first four terms of the AP, when the first term a and the common difference d are given as follows \small a=10,d=10

Answer:

It is given that
\small a=10,d=10
Now,
a_1= a =10
a_2= a_1 + d= 10 + 10 = 20
a_3= a_2 + d= 20 + 10 = 30
a_4= a_3 + d= 30 + 10 = 40
Therefore, the first four terms of the given series are 10,20,30,40

Q2 (ii) Write first four terms of the AP when the first term a and the common difference d are given as follows: \small a=-2,d=0
Answer:

It is given that
\small a=-2,d=0
Now,
a_1= a = -2
a_2= a_1 + d= -2 + 0 = -2
a_3= a_2 + d= -2 + 0 = -2
a_4= a_3 + d= -2 + 0 = -2
Therefore, the first four terms of the given series are -2,-2,-2,-2

Q2 (iii) Write first four terms of the AP when the first term a and the common difference d are given as follows \small a=4,d=-3

Answer:

It is given that
\small a=4,d=-3
Now,
a_1= a =4
a_2= a_1 + d= 4 - 3 = 1
a_3= a_2 + d= 1 - 3 = -2
a_4= a_3 + d= -2- 3 = -5
Therefore, the first four terms of the given series are 4,1,-2,-5

Q2 (iv) Write first four terms of the AP when the first term a and the common difference d are given as follows \small a=-1,d=\frac{1}{2}

Answer:

It is given that
\small a=-1,d=\frac{1}{2}
Now,
a_1= a =-1
a_2= a_1 + d= -1 + \frac{1}{2} = -\frac{1}{2}
a_3= a_2 + d= -\frac{1}{2} + \frac{1}{2} = 0
a_4= a_3 + d= 0+\frac{1}{2}= \frac{1}{2}
Therefore, the first four terms of the given series are -1,-\frac{1}{2},0, \frac{1}{2}

Q2 (v) Write first four terms of the AP when the first term a and the common difference d are given as follows \small a=-1.25,d=-0.25

Answer:

It is given that
\small a=-1.25,d=-0.25
Now,
a_1= a =-1.25
a_2= a_1 + d= -1.25 -0.25= -1.50
a_3= a_2 + d= -1.50-0.25=-1.75
a_4= a_3 + d= -1.75-0.25=-2
Therefore, the first four terms of the given series are -1.25,-1.50,-1.75,-2

Q3 (i) For the following APs, write the first term and the common difference: \small 3,1,-1,-3,...

Answer:

Given AP series is

\small 3,1,-1,-3,...

Now, first term of this AP series is 3

Therefore,

First-term of AP series (a) = 3

Now,

a_1=3 \ \ and \ \ a_2 = 1

And common difference (d) = a_2-a_1 = 1-3 = -2

Therefore, first term and common difference is 3 and -2 respectively

Q3 (ii) For the following APs, write the first term and the common difference: \small -5,-1,3,7,...

Answer:

Given AP series is

\small -5,-1,3,7,...

Now, the first term of this AP series is -5

Therefore,

First-term of AP series (a) = -5

Now,

a_1=-5 \ \ and \ \ a_2 = -1

And common difference (d) = a_2-a_1 = -1-(-5) = 4

Therefore, the first term and the common difference is -5 and 4 respectively

Q3 (iii) For the following APs, write the first term and the common difference: \small \frac{1}{3},\frac{5}{3},\frac{9}{3},\frac{13}{3},...

Answer:

Given AP series is

\small \frac{1}{3},\frac{5}{3},\frac{9}{3},\frac{13}{3},...
Now, the first term of this AP series is \frac{1}{3}

Therefore,

The first term of AP series (a) = \frac{1}{3}

Now,

a_1=\frac{1}{3} \ \ and \ \ a_2 = \frac{5}{3}

And common difference (d) = a_2-a_1 = \frac{5}{3}-\frac{1}{3} = \frac{5-1}{3} =\frac{4}{3}

Therefore, the first term and the common difference is \frac{1}{3} and \frac{4}{3} respectively

Q3 (iv) For the following APs, write the first term and the common difference: \small 0.6,1.7,2.8,3.9,...

Answer:

Given AP series is

\small 0.6,1.7,2.8,3.9,...

Now, the first term of this AP series is 0.6

Therefore,

First-term of AP series (a) = 0.6

Now,

a_1=0.6 \ \ and \ \ a_2 = 1.7

And common difference (d) = a_2-a_1 = 1.7-0.6 = 1.1

Therefore, the first term and the common difference is 0.6 and 1.1 respectively.

Q4 (i) Which of the following are APs? If they form an AP, find the common difference d and write three more terms. \small 2,4,8,12,...

Answer:

Given series is
\small 2,4,8,12,...
Now,
the first term to this series is = 2
Now,
a_1 = 2 \ \ and \ \ a_2 = 4 \ \ and \ \ a_3 = 8
a_2-a_1 = 4-2 = 2
a_3-a_2 = 8-4 = 4
We can clearly see that the difference between terms are not equal
Hence, given series is not an AP

Q4 (ii) Which of the following are APs ? If they form an AP, find the common difference d and write three more terms. \small 2,\frac{5}{2},3,\frac{7}{2},...

Answer:

Given series is
\small 2,\frac{5}{2},3,\frac{7}{2},...
Now,
first term to this series is = 2
Now,
a_1 = 2 \ \ and \ \ a_2 = \frac{5}{2} \ \ and \ \ a_3 = 3 \ \ and \ \ a_4 = \frac{7}{2}
a_2-a_1 = \frac{5}{2}-2 = \frac{5-4}{2}=\frac{1}{2}
a_3-a_2 = 3-\frac{5}{2} = \frac{6-5}{2} = \frac{1}{2}
a_4-a_3=\frac{7}{2}-3=\frac{7-6}{2} =\frac{1}{2}
We can clearly see that the difference between terms are equal and equal to \frac{1}{2}
Hence, given series is in AP
Now, the next three terms are
a_5=a_4+d = \frac{7}{2}+\frac{1}{2} = \frac{8}{2}=4
a_6=a_5+d = 4+\frac{1}{2} = \frac{8+1}{2}=\frac{9}{2}
a_7=a_6+d =\frac{9}{2} +\frac{1}{2} = \frac{10}{2}=5

Therefore, next three terms of given series are 4,\frac{9}{2} ,5

Q4 (iii) Which of the following are APs ? If they form an AP, find the common difference d and write three more terms. \small -1.2,-3.2,-5.2,-7.2,...

Answer:

Given series is
\small -1.2,-3.2,-5.2,-7.2,...
Now,
the first term to this series is = -1.2
Now,
a_1 = -1.2 \ \ and \ \ a_2 = -3.2 \ \ and \ \ a_3 = -5.2 \ \ and \ \ a_4 = -7.2
a_2-a_1 = -3.2-(-1.2) =-3.2+1.2=-2
a_3-a_2 = -5.2-(-3.2) =-5.2+3.2 = -2
a_4-a_3=-7.2-(-5.2)=-7.2+5.2=-2
We can clearly see that the difference between terms are equal and equal to -2
Hence, given series is in AP
Now, the next three terms are
a_5=a_4+d = -7.2-2 =-9.2
a_6=a_5+d = -9.2-2 =-11.2
a_7=a_6+d = -11.2-2 =-13.2
Therefore, next three terms of given series are -9.2,-11.2,-13.2

Q4 (iv) Which of the following are APs? If they form an AP, find the common difference d and write three more terms. \small -10,-6,-2,2,...

Answer:

Given series is
\small -10,-6,-2,2,...
Now,
the first term to this series is = -10
Now,
a_1 = -10 \ \ and \ \ a_2 = -6 \ \ and \ \ a_3 = -2 \ \ and \ \ a_4 = 2
a_2-a_1 = -6-(-10) =-6+10=4
a_3-a_2 = -2-(-6) =-2+6 = 4
a_4-a_3=2-(-2)=2+2=4
We can clearly see that the difference between terms are equal and equal to 4
Hence, given series is in AP
Now, the next three terms are
a_5=a_4+d = 2+4 =6
a_6=a_5+d = 6+4=10
a_7=a_6+d = 10+4=14
Therefore, next three terms of given series are 6,10,14

Q4 (v) Which of the following are APs? If they form an AP, find the common difference d and write three more terms. \small 3,3+\sqrt{2},3+2\sqrt{2},3+3\sqrt{2},...

Answer:

Given series is
\small 3,3+\sqrt{2},3+2\sqrt{2},3+3\sqrt{2},...
Now,
the first term to this series is = 3
Now,
a_1 = 3 \ \ and \ \ a_2 = 3+\sqrt2 \ \ and \ \ a_3 = 3+2\sqrt2 \ \ and \ \ a_4 = 3+3\sqrt2
a_2-a_1 = 3+\sqrt2-3= \sqrt2
a_3-a_2 = 3+2\sqrt2-3-\sqrt2 = \sqrt2
a_4-a_3 = 3+3\sqrt2-3-2\sqrt2 = \sqrt2
We can clearly see that the difference between terms are equal and equal to \sqrt2
Hence, given series is in AP
Now, the next three terms are
a_5=a_4+d = 3+3\sqrt2+\sqrt2=3+4\sqrt2
a_6=a_5+d = 3+4\sqrt2+\sqrt2=3+5\sqrt2
a_7=a_6+d = 3+5\sqrt2+\sqrt2=3+6\sqrt2
Therefore, next three terms of given series are 3+4\sqrt2, 3+5\sqrt2,3+6\sqrt2

Q4 (vi) Which of the following are APs? If they form an AP, find the common difference d and write three more terms. \small 0.2,0.22,0.222,0.2222,...

Answer:

Given series is
\small 0.2,0.22,0.222,0.2222,...
Now,
the first term to this series is = 0.2
Now,
a_1 = 0.2 \ \ and \ \ a_2 = 0.22 \ \ and \ \ a_3 = 0.222 \ \ and \ \ a_4 = 0.2222
a_2-a_1 = 0.22-0.2=0.02
a_3-a_2 = 0.222-0.22=0.002

We can clearly see that the difference between terms are not equal
Hence, given series is not an AP

Q4 (vii) Which of the following are APs? If they form an AP, find the common difference d and write three more terms. \small 0,-4,-8,-12,...

Answer:

Given series is
\small 0,-4,-8,-12,...
Now,
first term to this series is = 0
Now,
a_1 = 0 \ \ and \ \ a_2 = -4 \ \ and \ \ a_3 = -8 \ \ and \ \ a_4 = -12
a_2-a_1 = -4-0 =-4
a_3-a_2 = -8-(-4) =-8+4 = -4
a_4-a_3=-12-(-8)=-12+8=-4
We can clearly see that the difference between terms are equal and equal to -4
Hence, given series is in AP
Now, the next three terms are
a_5=a_4+d = -12-4 =-16
a_6=a_5+d = -16-4=-20
a_7=a_6+d = -20-4=-24
Therefore, the next three terms of given series are -16,-20,-24

Q4 (viii) Which of the following are APs? If they form an AP, find the common difference d and write three more terms. \small -\frac{1}{2},-\frac{1}{2},-\frac{1}{2},-\frac{1}{2},...

Answer:

Given series is
\small -\frac{1}{2},-\frac{1}{2},-\frac{1}{2},-\frac{1}{2},...
Now,
the first term to this series is = -\frac{1}{2}
Now,
a_1 = -\frac{1}{2} \ \ and \ \ a_2 = -\frac{1}{2} \ \ and \ \ a_3 = -\frac{1}{2} \ \ and \ \ a_4 = -\frac{1}{2}
a_2-a_1 = -\frac{1}{2}-\left ( -\frac{1}{2} \right ) = -\frac{1}{2}+\frac{1}{2}=0
a_3-a_2 = -\frac{1}{2}-\left ( -\frac{1}{2} \right ) = -\frac{1}{2}+\frac{1}{2}=0
a_4-a_3 = -\frac{1}{2}-\left ( -\frac{1}{2} \right ) = -\frac{1}{2}+\frac{1}{2}=0
We can clearly see that the difference between terms are equal and equal to 0
Hence, given series is in AP
Now, the next three terms are
a_5=a_4+d = -\frac{1}{2}+0=-\frac{1}{2}
a_6=a_5+d = -\frac{1}{2}+0=-\frac{1}{2}
a_7=a_6+d = -\frac{1}{2}+0=-\frac{1}{2}
Therefore, the next three terms of given series are -\frac{1}{2},-\frac{1}{2},-\frac{1}{2}

Q4 (ix) Which of the following are APs? If they form an AP, find the common difference d and write three more terms. \small 1,3,9,27,...

Answer:

Given series is
\small 1,3,9,27,...
Now,
the first term to this series is = 1
Now,
a_1 = 1 \ \ and \ \ a_2 = 3 \ \ and \ \ a_3 = 9 \ \ and \ \ a_4 = 27
a_2-a_1 = 3-1=2
a_3-a_2 =9-3=6


We can clearly see that the difference between terms are not equal
Hence, given series is not an AP

Q4 (x) Which of the following are APs ? If they form an AP, find the common difference d and write three more terms. \small a,2a,3a,4a,...

Answer:

Given series is
\small a,2a,3a,4a,...
Now,
the first term to this series is = a
Now,
a_1 = a \ \ and \ \ a_2 = 2a \ \ and \ \ a_3 = 3a \ \ and \ \ a_4 = 4a
a_2-a_1 = 2a-a =a
a_3-a_2 = 3a-2a =a
a_4-a_3=4a-3a=a
We can clearly see that the difference between terms are equal and equal to a
Hence, given series is in AP
Now, the next three terms are
a_5=a_4+d =4a+a=5a
a_6=a_5+d =5a+a=6a
a_7=a_6+d =6a+a=7a
Therefore, next three terms of given series are 5a,6a,7a

Q4 (xi) Which of the following are APs ? If they form an AP, find the common difference d and write three more terms. \small a,a^2,a^3,a^4,...
Answer:

Given series is
\small a,a^2,a^3,a^4,...
Now,
the first term to this series is = a
Now,
a_1 = a \ \ and \ \ a_2 = a^2 \ \ and \ \ a_3 = a^3 \ \ and \ \ a_4 = a^4
a_2-a_1 = a^2-a =a(a-1)
a_3-a_2 = a^3-a^2 =a^2(a-1)

We can clearly see that the difference between terms are not equal
Hence, given series is not in AP

Q4 (xii) Which of the following are APs ? If they form an AP, find the common difference d and write three more terms. \small \sqrt{2},\sqrt{8},\sqrt{18},\sqrt{32},...

Answer:

Given series is
\small \sqrt{2},\sqrt{8},\sqrt{18},\sqrt{32},...
We can rewrite it as
\sqrt2,2\sqrt2,3\sqrt2,4\sqrt2,....
Now,
first term to this series is = a
Now,
a_1 = \sqrt2 \ \ and \ \ a_2 = 2\sqrt2 \ \ and \ \ a_3 = 3\sqrt2 \ \ and \ \ a_4 = 4\sqrt2
a_2-a_1 = 2\sqrt2-\sqrt2 =\sqrt2
a_3-a_2 = 3\sqrt2-2\sqrt2 =\sqrt2
a_4-a_3=4\sqrt2-3\sqrt2=\sqrt2
We can clearly see that difference between terms are equal and equal to \sqrt2
Hence, given series is in AP
Now, the next three terms are
a_5=a_4+d =4\sqrt2+\sqrt2=5\sqrt2
a_6=a_5+d =5\sqrt2+\sqrt2=6\sqrt2
a_7=a_6+d =6\sqrt2+\sqrt2=7\sqrt2
Therefore, next three terms of given series are 5\sqrt2,6\sqrt2,7\sqrt2

That is the next three terms are \sqrt{50},\ \sqrt{72},\ \sqrt{98}

Q4 (xiii) Which of the following are APs? If they form an AP, find the common difference d and write three more terms. \small \sqrt{3},\sqrt{6},\sqrt{9},\sqrt{12},...

Answer:

Given series is
\small \sqrt{3},\sqrt{6},\sqrt{9},\sqrt{12},...
Now,
the first term to this series is = \sqrt3
Now,
a_1 = \sqrt3 \ \ and \ \ a_2 = \sqrt6 \ \ and \ \ a_3 = \sqrt9 \ \ and \ \ a_4 = \sqrt{12}
a_2-a_1 = \sqrt6-\sqrt3 =\sqrt3(\sqrt2-1)
a_3-a_2 = 3-\sqrt3 =\sqrt3(\sqrt3-1)
We can clearly see that the difference between terms are not equal
Hence, given series is not in AP

Q4 (xiv) Which of the following are APs ? If they form an AP, find the common difference d and write three more terms. \small 1^2,3^2,5^2,7^2,...

Answer:

Given series is
\small 1^2,3^2,5^2,7^2,...
we can rewrite it as
1,9,25,49,....
Now,
the first term to this series is = 1
Now,
a_1 =1 \ \ and \ \ a_2 = 9 \ \ and \ \ a_3 =25 \ \ and \ \ a_4 = 49
a_2-a_1 = 9-1 = 8
a_3-a_2 = 25-9=16
We can clearly see that the difference between terms are not equal
Hence, given series is not in AP

Q4 (xv) Which of the following are APs ? If they form an AP, find the common difference d and write three more terms. \small 1^2,5^2,7^2,73,...

Answer:

Given series is
\small 1^2,5^2,7^2,73,...
we can rewrite it as
1,25,49,73....
Now,
the first term to this series is = 1
Now,
a_1 =1 \ \ and \ \ a_2 = 25 \ \ and \ \ a_3 =49 \ \ and \ \ a_4 = 73
a_2-a_1 = 25-1 = 24
a_3-a_2 = 49-25=24
a_4-a_3 = 73-49=24
We can clearly see that the difference between terms are equal and equal to 24
Hence, given series is in AP
Now, the next three terms are
a_5=a_4+d = 73+24=97
a_6=a_5+d = 97+24=121
a_7=a_6+d = 121+24=145
Therefore, the next three terms of given series are 97,121,145

More About NCERT Solutions for Class 10 Maths Exercise 5.2

The problems in exercise 5.1 for Class 10 Maths are divided into four categories. Firstly we have a very basic question we have been given scenarios and we have to tell whether it is A.P or not then we have direct questions in which we have been given first term and common difference and we have to find the A.P series NCERT solutions for Class 10 Maths exercise 5.1 also consist of question in which we have to find first term and common difference and also we have question-based on finding whether series is AP or not and if it has to write more terms. All sorts of fundamental problems based on the AP general equation, first term, and common difference are covered in exercise 5.1 Class 10 Maths. Students can quickly go through the Arithmetic Progressions Class 10 Notes to revise all concepts all together.

Benefits of NCERT Solutions for Class 10 Maths Exercise 5.1

  • Class 10 Mathematics chapter 5 exercise 5.1 covers a broad range of topics that will be covered in greater depth in Class 10 Maths chapter 5 exercise 5.2.
  • NCERT Class 10 Maths chapter 5 exercise 5.1, will be helpful in Class 11 Maths chapter 9 sequences and series

  • Exercise 5.1 Class 10 Maths, is based on the general idea about arithmetic progression, which is important for building other concepts of the chapter.

Also see-

NCERT Solutions of Class 10 Subject Wise

Subject Wise NCERT Exemplar Solutions

Frequently Asked Questions (FAQs)

1. What do you mean by arithmetic progression according to NCERT solutions for Class 10 Maths 5 exercise 5.1?

Arithmetic progression is a series of numbers in which each term is obtained by adding a constant number to the next term except the first term.

2. What is general equation A.P series according to NCERT solutions for Class 10 Maths 5 exercise 5.1?

The general equation of the A.P series is a, a+d, a+2d and so on.

3. Give an example of A.P series

A very basic example of A.P series is 1,2,3,4 in which the first term is 1 and the common difference is also 1.

4. According to NCERT solutions for Class 10 Maths 5 exercise 5.1 what can be the value of common difference ?

Common differences can be zero, positive and even negative.

5. Is 2,2,2,2 is considered an A.P series.

Yes, it is considered A.P series as in exercise 5.1 Class 10 Maths it is clearly stated common difference can be zero so it is an AP series with a common difference zero.

6. What is the difference between finite and infinite A.P series?

The finite A.P series has limited-term while the infinite A.P series has unlimited terms and it never ends.

7. What is the number of solved examples before the exercise 5.1 Class 10 Maths?

Before the Class 10 Mathematics chapter 5 exercise 5.1, there are two primary questions that must be answered.

8. In the NCERT answers for Class 10 Maths chapter 5 exercise 5.1, how many questions are covered?

There are 4 questions in Class 10th Maths chapter exercise 5.1. Question 1 consist of four subparts.  

Question 2 consists of 5 subparts. Question 3 consists of four subparts. Question 1 consists of 15 subparts 

Also Read

Articles

Upcoming School Exams

Himachal Pradesh Board of Secondary Education 12th Examination

Application Date:09 September,2024 - 14 November,2024

Himachal Pradesh Board of Secondary Education 10th Examination

Application Date:09 September,2024 - 14 November,2024

National Institute of Open Schooling 10th examination

Admit Card Date:04 October,2024 - 29 November,2024

National Institute of Open Schooling 12th Examination

Admit Card Date:04 October,2024 - 29 November,2024

National Means Cum-Merit Scholarship

Application Date:05 October,2024 - 04 November,2024

View All School Exams

Certifications By Top Providers

Edx
 1113 courses
Coursera
 812 courses
Udemy
 394 courses
Futurelearn
 295 courses
Harvard University, Cambridge
 98 courses
IBM
 83 courses

Explore Top Universities Across Globe

University of Essex, Colchester
  Wivenhoe Park Colchester CO4 3SQ
University College London, London
  Gower Street, London, WC1E 6BT
The University of Edinburgh, Edinburgh
  Old College, South Bridge, Edinburgh, Post Code EH8 9YL
University of Bristol, Bristol
  Beacon House, Queens Road, Bristol, BS8 1QU
University of Delaware, Newark
  Newark, DE 19716
University of Nottingham, Nottingham
  University Park, Nottingham NG7 2RD

Related E-books & Sample Papers

Questions related to CBSE Class 10th

Have a question related to CBSE Class 10th ?
sadhu ashram ramgart road aligarh

If you're looking for directions or steps to reach Sadhu Ashram on Ramgart Road in Aligarh, here’s how you can get there:

Steps to Reach Sadhu Ashram, Ramgart Road, Aligarh:

  1. Starting Point:

    • Determine your starting point in Aligarh or the nearby area.

  2. Use Google Maps:

    • Open Google Maps on your phone or computer.
    • Enter "Sadhu Ashram, Ramgart Road, Aligarh" as your destination.
    • Follow the navigation instructions provided by Google Maps.

  3. By Local Transport:

    • Auto-rickshaw/Taxi: Hire an auto-rickshaw or taxi and inform the driver about your destination. Most local drivers should be familiar with Sadhu Ashram.
    • Bus: Check if there are any local buses that operate on Ramgart Road. Ask the bus conductor if the bus stops near Sadhu Ashram.

  4. Landmarks to Look For:

    • As you approach Ramgart Road, look for local landmarks that might help you confirm you’re on the right path, such as known shops, temples, or schools nearby.

  5. Ask for Directions:

    • If you're unsure, ask locals for directions to Sadhu Ashram on Ramgart Road. It's a known location in the area.

  6. Final Destination:

    • Once you reach Ramgart Road, Sadhu Ashram should be easy to spot. Look for any signage or ask nearby people to guide you to the exact location.

If you need detailed directions from a specific location or more information about Sadhu Ashram, feel free to ask

Read Complete Answer
sai raja 01 September,2024
show the previous year exams ?

Hello Aspirant,  Hope your doing great,  your question was incomplete and regarding  what exam your asking.

Read Complete Answer
SWETCHCHA MISKA 10 July,2024
as an icse student in class 10 is above 80 percent good enough to get into commerce in 11th cbse

Yes, scoring above 80% in ICSE Class 10 exams typically meets the requirements to get into the Commerce stream in Class 11th under the CBSE board . Admission criteria can vary between schools, so it is advisable to check the specific requirements of the intended CBSE school. Generally, a good academic record with a score above 80% in ICSE 10th result is considered strong for such transitions.

Read Complete Answer
Manasvini Gupta 23 July,2024
i had cleared 10th from cbse in 2022 2 years ago now i wish to apply for 12th as private candidate in 2025. can i do so?

hello Zaid,

Yes, you can apply for 12th grade as a private candidate .You will need to follow the registration process and fulfill the eligibility criteria set by CBSE for private candidates.If you haven't given the 11th grade exam ,you would be able to appear for the 12th exam directly without having passed 11th grade. you will need to give certain tests in the school you are getting addmission to prove your eligibilty.

best of luck!

Read Complete Answer
Ashvadha 03 July,2024
i had cleared 10th in 2022 from cbse can i apply for 12th as private candidate this year?

According to cbse norms candidates who have completed class 10th, class 11th, have a gap year or have failed class 12th can appear for admission in 12th class.for admission in cbse board you need to clear your 11th class first and you must have studied from CBSE board or any other recognized and equivalent board/school.

You are not eligible for cbse board but you can still do 12th from nios which allow candidates to take admission in 12th class as a private student without completing 11th.

Read Complete Answer
sadafreen356 03 July,2024
View All

Popular CBSE Class 10th Questions

A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

 Option 2)

0.16\; J
Option 3)

1.00\; J

 Option 4)

0.67\; J
Read More

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

 Option 2)

 6.45×10−3 kg
Option 3)

 9.89×10−3 kg

 Option 4)

12.89×10−3 kg

 
Read More

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

 Option 2)

200 \, \, J - 500 \, \, J
Option 3)

2\times 10^{5}J-3\times 10^{5}J

 Option 4)

20,000 \, \, J - 50,000 \, \, J
Read More

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

 Option 2)

\; K\;
Option 3)

zero\;

 Option 4)

K/4
Read More

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

 Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced
Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

 Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .
Read More

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

 Option 2)

3.125 × 10-2
Option 3)

1.25 × 10-2

 Option 4)

2.5 × 10-2
Read More

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

 Option 2)

increase two fold
Option 3)

remain unchanged

 Option 4)

be a function of the molecular mass of the substance.
Read More

With increase of temperature, which of these changes?

Option 1)

Molality

 Option 2)

Weight fraction of solute
Option 3)

Fraction of solute present in water

 Option 4)

Mole fraction.
Read More

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

 Option 2)

6.023 × 1022
Option 3)

half that in 8 g He

 Option 4)

558.5 × 6.023 × 1023
Read More

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

 Option 2)

more than 3 but less than 6
Option 3)

more than 6 but less than 9

 Option 4)

more than 9
Read More

Colleges After 12th

Applications for Admissions are open.

VMC VIQ Scholarship Test
VMC VIQ Scholarship Test
Apply

Register for Vidyamandir Intellect Quest. Get Scholarship and Cash Rewards.

JEE Main Important Physics formulas
JEE Main Important Physics formulas
Apply

As per latest 2024 syllabus. Physics formulas, equations, & laws of class 11 & 12th chapters

JEE Main Important Chemistry formulas
JEE Main Important Chemistry formulas
Apply

As per latest 2024 syllabus. Chemistry formulas, equations, & laws of class 11 & 12th chapters

TOEFL ® Registrations 2024
TOEFL ® Registrations 2024
Apply

Accepted by more than 11,000 universities in over 150 countries worldwide

Pearson | PTE
Pearson | PTE
Apply

Register now for PTE & Unlock 20% OFF : Use promo code: 'C360SPL20'. Valid till 15th NOV'24! Trusted by 3,500+ universities globally

JEE Main high scoring chapters and topics
JEE Main high scoring chapters and topics
Apply

As per latest 2024 syllabus. Study 40% syllabus and score upto 100% marks in JEE

View All Application Forms
News and Notifications
7 days ago :
October 11, 2024 - 09:11 PM IST :
October 03, 2024 - 11:28 AM IST :
CBSE Board Exam 2025: Class 10, 12 practical exam dates out at cbse.gov.in; passing marks
October 24, 2024 - 09:54 PM IST
CBSE urges schools to complete 100% registration of Class 9, 11 students by October 16; late fee for delays
October 11, 2024 - 08:56 AM IST
CBSE sample papers 2024-25 for Class 10, 12 out at cbseacademic.nic.in; marking scheme
September 05, 2024 - 07:38 PM IST
CBSE conducts surprise inspections at 27 schools in Rajasthan, Delhi to curb menace of dummy schools
September 03, 2024 - 03:02 PM IST
CBSE Class 10 compartment result 2024 declared; 48.83% pass
August 05, 2024 - 03:14 PM IST
CBSE Class 10 compartment result 2024 soon; passing marks, websites to check
August 05, 2024 - 02:44 PM IST
CBSE Class 10 compartment result 2024 expected soon at cbseresults.nic.in; grading system, passing marks
August 04, 2024 - 04:20 PM IST

Stay up-to date with CBSE Class 10th News

Back to top