NCERT Solutions For Class 10 Maths Chapter 5 Arithmetic Progressions
NCERT solutions for class 10 maths chapter 5 Arithmetic Progressions  This chapter covers the type of questions in which succeeding terms are obtained by adding the same number to the preceding terms. The NCERT solutions for class 10 maths chapter 5 Arithmetic Progressions covers all the questions given in the NCERT books for Class 10. These solutions will be helpful in your academics as well as in your higher studies. These NCERT solutions for class 10 provides a proper to the point explanation to each question. You must refer to the NCERT solutions for class 10 maths chapter 5 Arithmetic Progressions while preparing for the exams.
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NCERT solutions for class 10 maths chapter 5 Topics
Checking whether the series is an arithmetic progression or not.
To find the nth term or last term or an arithmetic progression.
Questions based on the sum of an arithmetic progression
Use of linear equation concept in an arithmetic progression
Application of nth term and sum formula
NCERT solutions for class 10 maths chapter 5 Arithmetic Progressions Excercise: 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 for the first km and for each additional .
Answer:
It is given that
Fare for = Rs. 15
And after that for each additional
Now,
Fare for = Fare of first km + Additional fare for 1 km
= Rs. 15 + 8 = Rs 23
Fare for = Fare of first km + Fare of additional second km + Fare of additional third km
= Rs. 23 + 8= Rs 31
Fare of n km =
( We multiplied by n  1 because the first km was fixed and for 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 of the air remaining in the cylinder at a time.
Answer:
It is given that
vacum pump removes 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
Similarly, Quantity removed after 2 ^{ nd } step = Quantity removed in first step Remaining quantity after 1 ^{ st } step
Now,
Remaining quantity after 2 ^{ nd } step would be = Remaining quantity after 1 ^{ st } step  Quantity removed after 2 ^{ nd } step
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 for the first metre and rises by for each subsequent meter.
Answer:
It is given that
Cost of digging of 1st meter = Rs 150
and
rises by 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 is deposited at compound interest at per annum .
Answer:
Amount in the beginning = Rs. 10000
Interest at the end of 1st year at the rate of
is of 10000 =
Therefore, amount at the end of 1st year will be
= 10000 + 800
= 10800
Now,
Interest at the end of 2nd year at rate of
is of 10800 =
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
Answer:
It is given that
Now,
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:
Answer:
It is given that
Now,
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
Answer:
It is given that
Now,
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
Answer:
It is given that
Now,
Therefore, the first four terms of the given series are
Q2 (v) Write first four terms of the AP when the first term a and the common difference d are given as follows
Answer:
It is given that
Now,
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:
Answer:
Given AP series is
Now, first term of this AP series is 3
Therefore,
Firstterm of AP series (a) = 3
Now,
And common difference (d) =
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:
Answer:
Given AP series is
Now, the first term of this AP series is 5
Therefore,
Firstterm of AP series (a) = 5
Now,
And common difference (d) =
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:
Answer:
Given AP series is
Now, the first term of this AP series is
Therefore,
The first term of AP series (a) =
Now,
And common difference (d) =
Therefore, the first term and the common difference is and respectively
Q3 (iv) For the following APs, write the first term and the common difference:
Answer:
Given AP series is
Now, the first term of this AP series is 0.6
Therefore,
Firstterm of AP series (a) = 0.6
Now,
And common difference (d) =
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.
Answer:
Given series is
Now,
the first term to this series is = 2
Now,
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.
Answer:
Given series is
Now,
first term to this series is = 2
Now,
We can clearly see that the difference between terms are equal and equal to
Hence, given series is in AP
Now, the next three terms are
Therefore, next three terms of given series are
Q4 (iii) Which of the following are APs ? If they form an AP, find the common difference d and write three more terms.
Answer:
Given series is
Now,
the first term to this series is = 1.2
Now,
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
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.
Answer:
Given series is
Now,
the first term to this series is = 10
Now,
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
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.
Answer:
Given series is
Now,
the first term to this series is = 3
Now,
We can clearly see that the difference between terms are equal and equal to
Hence, given series is in AP
Now, the next three terms are
Therefore, next three terms of given series are
Q4 (vi) Which of the following are APs? If they form an AP, find the common difference d and write three more terms.
Answer:
Given series is
Now,
the first term to this series is = 0.2
Now,
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.
Answer:
Given series is
Now,
first term to this series is = 0
Now,
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
Therefore, the next three terms of given series are 16,20,24
Answer:
Given series is
Now,
the first term to this series is =
Now,
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
Therefore, the next three terms of given series are
Q4 (ix) Which of the following are APs? If they form an AP, find the common difference d and write three more terms.
Answer:
Given series is
Now,
the first term to this series is = 1
Now,
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.
Answer:
Given series is
Now,
the first term to this series is = a
Now,
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
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.
Answer:
Given series is
Now,
the first term to this series is = a
Now,
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.
Answer:
Given series is
We can rewrite it as
Now,
first term to this series is = a
Now,
We can clearly see that difference between terms are equal and equal to
Hence, given series is in AP
Now, the next three terms are
Therefore, next three terms of given series are
That is the next three terms are
Q4 (xiii) Which of the following are APs? If they form an AP, find the common difference d and write three more terms.
Answer:
Given series is
Now,
the first term to this series is =
Now,
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.
Answer:
Given series is
we can rewrite it as
Now,
the first term to this series is = 1
Now,
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.
Answer:
Given series is
we can rewrite it as
Now,
the first term to this series is = 1
Now,
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
Therefore, the next three terms of given series are 97,121,145
NCERT solutions for class 10 maths chapter 5 Arithmetic Progressions Excercise: 5.2
a  d  n 
 
(i) (ii) (iii) (iv) (v)  7
 3
0  8 10 18
105 
0

Answer:
(i)
It is given that
Now, we know that
Therefore,
(ii) It is given that
Now, we know that
(iii) It is given that
Now, we know that
Therefore,
(iv) It is given that
Now, we know that
Therefore,
(v) It is given that
Now, we know that
Therefore,
Q2 (i) Choose the correct choice in the following and justify: th term of the AP: is
Answer:
Given series is
Here,
and
Now, we know that
It is given that
Therefore,
Therefore, th term of the AP: is 77
Hence, Correct answer is (C)
Q2 (ii) Choose the correct choice in the following and justify : 11th term of the AP: is
Answer:
Given series is
Here,
and
Now, we know that
It is given that
Therefore,
Therefore, 11th term of the AP: is 22
Hence, the Correct answer is (B)
Q3 (i) In the following APs, find the missing terms in the boxes :
Answer:
Given AP series is
Here,
Now, we know that
Now,
Therefore, the missing term is 14
Q3 (ii) In the following APs, find the missing terms in the boxes:
Answer:
Given AP series is
Here,
Now,
Now, we know that
Now,
And
Therefore, missing terms are 18 and 8
AP series is 18,13,8,3
Q3 (iii) In the following APs, find the missing terms in the boxes :
Answer:
Given AP series is
Here,
Now, we know that
Now,
And
Therefore, missing terms are and 8
AP series is
Q3 (iv) In the following APs, find the missing terms in the boxes :
Answer:
Given AP series is
Here,
Now, we know that
Now,
And
And
And
Therefore, missing terms are 2,0,2,4
AP series is 4,2,0,2,4,6
Q3 (v) In the following APs, find the missing terms in the boxes :
Answer:
Given AP series is
Here,
Now,
Now, we know that
Now,
And
And
And
Therefore, missing terms are 53,23,8,7
AP series is 53,38,23,8,7,22
Q4 Which term of the AP : is ?
Answer:
Given AP is
Let suppose that nth term of AP is 78
Here,
And
Now, we know that that
Therefore, value of 16th term of given AP is 78
Q5 (i) Find the number of terms in each of the following APs :
Answer:
Given AP series is
Let's suppose there are n terms in given AP
Then,
And
Now, we know that
Therefore, there are 34 terms in given AP
Q5 (ii) Find the number of terms in each of the following APs :
Answer:
Given AP series is
suppose there are n terms in given AP
Then,
And
Now, we know that
Therefore, there are 27 terms in given AP
Q6 Check whether is a term of the AP :
Answer:
Given AP series is
Here,
And
Now,
suppose 150 is nth term of the given AP
Now, we know that
Value of n is not an integer
Therefore, 150 is not a term of AP
Q7 Find the st term of an AP whose th term is and the th term is .
Answer:
It is given that
th term of an AP is and the th term is
Now,
And
On solving equation (i) and (ii) we will get
Now,
Therefore, 31st terms of given AP is 178
Q8 An AP consists of terms of which rd term is and the last term is . Find the th term.
Answer:
It is given that
AP consists of terms of which rd term is and the last term is
Now,
And
On solving equation (i) and (ii) we will get
Now,
Therefore, 29th term of given AP is 64
Q9 If the rd and the th terms of an AP are and respectively, which term of this AP is zero?
Answer:
It is given that
rd and the th terms of an AP are and respectively
Now,
And
On solving equation (i) and (ii) we will get
Now,
Let nth term of given AP is 0
Then,
Therefore, 5th term of given AP is 0
Q10 The th term of an AP exceeds its th term by . Find the common difference.
Answer:
It is given that
th term of an AP exceeds its th term by
i.e.
Therefore, the common difference of AP is 1
Q11 Which term of the AP : will be more than its th term?
Answer:
Given AP is
Here,
And
Now, let's suppose nth term of given AP is more than its th term
Then,
Therefore, 65th term of given AP is more than its th term
It is given that
Two APs have the same common difference and difference between their th terms is
i.e.
Let common difference of both the AP's is d
Now, difference between 1000th term is
Therefore, difference between 1000th term is 100
Q 13 How many threedigit numbers are divisible by ?
Answer:
We know that the first three digit number divisible by 7 is 105 and last threedigit number divisible by 7 is 994
Therefore,
Let there are n three digit numbers divisible by 7
Now, we know that
Therefore, there are 128 threedigit numbers divisible by 7
Q14 How many multiples of lie between and ?
Answer:
We know that the first number divisible by 4 between 10 to 250 is 12 and last number divisible by 4 is 248
Therefore,
Let there are n numbers divisible by 4
Now, we know that
Therefore, there are 60 numbers between 10 to 250 that are divisible by 4
Q15 For what value of , are the th terms of two APs: and equal?
Answer:
Given two AP's are
and
Let first term and the common difference of two AP's are a , a' and d , d'
And
Now,
Let nth term of both the AP's are equal
Therefore, the 13th term of both the AP's are equal
Q16 Determine the AP whose third term is and the th term exceeds the th term by .
Answer:
It is given that
3rd term of AP is and the th term exceeds the th term by
i.e.
And
Put the value of d in equation (i) we will get
Now, AP with first term = 4 and common difference = 6 is
4,10,16,22,.....
Q17 Find the th term from the last term of the AP : .
Answer:
Given AP is
Here,
And
Let suppose there are n terms in the AP
Now, we know that
So, there are 51 terms in the given AP and 20th term from the last will be 32th term from the starting
Therefore,
Therefore, 20th term from the of given AP is 158
It is given that
sum of the th and th terms of an AP is and the sum of the th and th terms is
i.e.
And
On solving equation (i) and (ii) we will get
Therefore,first three of AP with a = 13 and d = 5 is
13,8,3
Answer:
It is given that
Subba Rao started work at an annual salary of Rs 5000 and received an increment of Rs 200 each year
Therefore,
Let's suppose after n years his salary will be Rs 7000
Now, we know that
Therefore, after 11 years his salary will be Rs 7000
after 11 years, starting from 1995, his salary will reach to 7000, so we have to add 10 in 1995, because these numbers are in years
Thus , 1995+10 = 2005
Answer:
It is given that
Ramkali saved Rs 5 in the first week of a year and then increased her weekly savings by Rs
Therefore,
after th week, her weekly savings become Rs
Now, we know that
Therefore, after 10 weeks her saving will become Rs 20.75
NCERT solutions for class 10 maths chapter 5 Arithmetic Progressions Excercise: 5.3
Q1 (i) Find the sum of the following APs: to terms.
Answer:
Given AP is
to terms
Here,
And
Now, we know that
Therefore, the sum of AP to terms is 245
Q1 (ii) Find the sum of the following APs: to terms.
Answer:
Given AP is
to terms.
Here,
And
Now, we know that
Therefore, the sum of AP to terms. is 180
Q1 (iii) Find the sum of the following APs: to terms.
Answer:
Given AP is
to terms..
Here,
And
Now, we know that
Therefore, the sum of AP to terms. is 5505
Q1 (iv) Find the sum of the following APs: to terms.
Answer:
Given AP is
to terms.
Here,
And
Now, we know that
Therefore, the sum of AP to terms. is
Q2 (i) Find the sums given below :
Answer:
Given AP is
We first need to find the number of terms
Here,
And
Let suppose there are n terms in the AP
Now, we know that
Now, we know that
Therefore, the sum of AP is
Q2 (ii) Find the sums given below :
Answer:
Given AP is
We first need to find the number of terms
Here,
And
Let suppose there are n terms in the AP
Now, we know that
Now, we know that
Therefore, the sum of AP is 286
Q2 (iii) Find the sums given below :
Answer:
Given AP is
We first need to find the number of terms
Here,
And
Let suppose there are n terms in the AP
Now, we know that
Now, we know that
Therefore, the sum of AP is 8930
Q3 (i) In an AP: given , , , find and .
Answer:
It is given that
Let suppose there are n terms in the AP
Now, we know that
Now, we know that
Therefore, the sum of the given AP is 440
Q3 (ii) In an AP: given , , find and .
Answer:
It is given that
Now, we know that
Therefore, the sum of given AP is 273
Q3 (iii) In an AP: given find and .
Answer:
It is given that
Now, we know that
Therefore, the sum of given AP is 246
Q3 (iv) In an AP: given find and
Answer:
It is given that
Now, we know that
On solving equation (i) and (ii) we will get
Now,
Therefore, the value of d and 10th terms is 1 and 8 respectively
Q3 (vi) In an AP: given find and .
Answer:
It is given that
Now, we know that
n can not be negative so the only the value of n is 5
Now,
Therefore, value of n and nth term is 5 and 34 respectively
Q3 (vii) In an AP: given find and .
Answer:
It is given that
Now, we know that
Now, we know that
Now, put this value in (i) we will get
Therefore, value of n and d are respectively
Q3 (viii) In an AP: given find and .
Answer:
It is given that
Now, we know that
Now, we know that
Value of n cannot be negative so the only the value of n is 7
Now, put this value in (i) we will get
a = 8
Therefore, the value of n and a are 7 and 8 respectively
Q3 (ix) In an AP: given find .
Answer:
It is given that
Now, we know that
Therefore, the value of d is 6
Q3 (x) In an AP: given and there are total terms. Find .
Answer:
It is given that
Now, we know that
Now, we know that
Therefore, the value of a is 4
Q4 How many terms of the AP: must be taken to give a sum of ?
Answer:
Given AP is
Here,
And
Now , we know that
Value of n can not be negative so the only the value of n is 12
Therefore, the sum of 12 terms of AP must be taken to give a sum of .
Answer:
It is given that
Now, we know that
Now, we know that
Now, put this value in (i) we will get
Therefore, value of n and d are respectively
Answer:
It is given that
Now, we know that
Now, we know that
Therefore, there are 38 terms and their sun is 6973
Q7 Find the sum of first terms of an AP in which and nd term is .
Answer:
It is given that
Now, we know that
Now, we know that
Therefore, there are 22 terms and their sum is 1661
Q8 Find the sum of first terms of an AP whose second and third terms are and respectively.
Answer:
It is given that
And
Now,
Now, we know that
Therefore, there are 51 terms and their sum is 5610
Q9 If the sum of first terms of an AP is and that of terms is , find the sum of first terms.
Answer:
It is given that
Now, we know that
Similarly,
On solving equation (i) and (ii) we will get
a = 1 and d = 2
Now, the sum of first n terms is
Therefore, the sum of n terms is
Q10 (i) Show that form an AP where an is defined as below : Also find the sum of the first terms.
Answer:
It is given that
We will check values of for different values of n
and so on.
From the above, we can clearly see that this is an AP with the first term(a) equals to 7 and common difference (d) equals to 4
Now, we know that
Therefore, the sum of 15 terms is 525
Q10 (ii) Show that form an AP where an is defined as below : . Also find the sum of the first terms in each case.
Answer:
It is given that
We will check values of for different values of n
and so on.
From the above, we can clearly see that this is an AP with the first term(a) equals to 4 and common difference (d) equals to 5
Now, we know that
Therefore, the sum of 15 terms is 465
Answer:
It is given that
the sum of the first terms of an AP is
Now,
Now, first term is
Therefore, first term is 3
Similarly,
Therefore, sum of first two terms is 4
Now, we know that
Now,
Similarly,
Q12 Find the sum of the first positive integers divisible by .
Answer:
Positive integers divisible by 6 are
6,12,18,...
This is an AP with
Now, we know that
Therefore, sum of the first positive integers divisible by is 4920
Q13 Find the sum of the first multiples of .
Answer:
First 15 multiples of 8 are
8,16,24,...
This is an AP with
Now, we know that
Therefore, sum of the first 15 multiple of 8 is 960
Q14 Find the sum of the odd numbers between and .
Answer:
The odd number between 0 and 50 are
1,3,5,...49
This is an AP with
There are total 25 odd number between 0 and 50
Now, we know that
Therefore, sum of the odd numbers between and 625
Answer:
It is given that
Penalty for delay of completion beyond a certain date is Rs for the first day, Rs for the second day, Rs for the third day and penalty for each succeeding day being Rs more than for the preceding day
We can clearly see that
200,250,300,..... is an AP with
Now, the penalty for 30 days is given by the expression
Therefore, the penalty for 30 days is 27750
Answer:
It is given that
Each price is decreased by 20 rupees,
Therefore, d = 20 and there are total 7 prizes so n = 7 and sum of prize money is Rs 700 so
Let a be the prize money given to the 1st student
Then,
Therefore, the prize given to the first student is Rs 160
Now,
Let is the prize money given to the next 6 students
then,
Therefore, prize money given to 1 to 7 student is 160,140,120,100,80,60.40
Answer:
First there are 12 classes and each class has 3 sections
Since each section of class 1 will plant 1 tree, so 3 trees will be planted by 3 sections of class 1. Thus every class will plant 3 times the number of their class
Similarly,
No. of trees planted by 3 sections of class 1 = 3
No. of trees planted by 3 sections of class 2 = 6
No. of trees planted by 3 sections of class 3 = 9
No. of trees planted by 3 sections of class 4 = 12
Its clearly an AP with first term (a) = 3 and common difference (d) = 3 and total number of classes (n) = 12
Now, number of trees planted by 12 classes is given by
Therefore, number of trees planted by 12 classes is 234
[ Hint : Length of successive semicircles is with centres at respectively.]
Answer:
From the abovegiven figure
Circumference of 1st semicircle
Similarly,
Circumference of 2nd semicircle
Circumference of 3rd semicircle
It is clear that this is an AP with
Now, sum of length of 13 such semicircles is given by
Therefore, sum of length of 13 such semicircles is 143 cm
Answer:
As the rows are going up, the no of logs are decreasing,
We can clearly see that 20, 19, 18, ..., is an AP.
and here
Let suppose 200 logs are arranged in 'n' rows,
Then,
Now,
case (i) n = 25
But number of rows can not be in negative numbers
Therefore, we will reject the value n = 25
case (ii) n = 16
Therefore, the number of rows in which 200 logs are arranged is equal to 5
Answer:
Distance travelled by the competitor in picking and dropping 1st potato
Distance travelled by the competitor in picking and dropping 2nd potato
Distance travelled by the competitor in picking and dropping 3rd potato
and so on
we can clearly see that it is an AP with first term (a) = 10 and common difference (d) = 6
There are 10 potatoes in the line
Therefore, total distance travelled by the competitor in picking and dropping potatoes is
Therefore, the total distance travelled by the competitor in picking and dropping potatoes is 370 m
NCERT solutions for class 10 maths chapter 5 Arithmetic Progressions Excercise: 5.4
Q1 Which term of the AP: is its first negative term? [ Hint : Find for ]
Answer:
Given AP is
Here
Let suppose nth term of the AP is first negative term
Then,
If nth term is negative then
Therefore, first negative term must be 32nd term
Answer:
It is given that sum of third and seventh terms of an AP are and their product is
Now,
And
put value from equation (i) in (ii) we will get
Now,
case (i)
Then,
case (ii)
Then,
Q3 A ladder has rungs cm apart. (see Fig. ). The rungs decrease uniformly in length from cm at the bottom to cm at the top. If the top and the bottom rungs are m apart, what is the length of the wood required for the rungs? [ Hint: Number of rungs
Answer:
It is given that
The total distance between the top and bottom rung
Distance between any two rungs = 25 cm
Total number of rungs =
And it is also given that bottommost rungs is of 45 cm length and topmost is of 25 cm length.As it is given that the length of rungs decrease uniformly, it will form an AP with
Now, we know that
Now, total length of the wood required for the rungs is equal to
Therefore, the total length of the wood required for the rungs is equal to 385 cm
Answer:
It is given that the sum of the numbers of the houses preceding the house numbered is equal to the sum of the numbers of the houses following it
And 1,2,3,.....,49 form an AP with a = 1 and d = 1
Now, we know that
Suppose their exist an n term such that ( n < 49)
Now, according to given conditions
Sum of first n  1 terms of AP = Sum of terms following the nth term
Sum of first n  1 term of AP = Sum of whole AP  Sum of first m terms of AP
i.e.
Given House number are not negative so we reject n = 35
Therefore, the sum of no of houses preceding the house no 35 is equal to the sum of no of houses following the house no 35
Answer:
It is given that
football ground comprises of steps each of which is m long and Each step has a rise of and a tread of
Now,
The volume required to make the first step =
Similarly,
The volume required to make 2nd step =
And
The volume required to make 3rd step =
And so on
We can clearly see that this is an AP with
Now, the total volume of concrete required to build the terrace of 15 such step is
Therefore, the total volume of concrete required to build the terrace of 15 such steps is
About NCERT solutions for class 10 maths chapter 5 Arithmetic Progressions
We can see many patterns in nature such as the arrangement of petals of a flower, the pattern of honeycomb, etc. We follow certain patterns in daily life. For example, Seema puts rupees 1000 into her daughter’s money box when she was one year old and increased the amount by 100 every year. Then the pattern will be 1000,1100,1200,...........for 1st, 2nd, 3rd,................years.
The sum of first n terms is given by
where a is the first term, n is the number of terms in the given A.P and d is a common difference.
NCERT Solutions for Class 10 Maths for Other Chapters
Chapter No.  Chapter Name 
Chapter 1  
Chapter 2  
Chapter 3  NCERT solutions for class 10 maths chapter 3 Pair of Linear Equations in Two Variables 
Chapter 4  NCERT solutions for class 10 maths chapter 4 Quadratic Equations 
Chapter 5  NCERT solutions for class 10 chapter 5 Arithmetic Progressions 
Chapter 6  
Chapter 7  NCERT solutions for class 10 maths chapter 7 Coordinate Geometry 
Chapter 8  NCERT solutions for class 10 maths chapter 8 Introduction to Trigonometry 
Chapter 9  NCERT solutions for class 10 maths chapter 9 Some Applications of Trigonometry 
Chapter 10  
Chapter 11  
Chapter 12  NCERT solutions for class 10 chapter maths chapter 12 Areas Related to Circles 
Chapter 13  NCERT solutions class 10 maths chapter 13 Surface Areas and Volumes 
Chapter 14  
Chapter 15 
Features of NCERT solutions for class 10 maths chapter 5 Arithmetic Progressions
The NCERT solutions cover all the questions given in the exercise and in between the chapters. This makes it really helpful in homework as well as in the board exam preparations.
The Class 10 Maths Chapter 5 NCERT solutions offer a step by step description of the solutions.
These NCERT solutions for class 10 maths chapter 5 Arithmetic Progressions is given in a simple language and are easy to understand.
For NCERT Solutions for Class 10 Maths Chapter 5 PDF download, you can save the webpage of this article.
How to use NCERT solutions for class 10 maths chapter 5 Arithmetic Progressions?
Before practising the NCERT Class 10 maths solutions chapter 5, go through the conceptual theory given in the NCERT textbook.
After going through the conceptual part, jump on to practice exercises available.
While solving the exercises if you are facing problems in any specific question, then take the help of NCERT solutions for class 10 maths chapter 5 Arithmetic Progressions.
Once you have done the practice exercises you can move to previous year questions.
Frequently Asked Question (FAQs)  NCERT Solutions For Class 10 Maths Chapter 5 Arithmetic Progressions
Question: What are the important topics of NCERT class 10 maths chapter 5 arithmetic progressions?
Answer:
Checking whether the series is an arithmetic progression or not, nth term or last term of arithmetic progression, sum of an arithmetic progression, use of linear equation concept in an arithmetic progression and application of nth term and sum of a.p. formulae are the important topics in this chapter.
Question: Can I get the solutions of NCERT from the official website of CBSE?
Answer:
No, CBSE doesn't provide the NCERT solutions for any class or subject.
Question: Is NCERT is enough to prepare for 10th board exams?
Answer:
Yes, NCERT Books are enough to prepare for the board exams.
Question: What is the weightage of the chapter arithmetic progression for CBSE board exam?
Answer:
CBSE does not provide the marks distributions chapterwise but it provides the total weightage of a unit( up to 45 chapters). As per CBSE the total weightage of algebra ( 4 chapters) is 20 marks in the final board.
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Questions related to CBSE Class 10th
Board cbse class 10 Is cancelled and promote all student please sir because of corona do not study
Hello aspirant,
No CBSE and ICSE board didn't not cancelled the 10th and 12th exam.
As they had declared that 10th and 12th class exam will not be cancelled and students will have to give it in offline mode in this pandemic also.
Hope this helps you
All the best for your future
CBSE class 10th malayalam can someone please tell me a good one
Hello Dear,
Malayalam is the mother tongue of those who have their roots in Kerala. It is one of the Dravidian languages that has similarities in words with its other contemporary languages such as Tamil and Telugu. People who speak Malayalam are referred to as Mallus. You can download the previous year papers from the link below:
https://school.careers360.com/download/samplepapers/cbse10thclassmalayalamsolvedsamplepaper2021
Good Luck
When will the Karnataka NTSE Stage 1 Merit list(i.e people who have cleared stage1) and the cut off come out?
DSERT, Karnataka has released the NTSE Karnataka 2021 Result stage 1 on March 9, 2021 on their official website ( dsert.kar.nic.in ). It is available in the form of districtwise marks lists , consisting of registration numbers and marks of all students.
Along with the NTSE stage 1 result, cutoff and merit list of selected candidates is also released.
The students those who have cleared the stage 1 exam are eligible to seat for the stage 2 exam which is going to be held on June 13, 2021 . The scholarship will be given to the students who will pass the stage 2 examination.
To know more about NTSE Karnataka Important Dates, Cutoff, Scholarship visit :
https://www.google.com/amp/s/school.careers360.com/articles/ntsekarnatakaresult/amp
Hope it helps
Thank You !!!
Age criteria to appear in class 10 CBSE exam
Hello Bishal
According to guidelines of CBSE, minimum age to appear for class x must be 14 years. There is no upper limit to appear for class x cbse board.
A candidate can appear for maximum three attempts.
Some candidates give private exams or sometimes students fail in standard ix then they privately appear for class x then their age must be more than 14 years. Sometimes students appear for x class after one year gap of passing class ix then also their age would be 15 or 16 as there is no upper limit age.
My child is supposed to make another attempt at compartment exam of 10th cbse Missed last date for applying. since the boards are postponed to May 4th can I pay her fees and apply now.
Hello sir I'm sorry to inform you but now you're not eligible to fill the registration form as last date to fill the registration Form was 9th December 2020. Yes examination gets postponed to May but portal to fill the registration form is not open , in case if the registration form portal will open again you can fill the registration form and make the payment.
Feel free to comment if you've any doubt
Good luck