# NCERT Solutions For Class 10 Maths Chapter 5 Arithmetic Progressions

##### NCERT Solutions For Class 10 Maths Chapter 5 Arithmetic Progressions

Class 10 Maths NCERT Solutions Chapter 5 Arithmetic Progressions – This Class 10 Maths Chapter 5 contains issues where the subsequent terms are derived by adding the same number to the preceding terms. The NCERT solutions for Class 10 Maths chapter 5 Arithmetic Progressions cover all of the class 10 exercise questions. Arithmetic Progression Class 10 solutions can aid you in both your academics and your future pursuits. These NCERT solutions for Class 10 provide a concise explanation for each question. While studying for the tests, you should consult the Arithmetic Progression Class 10 NCERT solutions.

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  .

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.

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.

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 .

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

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:

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

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

It is given that

Now,

Therefore, the first four terms of the given series are

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:

Given AP series is

Now, first term of this AP series is 3

Therefore,

First-term 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:

Given AP series is

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

Therefore,

First-term of AP series (a) = -5

Now,

And common difference (d) =

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

Q3 (iii) For the following APs, write the first term and the common difference:

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:

Given AP series is

Now, the first term of this AP series is 0.6

Therefore,

First-term 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.

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

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.

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.

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.

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.

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.

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

Q4 (viii) Which of the following are APs? If they form an AP, find the common difference d and write three more terms.

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.

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.

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.

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.

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.

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.

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.

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

Q1 Fill in the blanks in the following table, given that a is the first term, d the common difference and  the  th term of the AP:

(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

(A)  (B)  (C)  (D)

Given series is

Here,
and

Now, we know that

It is given that
Therefore,

Therefore,  th term of the AP:  is -77

Q2 (ii) Choose the correct choice in the following and justify : 11th term of the AP:  is

(A)  (B)  (C)  (D)

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 :

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:

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 :

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 :

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 :

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  ?

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 :

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 :

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 :

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  .

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.

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?

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.

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?

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

Q12 Two APs have the same common difference. The difference between their  th terms is  , what is the difference between their  th terms?

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 three-digit numbers are divisible by  ?

We know that the first three digit number divisible by 7 is 105 and last three-digit 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 three-digit numbers divisible by 7

Q14 How many multiples of  lie between  and  ?

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?

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  .

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 :  .

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

Q18 The sum of the  th and  th terms of an AP is  and the sum of the  th and  th terms is  Find the first three terms of the AP.

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

Q19 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?

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

Q20 Ramkali saved Rs 5 in the first week of a year and then increased her weekly savings by Rs  . If in the  th week, her weekly savings become Rs  , find

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.

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.

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.

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.

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 :

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 :

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 :

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  .

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  .

It is given that

Now, we know that

Therefore, the sum of given AP is 273

Q3 (iii) In an AP: given  find  and  .

It is given that

Now, we know that

Therefore, the sum of given AP is 246

Q3 (iv) In an AP: given  find  and

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 (v) In an AP: given  , find  and  .

It is given that

Now, we know that

Now,

Q3 (vi) In an AP: given  find  and  .

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  .

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  .

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 and -8 respectively

Q3 (ix) In an AP: given  find  .

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  .

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  ?

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  .

Q5 The first term of an AP is  , the last term is  and the sum is  . Find the number of terms and the common difference.

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

Q6 The first and the last terms of an AP are  and  respectively. If the common difference is  , how many terms are there and what is their sum?

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  .

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.

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.

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.

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 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.

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 and common difference (d) equals to -5
Now, we know that

Therefore, the sum of 15 terms is -465

Q11 If the sum of the first  terms of an AP is  , what is the first term (that is  )? What is the sum of first two terms? What is the second term? Similarly, find the  rd, the  th and the  th terms

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  .

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  .

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  .

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

Q15 A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows: Rs  for the first day, Rs  for the second day, Rs  for the third day, etc., the penalty for each succeeding day being Rs  more than for the preceding day. How much money the contractor has to pay a penalty, if he has delayed the work by  days?

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

Q16 A sum of Rs  is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is Rs  less than its preceding prize, find the value of each of the prizes.

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

Q17 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  tree, a section of Class II will plant  trees and so on till Class XII. There are three sections in each class. How many trees will be planted by the students?

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) and common difference (d) 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

Q18 A spiral is made up of successive semicircles, with centres alternately at  and  ??????, starting with centre at  , of radii  as shown in Fig.  . What is the total length of such a spiral made up of thirteen consecutive semicircles? (Take  )

Hint : Length of successive semicircles is  with centres at  respectively.]

From the above-given 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

Q19  logs are stacked in the following manner:  logs in the bottom row,  in the next row,  in the row next to it and so on (see Fig.  ). In how many rows are the  logs placed and how many logs are in the top row?

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

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  ]

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  ]

Given AP is

Here
Let suppose nth term of the AP is first negative term
Then,

If nth term is negative then

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Therefore, first negative term must be 32nd term

Q2 The sum of the third and the seventh terms of an AP is  and their product is  . Find the sum of first sixteen terms of the AP.

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

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 bottom-most 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

Q4 The houses of a row are numbered consecutively from  to  . Show that there is a value of  such 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. Find this value of  . [ Hint :  ]

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

Q5 A small terrace at a football ground comprises of  steps each of which is  m long and built of solid concrete. Each step has a rise of  and a tread of  . (see Fig.  ). Calculate the total volume of concrete required to build the terrace.
Hint: Volume of concrete required to build the first step  ]

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