5. ARITHMETIC PROGRESSION (SCERT)

SEBA Class10 Chapter 5. ARITHMETIC PROGRESSION

Chapter 5. ARITHMETIC PROGRESSIONS

Chapter 5. Arithmetic Progressions

Exercise 5.1 complete solution

Exercise 5.2 complete solution

Exercise 5.3 complete solution

Exercise 5.4 (Optional) complete solution

1. If are in AP, then

and is called the arithmetic mean of and .

2. The general form of an AP is . where is the first term and is the common difference .

3. Let the first term of an AP by , second term by ,…………, term by and the common difference by . Then the AP becomes .

So, commom difference .

4. The term of the AP with first term and common difference is given by

5. The sum of first positive integers is given by .

6. The sum of the first terms of an AP is given by

If is the last term of the finite AP, say the term, then the sum of all terms of the AP is given by :

7. The term of an AP is the difference of the sum to first terms and the sum to first terms of it, i.e., .

Class 10 Maths Chapter 5. ARITHMETIC PROGRESSIONS Exercise 5.1 Solutions

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

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

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

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

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

Solution : (i) In this statement :

Here , first term and Common difference

So, ,, . Because , the additional term continue increase by the number 8 .Hence , the list forms an A.P.

(ii) In this statement : let be the amount of air present in a cylinder .

When a vacuum pump removes of the air remaining in the cylinder .

First terms

Common difference

Therefore , the list of the form is not an A.P .

(iii) In this statement :

First term and Common difference

Therefore , the list of the form is an AP .

(iv) In this statement : Amount = Rs 10000

First years

Second years

Therefore , the list does not form an A.P .

2. Write first four terms of the AP, when the first term and the common difference are given as follows : (i) (ii) (iii) (iv) (v)

Solution: (i)

terms

(ii)

terms

(iii)

terms

(iv)

terms

(v)

terms

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

(i)

(ii)

(iii)

(iv)

Solution : (i)

First term

Common difference

(ii)

First term

Common difference

(iii)

First term

Common difference

(iv)

First term

Common difference

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

(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

(viii)

(ix)

(x)

(xi)

(xii)

(xiii)

(xiv)

(xv)

Solution : (i)

First term

Common difference

It is not an A.P.

(ii)

First term

Common difference

It is an A.P.

terms

(iii)

First term

Common difference

It is an A.P.

terms

(iv)

First term

Common difference

It is an A.P.

terms

(v)

First term

Common difference

It is an A.P.

terms

(vi)

First term

Common difference

It is not an A.P.

(vii)

First term

Common difference

It is an A.P.

terms

(viii)

First term

Common difference

It is an A.P.

terms

(ix)

First term

Common difference

It is not an A.P.

(x)

First term

Common difference

It is an A.P.

terms

(xi)

First term

Common difference

It is not an A.P.

(xii)

First term

Common difference

It is an A.P.

terms

(xiii)

First term

Common difference

It is not an A.P.

(xiv)

First term

Common difference

It is not an A.P.

(xv)

First term

Common difference

It is an A.P.

terms

Class 10 Maths Chapter 5. ARITHMETIC PROGRESSIONS Exercise 5.2 Solutions

1. Fill in the blanks in the following table , given that is the first term , the common difference and the term of the AP :

(i)

(ii)

(iii)

(iv)

(v)

– 18

…….

– 18.9

3.5

…….

– 3

2.5

…….

105

……

– 5

3.6

……

Solution : (i) Here ,

We know that ,

(ii) Here ,

We know that ,

Therefore, the common difference is 2 .

(iii) Here ,

We know that ,

Therefore, the first teram is 46 .

(iv) Here ,

We know that ,

(v) Here ,

We know that ,

2. Choose the correct choice in the following and justify :

(i) term of the AP : is :

(A) 97 (B) 77 (C) – 77 (D) – 87

Solution : (C) – 77

[ Here ,

We know that ,

]

(ii) term of the A.P : is :

(A) 28 (B) 22 (C) – 38 (D)

Solution : (B) 22

[ Here ,

We know that ,

]

3. In the following APs , find the missing terms in the boxes :

(i) (ii) (iii) (iv) (v)

Solution : (i)

Here ,

term

(ii)

Let and be the first term and common difference of an AP respectively .

And

Putting in , we have

First term and terms

(iii)

Here,

terms and terms

(iv)

Here,

terms , terms ,

terms and terms

(v)

Let and be the first term and common difference of an AP respectively .

And

Putting in , we have

First term ,

terms ,

and terms

4. Which term of the AP : is 78 ?

Solution : Here ,

We know that ,

5. Find the number of terms in each of the following APs:

(i) (ii)

Solution : (i)

Here ,

We know that,

(ii)

Solution : Here,

We know that,

6. Check whether is a term of the AP :

Solution : Here,

We know that,

(Impossible , because is not a fraction number)

So, is not a term of the AP .

7. Find the term of an AP whose term is 38 and the term is 73 .

Solution : Let and be the first term and common difference of an AP respectively .

A/Q ,

And

Putting in equation , we have

8. An AP consists of 50 terms of which term is 12 and the last term is 106 . Find the term .

Solution : Let and be the first term and common difference of an AP respectively .

A/Q ,

and

Putting in equation , we have

9. If the and the terms of an AP are 4 and – 8 respectively, which term of this AP is zero ?

Solution : Let and be the first term and common difference of an AP respectively .

A/Q ,

and

Putting in equation , we have

Therefore, term of an AP is zero .

10. The term of an AP exceeds its term by 7 . Find the common difference .

Solution : Let and be the first term and common difference of an AP respectively .

A/Q ,

Therefore, the common difference is 1 .

11. Which term of the AP : will be 132 more than its term ?

Solution : Here ,

A/Q ,

Therefore, term of the AP is 132 more than its term .

12. Two APs have the same common difference . The difference between their term is 100 , what is the difference between their terms ?

Solution : Let and' are two First terms and is the common difference of the two AP respectively .

A/Q ,

Therefore, the difference between their terms is 100 .

13. How many three-digit numbers are divisible by 7 ?

Solution : The list of two-digit numbers divisible by 7 is :

Here ,

We know that ,

So , there are 128 three-digit numbers divisible by 7 .

14. How many multiples of 4 lie between 10 and 250 ?

Solution : The list of the multiples of 4 between 10 and 250 are :

Here ,

We know that ,

So , there are 60 numbers multiple of 4 lie between 10 and 250 .

15. For what value of , are the terms of two APs : and equal ?

Solution :

16. Determine the AP whose third term is 16 and the term exceeds the term by 12 .

Solution : Let and be the first term and common difference of an AP respectively .

A/Q ,

and

Putting in equation i , we have

Required the AP is :

i.e.,

17. Find the term from the last term of the AP : .

Solution : Here,

term from the end

[ Note : term from the end ]

18. The sum of the and terms of an AP is 24 and the sum of the and terms is 44 . Find the first three terms of the AP.

Solution : Let and be the first term and common difference of an AP respectively .

A/Q ,

and

Putting in equation , we have

Therefore, three terms of an AP’s are : .

19. 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 ?

Solution : Here ,

We know that ,

20 . Ramkali saved Rs 5 in the first week of a year and then increased her weekly savings by Rs 1.75 . If in the week , her weekly saving become Rs 20.75 , find .

Solution : Here ,

We know that,

Class 10 Maths Chapter 5. ARITHMETIC PROGRESSIONS Exercise 5.3 Solutions

1. Find the sum of the following APs :

(i) to 10 terms .

Solution : Here ,

We know that,

(ii) to 12 terms .

Solution : Here ,

We know that,

(iii) to 100 terms .

Solution : Here ,

We know that,

(iv) to 11 terms .

Solution : Here ,

We know that,

2. Find the sums given below :

(i)

Solution : Here,

and

We know that ,

We have,

(ii)

Solution :

Here ,

We know that ,

and

(iii)

Solution : Here ,

We know that ,

and

3. In an AP :

(i) given , find and .

Solution : We know that ,

and

(ii) given, , find and .

Solution : We know that ,

and

(iii) given , , find and .

Solution : We know that ,

and

(iv) given, , find and .

Solution : We have ,

and

[ From ]

Putting the value of in , we get

(v) given,, find and .

Solution: We know that,

Given ,

(vi) given, , find and .

Solution : We have ,

[Impossible]

and

(vii) given, , find and .

Solution : We have ,

and

[ From ]

From , we get,

(viii) given, , find and .

Solution : We have ,

and

[impossible]

From , we get

(ix) given, , find .

Solution : We have,

(x) given, , and there are total 9 terms . Find .

Solution : We know that,

4. How many terms of the AP : must be taken to give a sum of 636 ?

Solution: Here, , ,

We know that,

(Impossible)

Therefore, the number of terms is 12 .

5. The first term of an AP is 5 , the last term is 45 and the sum is 400 . Find the number of terms and the common difference .

Solution : Let and be the number of terms and the common difference respectively .

and

[ From ]

From , we get

6. The first and the last terms of an AP are 17 and 350 respectively . If the common difference is 9, how many terms are there and what is their sum ?

Solution : Here ,

7. Find the sum of first 22 terms of an AP in which and 22nd term is 149 .

Solution : Here ,

8. Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively .

Solution : Let and be the first term and common difference respectively .

A/Q ,

and

Putting he value of in , we get

Again,

9. If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289 , find the sum of first terms .

Solution : Let and be the first term and common difference of an AP respectively .

We know that,

and

Putting the value of in , we get

Again,

10. Show that form an AP where is defined as below :

(i) (ii)

Also find the sum of the first 15 terms in each case .

Solution : (i) We have ,

Here ,

(ii) We have,

Here,

Now,

11. 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 3rd , the 10th and the terms .

Solution : We have ,

We have ,

12. Find the sum of the first 40 positive integers divisible by 6 .

Solution : The A.P’s are : 6 , 12 , 18 , 24 ,………….

Here ,

13. Find the sum of the first 15 multiples of 8 .

Solution: The AP’s are : 8 , 16 , 24 , …………………. .

Here, , ,

We know that,

14. Find the sum of the odd numbers between 0 and 50 .

Solution : The A.P’s are : 1 , 3 , 5 , 7 , …………… , 49

Here,

A/Q ,

15. A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows : Rs 200 for the day , Rs 250 for the second day , Rs 300 for the third day , etc., the penalty for each succeeding day being Rs 50 more than for the preceding day . How much money the contractor has to pay as penalty , if he has delayed the work by 30 days ?

Solution : Here, , and

We know that ,

Therefore, if the contractor has delayed the work by 30 days, the total penalty amount to be paid would be 27,750 rupees.

16. A sum of Rs 700 is to be used to give seven cash prizes to students of a school for their overall academic performance . If each prize is Rs 20 less than its preceding prize, find the value of each of the prizes .

Solution : Let the first prize be .

Then the list of prize of an A.P’s. are :

We know that ,

So, the first prize () = Rs 160.

The subsequent prizes are:

160 – 20 = 140 , 140 – 20 = 120 , 120 – 20 =100 , 100 – 20 = 80 , 80 – 20 = 60 , 60 – 20 = 40

The values of the seven prizes are: 160, 140, 120, 100, 80, 60, and 40 rupees.

17. 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 1 tree , a section of Class II will plant 2 trees and so on till Class XII . There are three sections of each class . How many trees will be planted by the students ?

Solution : The total number of trees planted by each class will be the sum of the arithmetic series from 1 to 12 (as the classes range from Class 1 to Class 12).

Here, First term (for Class 1)

Last term (for Class 12)

Number of terms (classes) ,

We know that ,

This represents the number of trees planted by each class. Since there are three sections in each class, then

The total trees planted by each class = 78 × 3 = 234

Therefore, all the students in the school will plant 234 trees per class, leading to a collective planting of 234 trees by all sections in each class from Class 1 to Class 12

18. A spiral is made up of successive semicircles , with centres alternately at A and B ,starting with centre at A , of radii 0.5 cm, 1.0 cm , 1.5 cm , 2.0 cm , ……….. as shown in Fig. 5.4 . What is the total length of such a spiral made up of thirteen consecutive semicircle ? (Take )

[ Hint : Length of successive semicircles is with centres at A , B , A , B , ……., respectively]

Solution : The length of a semi circle's circumference

, , , .

Here, , and

We know that ,

Therefore, the total length of the spiral made up of thirteen consecutive semicircles with radii increasing by 0.5 cm each is 143 centimeters.

19. 200 logs are stacked in the following manner : 20 logs in the bottom row ,19 in the next row , 18 in the row next to it and so on (see Fig. 5.5) . In how many rows are the 200logs placed and how many logs are in the top row ?

Solution : Here, , and

We know that ,

For

(impossible)

For

So, there are 16 rows of logs, and the top row has 5 logs.

20. In a potato race ,a bucket is placed at the starting point , which is 5 m from the first potato , and the other potatoes are placed 3 m apart in a straight line . There are ten potatoes in the line (see Fig. 5.6)

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 2×5 + 2×(5+3)]

Solution : For the first potato:

The competitor runs 5 meters to pick it up and then 5 meters back to the bucket.

Total distance = 2 × 5 .

For the second potato:

The competitor runs 5 meters to pick it up, then 5 meters back to the bucket for the first potato, and an additional 3 meters to pick up the second potato.

So, the total distance is 2 × (5 + 3) .

All subsequent potatoes are : , , , …………. ,

i.e., 10 , 16 , 22 , …………. , 64

Here , , and

We know that,

Therefore, the total distance the competitor has to run to pick up all ten potatoes and return them to the bucket is 370 meters.

**Class 10 Maths Chapter 5. ARITHMETIC PROGRESSIONS Exercise 5.4 (Optional)* Solutions**

1. Which term of the AP : 121, 117, 113, . . ., is its first negative term? [Hint : Find for ]

Solution : Here, and

We know that ,

∴

(n always whole number, except 0)

The first negative term, therefore, occurs at n = 32 . This is the 32nd term of the sequence.
2. The sum of the third and the seventh terms of an AP is 6 and their product is 8. Find the sum of first sixteen terms of the AP.

Solution : let , and are first term and common difference of an A.P respectively .

A/Q,

and

From and , we get

Putting in (i) Eq., then

3. A ladder has rungs 25 cm apart. (see Fig. 5.7). The rungs decrease uniformly in length from 45 cm at the bottom to 25 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 ]

Solution : Given ,the top and bottom rungs are 2.5 meters apart.

The rungs decrease uniformly from 45 cm at the bottom to 25 cm at the top.

The total length spanned by the ladder in centimeters

The gap between each rung is 25 cm.

We have , Number of rungs

Here, and

We know that,

Therefore, the total length of wood required for the rungs on the ladder is 385 cm .

4. The houses of a row are numbered consecutively from 1 to 49. Show that there is a value of x such that the sum of the numbers of the houses preceding the house numbered x is equal to the sum of the numbers of the houses following it. Find this value of . [Hint : ]

Solution: Given , the houses are numbered from 1 to 49 .

We know that ,

Here, ,

And

A/Q,

(only positive value)

Therefore, the value of such that the sum of the numbers of the houses preceding is equal to the sum of the numbers of the houses following it is 35 .

5. A small terrace at a football ground comprises of 15 steps each of which is 50 m long and built of solid concrete. Each step has a rise of m and a tread of m. (see Fig. 5.8). Calculate the total volume
of concrete required to build the terrace. [Hint : Volume of concrete required to build the first step ]

Solution : Volume of concrete required to build the first step

Volume of concrete required to build the second step

Volume of concrete required to build the 3rd step

The list of the volume of concrete required to build the terrace an A.P.’s :

Here, , and

We know that ,

Hence, the correct total volume of concrete required to build the terrace of 15 steps is