(1) An Arithmetic Progression (AP) is a list of numbers in which each term is obtained by adding a fixed number ‘d’ to the preceding term, except the first term. The fixed number is called the common difference.

The general term of an AP is

a, a+d, a+2d, a +3d, ……….

(2) A given list of numbers a1, a2, a3 ………………… is an AP if the differences

a2 - a1, a3 – a2, a4 – a3, ………………… give the same value.

In other words, if a k+1 – ak is the same for different values of k, the term is an AP

(3) In an AP with first term a and common difference d, the nth term is given by

an = a + (n – 1 ) d

(4) The sum of the first n terms of an AP:

(5) If l is the last term of the finite AP, say the nth term, then the sum of all terms of the AP is given by:

## Solution of Exercise 5.1 (NCERT)

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

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

Solution:

Fare for 1st km = Rs. 15

Fare for 2nd km = Rs. 15 + 8 = Rs 23

Fare for 3rd km = Rs. 23 + 8 = 31

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

Hence, it is an AP

Fare for 1st km = Rs. 15

Fare for 2nd km = Rs. 15 + 8 = Rs 23

Fare for 3rd km = Rs. 23 + 8 = 31

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

Hence, it is an AP

(ii) The amount of air present in the cylinder when a vacuum pump removes ¼ of the air remaining in the cylinder at a time.

Solution:

Let us assume, initial quantity of air = 1

Here each subsequent term is not obtained by adding a fixed number to the previous term.

Hence, it is not an AP.

Let us assume, initial quantity of air = 1

Here each subsequent term is not obtained by adding a fixed number to the previous term.

Hence, it is not an AP.

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

Solution:

Cost of digging of 1st meter = 150

Cost of digging of 2nd meter = 150 + 50 = 200

Cost of digging of 3rd meter = 200 + 50 = 250

Here, each subsequent term is obtained by adding a fixed number (50) to the previous term.

Hence, it is an AP.

Cost of digging of 1st meter = 150

Cost of digging of 2nd meter = 150 + 50 = 200

Cost of digging of 3rd meter = 200 + 50 = 250

Here, each subsequent term is obtained by adding a fixed number (50) to the previous term.

Hence, it is an AP.

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

Solution:

Amount in the beginning = Rs. 10000

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

Question: 2 – Write first four terms of AP, when the first term a and the common difference d are given as follows:

(i) a = 10, d = 10

Solution:

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

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

## Solution of Exercise 5.2 (NCERT)

Question: 1 – Fill in the blanks in the following table, given that a is the first term, d the common difference and an is the nth term of the AP.

Question: 2 – Choose the correct choice in the following and justify:

(i) 30th term of the AP: 10, 7, 4, ……….

(A) 97 (B) 77 (C) – 77 (D) – 87

Solution:

Here, a = 10, d = – 3 and n = 30

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

Question: 4 – Which term of the AP: 3, 8, 13, 18, ……………… is 78?

Solution:

Question: 5 – Find the number of terms in each of the following APs

Question: 6 – Check whether – 150 is a term of the AP; 11, 8, 5, 2, ……………

Quesiton: 7 – Find the 31st term of an AP whose 11th term is 38 and 16th term is 73

Question: 8 – An AP consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term.

Question: 9 – If the 3rd and the 9th term of an AP are 4 and – 8 respectively. Which term of this AP is zero?

Question: 10 – The 17th term of an AP exceeds its 10th term by 7. Find the common difference

Question: 11 – Which term of the AP: 3, 15, 27, 39, …. Will be 132 more than its 54th term

Question: 12 – Two APs have the same common difference. The difference between their 100th terms is 100, what is the difference between their 1000th terms?

Solution:

Since, both the APs have same common difference, thus in this case the difference between each corresponding terms will be 100.

Question: 13 – How many three digit numbers are divisible by 7?

Solution:

Since, 100 is the smallest three digit number and it gives a reminder of 2 when divided by 7, therefore, 105 is the smallest three digit number which is divisible by 7.

Since, 999 is greatest three digit number, and it gives a reminder of 5, thus 999 – 5 = 994 will be the greatest three digit number which is divisible by 7.

Therefore, here we have,

First term (a) = 105,

The last term (n) = 994

The common difference = 7

Thus, there are 128 three digit numbers which are divisible by 7.

Since, 100 is the smallest three digit number and it gives a reminder of 2 when divided by 7, therefore, 105 is the smallest three digit number which is divisible by 7.

Since, 999 is greatest three digit number, and it gives a reminder of 5, thus 999 – 5 = 994 will be the greatest three digit number which is divisible by 7.

Therefore, here we have,

First term (a) = 105,

The last term (n) = 994

The common difference = 7

Thus, there are 128 three digit numbers which are divisible by 7.

Question: 14 – How many multiples of 4 lie between 10 and 250?

Solution:

12 is the first number after 10 which is divisible by 4.

Since, 250 gives a remainder of 2 when divided by 4, thus 250 – 2 = 248 is the greatest number less than 250 which is divisible by 4

Here, we have first term (a) = 12, last term (n) = 248 and common difference (d) = 4

Thus, number of terms (n) =?

Thus, there are 60 numbers between 10 and 250 that are divisible by 4.

12 is the first number after 10 which is divisible by 4.

Since, 250 gives a remainder of 2 when divided by 4, thus 250 – 2 = 248 is the greatest number less than 250 which is divisible by 4

Here, we have first term (a) = 12, last term (n) = 248 and common difference (d) = 4

Thus, number of terms (n) =?

Thus, there are 60 numbers between 10 and 250 that are divisible by 4.

Question: 15 – For what value of n, are the nth terms of two APs; 63, 65, 67, ………. and 3, 10, 17, ……. equal.

Solution:

In first AP: a = 63, d = 2

In second AP: a = 3, d = 7

As per question:

63 + (n – 1) 2 = 3 + (n – 1) 7

⇒ 63 – 3 + (n – 1) 2 = (n – 1) 7

⇒ 60 + 2n – 2 = 7n – 7

⇒ 2n + 58 = 7n – 7

⇒ 2n + 58 + 7 = 7n

⇒ 2n + 65 = 7n

⇒ 7n – 2n = 65

⇒ 5n = 65

⇒ n = 65/5 = 13

Thus, for the 13 value of n, nth term of given two APs will be equal

## Solution of Exercise 5.2 (NCERT) Part -6

16. Determine the AP whose third term is 16 and the 7th term exceeds the 5th term by 12.

Solution:

Therefore, AP: 4, 16, 28, 40, ………………….

17. Find the 20th term from the last term of the AP; 3, 8, 13, …………….253.

Solution: a = 3, d = 5

Now, 253 = a + (n + 1) d

⇒ 253 = 3 + (n -1) x 5

⇒ 253 = 3 + 5n – 5 = – 2

⇒ 5n = 253 + 2 = 255

⇒ n = 255/5 = 51

There 20th terms from the last term = 51 – 19 = 32

Thus, 20th term is 158

Now, 253 = a + (n + 1) d

⇒ 253 = 3 + (n -1) x 5

⇒ 253 = 3 + 5n – 5 = – 2

⇒ 5n = 253 + 2 = 255

⇒ n = 255/5 = 51

There 20th terms from the last term = 51 – 19 = 32

Thus, 20th term is 158

Question: 18 – The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms of the AP.

Solution:

Hence, first three terms of AP: – 13, – 8, – 3

Question: 19 – Subha 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:

Question: 20 – Ramkali save Rs. 5 in the first week of a year and then increased her weekly savings by Rs. 1.75. If in the nth week, her weekly savings become Rs. 20.75, find n.

Solution:

## Solution of Exercise 5.3 (NCERT

Question: 1 – Find the sum of the following APs:

(i) 2, 7, 1, 2, ………………….. to 10 terms

Solution:

We know that sum of the nth term,

Thus, sum of the 10 terms of given AP (Sn)=245

(ii) – 37, – 33, – 29, ………………..to 12 terms

Solution:

We know that sum of the nth term,

Thus, sum of the 12 terms of given AP (Sn)= – 180

(iii) 0.6, 1.7, 2.8, ……………… to 100 terms

Solution:

We know that sum of the nth term,

Thus, sum of the 100 term of given AP (Sn)= 5505

Question: 2 – Find the sums given below:

Question: 3 - In an AP:

(a) Given a = 5, d = 3, an = 50, find n and Sn.

Answer: We know;

Now, sum of n terms can be calculated as follows:

(b) Given a = 7, a13 = 35, find d and S13.

Answer: We know;

Now, sum of n terms can be calculated as follows:

(c) Given a12 = 37, d = 3, find a and S12.

Answer: We can find a as follows:

Now, sum of n terms can be calculated as follows:

(d) Given a3 = 15, S10 = 125, find d and a10.

Answer: We know;

As per question; the 3rd term is 15, which means;

(e) Given d = 5, S9 = 75, find a and a9.

Answer: We know;

Now, the 9th term can be calculated as follows:

(f) Given a = 2, d = 8, Sn = 90, find n and an.

Answer: We know;

Now, 5th term can be calculated as follows:

(g) Given a = 8, an = 62, Sn = 210, find n and d.

Answer: We know;

We have,

(h) Given an = 4, d = 2 Sn = -14, find n and a.

Answer: We know;

We know;

From equation (1) and (2);

Since a

_{n}and d are positive; so need to take the negative value of a, i.e. – 8.
Using this, we can find the number of terms as follows:

(i) Given a = 3, n = 8, S = 192, find d.

Answer: We know;

(j) Given l = 28, S = 144, and there are total 9 terms, find a.

Answer: We know;

4. How many terms of the AP : 9, 17, 25, . . . must be taken to give a sum of 636?

Answer: Here; a = 9, d = 8 and Sn = 636

We know;

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.

Answer: We know;

We know;

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?

Answer: We have; a = 17, an = 350 and d = 9

We know;

The sum can be calculated as follows:

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

Answer: We have; n = 22, d = 7 and a22 = 149

We know;

The sum can be calculated as follows:

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

Answer: We have; a2 = 14, a3 = 18 and n = 51

Here;

We can find the sun of 51 terms as follows:

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 n terms.

Answer: Here;

Similarly;

Subtracting the 4th term from 9th term; we get

Using the value of d in the equation for 4th term; we can find a as follows:

Using the values of a and d; we can find the sum of first n terms as follows:

10. Show that a1, a2, …., an, …. Form an AP where an is defined as below:

(i) an = 3 + 4n

Answer: Let us take different values for a, i.e. 1, 2, 3 and so on

Here; each subsequent member of the series is increasing by 4 and hence it is an AP.

(ii) an = 9 – 5n

Answer: Let us take different values for a, i.e. 1, 2, 3 and so on

Here; each subsequent member of the series is decreasing by – 5 and hence it is an AP.

11. If the sum of the first n terms of an AP is 4n – n2, what is the first term (that is S1)? What is the sum of first two terms? What is the second term? Similarly, find the 3rd, the 10th and the nth terms.

Answer: We can find the first term as follows:

Now; sum of first two terms can be calculated as follows:

Hence; second term = 4 – 3 = 1

And d = 1 – 3 = - 2

So, 3rd term can be calculated as follows:

Similarly, 10th term can be calculated as follows:

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

Answer: The smallest positive integer which is divisible by 6 is 6 itself and its 40th multiple will be 6 x 40 = 240

So, we have a = 6, d = 6, n = 40 and 40th term = 240.

Sum of first 40 positive integers divisible by 6 can be calculated as follows:

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

Answer: We have a = 8, d = 8 and n = 15

Sum of first 15 multiples of 8 can be calculated as follows:

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

Answer: There are 25 odd numbers between 0 and 50

Sum of first n odd numbers = n2.

=25

^{2}=625
15. A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows: Rs 200 for the first 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?

Answer: Here; a = 200, d = 50 and n = 30

We can find the penalty by using the sum of 30 terms;

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.

Answer: Here; d = 20 n = 7 and Sn = 700

We know;

So the prizes in ascending order are as follows:

Rs. 50, Rs. 60, Rs. 80, Rs. 100, Rs. 120, Rs. 140 and Rs. 160

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?

Answer: Since each section of class 1 will plant 1 tree, so 3 trees will be planted by 3 sections of class 1.

Similarly,

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

We have; a = 3, d = 3 and n = 12

We can find the total number of trees as follows:

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 semicircles? (Take )

[Hint : Length of successive semicircles is l1, l2, l3, l4, . . . with centres at A, B, A, B, . . ., respectively.]

Answer: Circumference of first semicircle

It is clear that a = 0.5 π, d = 0.5π and n = 13

Hence; length of spiral can be calculated as follows:

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. In how many rows are the 200 logs placed and how many logs are in the top row?

Answer: We have; a = 20, d = - 1 and Sn = 200

We know;

If number of rows is 25 then;

Since; negative value for number of logs is not possible hence; number of rows = 16

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)]

Answer: Distance covered in picking and dropping the 1st potato = 2 x 5 = 10 m

Distance covered in picking and dropping the 2nd potato = 2 x (5+3) = 16 m

Distance covered in picking and dropping the 3rd potato = 2 x (5 + 3 + 3) = 22 m

So, we have a = 10, d = 6 and n = 10

Total distance can be calculated as follows:

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