The document defines arithmetic progressions and provides examples. Some key points: - An arithmetic progression is a list of numbers where each term is obtained by adding a fixed number (called the common difference) to the preceding term. - The general formula for the nth term of an AP is an = a + (n-1)d, where a is the first term and d is the common difference. - Examples show how to determine if a list of numbers forms an AP, write the next terms, and find a specific term by using the general formula. - There are finite APs, which have a last term, and infinite APs, which go on forever without a last term.

1. Arithmetic Progressions

3. Definition:
An arithmetic progression is a list of numbers in
which each term is obtained by adding a fixed number to
the preceding term except the first term.
• This fixed number is called the common difference of the
AP. Remember that common difference can be positive,
negative or zero.
Consider some example:
1. The heights ( in cm ) of some students of a school
standing in a queue in the morning assembly are 147 ,
148, 149, . . ., 157.
AP: 147,148,149,……,157
=> a1, a2, a3,…….an
=> a1 - first term, a2 – second term, an nth term of AP
=> Common difference is represented as d

4. General form of an AP.
2,4,6,8,…….. 168
a, a + d, a + 2d, a+3d,…….. a + (n-1)d
2nd term = a + (2-1)d = a + d
3rd term = a + (3-1) d = a+2d
nth term = a + (n-1)d
an = a + (n-1)d
Example: Which of the following list of numbers form an
AP? If they form an AP, write the next two terms :
1. 4, 10, 16, 22, . . .
2. 1, – 1, – 3, – 5, . . .
3. – 2, 2, – 2, 2, – 2, . . .
4. 1, 1, 1, 2, 2, 2, 3, 3, 3, . .

5. (i) 4, 10, 16, 22, . . .
We have a2 – a1 = 10 – 4 = 6
a3 – a2 = 16 – 10 = 6
a4 – a3 = 22 – 16 = 6
i.e., an – a n-1 is the same every time.
So, the given list of numbers forms an AP with the common
difference d = 6.
The next two terms are:
a4 +d = 22 + 6 = 28 => a5 = 28
a5 +d = 28 + 6 = 34 => a6 = 34

6. Types of AP
• There are two types of AP, they are
a) Finite AP:
the series of the numbers are having finite
terms.[ these are having last term i.e. an]
Exa: 3,6,9,…….., 42
b) Infinite AP:
the series of the numbers are having infinite
terms/ no last term
Exa: 3,6,9,……………., am
 an is also called the general term of the AP. If there
are m terms in the AP, then
 am represents the last term which is sometimes also
denoted by l.

7. Let us consider some examples.
1. Find the 10th term of the AP : 2, 7, 12, . . .
Ans. Here, a = 2, d = 7 – 2 = 5 and n = 10.
We have an = a + (n – 1) d
a10 = 2 + (10-1)5
= 2 +9 x 5
= 2 + 45
= 47
Therefore, the 10th term of the given AP is 47.
2. Find the 29th term of the AP : 3, 6, 9, . . .
3. Find the 47th term of the AP : 270, 263, 256, . . .

8. 4. Which term of the AP : 21, 18, 15, . . . is – 81? Also, is any
term 0? Give reason for your answer.
Ans. Here, a = 21, d = 18 – 21 = -3 and an = - 81.
We have an = a + (n – 1) d
- 81 = 21 + (n-1)-3
- 81= 21 - 3n + 3
- 81 = 24 - 3n
- 81 – 24 = - 3n
- 3n = - 105
n = 35
Therefore, the 35th term of the given AP is – 81.
Next, we want to know if there is any n for which an = 0. If
such an n is there, then
21 + (n – 1) (–3) = 0,
i.e., 3(n – 1) = 21 [ 3n – 3 = 21]
i.e., n = 8
So, the 8th term of AP is 0.

9. 4. Determine the AP whose 3rd term is 5 and the 7th term is 9.
Solution : We have
a3 = a + (3 – 1) d = 5
a + 2d = 5 (1)
And a7 = a + (7 – 1) d =
a + 6d = 9 (2)
By subtracting equation number 2 from 1, we get
a + 6d = 9
- [a + 2d = 5]
--------------
4d = 4 => d = 1
By equation 1, we get a + 2d = 5
a + 2 x 1 = 5 => a = 5 – 2 => a = 3
Therefor AP : 3, 4,5,6,......
5. Find the 31st term of an AP whose 11th term is 38
and the 16th term is 73.

10. Exa. 6. Check whether 301 is a term of the list of numbers 5, 11, 17, 23,…..,
Exa. 7. How many two-digit numbers are divisible by 3?
Exa. 8. Find the 11th term from the last term (towards the first term)
of the AP : 10, 7, 4, . . ., – 62.
Ans. Here, a = 10, d = 7 – 10 = – 3, l = – 62,
Since we need to find last 11th term
-62, -59, -56, …….. 10 [ reverse form of AP]
 a = - 62, d = 3 and n = 11
 11th term of AP
 a11 = a + (n – 1 ) d
 a11 = -62 + (11-1)3
 = - 62 + 10 x 3
 = - 62 + 30
 = - 32
Therefore the 11th term of AP is - 32 .

11. Exa. 9. A sum of ` 1000 is invested at 8% simple interest per year. Calculate
the interest at the end of each year. Do these interests form an AP? If so,
find the interest at the end of 30 years making use of this fact.
Ans. W.k.t.
the SI = PTR/100
=> SI of one year is 80
=> SI of second year is 160
=> SI of third year is 240
AP: 80, 160, 240, ……….. a30
Can we find the simple interest for 30 years by using nth term of AP?
Yes.
 a = 80, d = 80 and n = 30
 a30 = a + ( n- 1 ) d
 = 80 + ( 30 – 1 ) 80
 = 80 + 29 x 80
 = 80 + 2320
 = 2400
Therefore the simple interest for 30 years is Rs. 2400/-.

12. 10. If the 3rd and the 9th terms of an AP are 4 and – 8
respectively, which term of this AP is zero?
Ans. Given,
3rd term => a + 2d = 4 (1)
9th term => a + 8d = -8 (2)
By subtracting equation 1 from 2 we get
a + 8d – a – 2d = - 8 – 4
6d = - 12
d = - 2
 By equation 1, we get.
 a + 2d = 4
 a + 2 x (-2) = 4
 a – 4 = 4
 a = 4 + 4
 a = 8
 The AP : 8, 6,4,2,0 => the 5th term an AP is 0

13. 11. The 17th term of an AP exceeds its 10th term by 7. Find
the common difference.
Ans. Given,
17th => a + 16d
10th term => a + 9d
a17 – a10 = 7
a + 16d – a – 9d = 7
7d = 7
d= 1
Therefore common difference is 1.
12. Which term of the AP : 3, 15, 27, 39, . . . will be 132 more than its 54th term?
Ans. Find the 54th term
a54 = 3 + 53 *12
= 3 + 636
= 639
a54 + 132 = 3 + (n – 1 ) 12
771= 3 + 12n – 12
771 = - 9 + 12n
771+9 = 12n
n = 780/12
n = 65
There fore 65th term of AP is 132 more than 54th term

14. 13. Determine the AP whose third term is 16 and the
7th term exceeds the 5th term by 12.
Ans. Given,
3rd term => a + 2d = 16 (1)
7th term – 5th term = 12
a + 6d – a – 4d = 12 (2)
2d = 12
d = 6
By equation 1, we ge
 a + 2d = 16
 a + 2 * 6 = 16
 a + 12 = 16
 a = 4
The AP is : 4, 10, 16, …….
14. If the 10th of an AP is 52 and 17th term is 20 more than
its 13th term, find the AP.

15. 15. Determine the AP whose 3rd term is 5 and the 7th term is 9.
16. Find the 20th term from the last term of the AP : 3, 8, 13, . . ., 253.
17. 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.
Ans. Given,
a4 + a8 = 24 => a+3d +a +7d = 24
=> 2a + 10d = 24 (1)
a6 + a10 = 44 => a + 5d + a +9d = 44
=> 2a + 14d = 44 (2)
By subtracting eq 1 from eq 2, we get
2a + 14d – 2a - 10d = 44 – 24
4d = 20
d = 5
 By equation 1, we get
 2a + 10d = 24
 2a + 10 * 5 = 24
 2a + 50 = 24 => 2a = - 26
 a = - 13

16. 18. The sum of the 2nd and the 7th term of an AP is 30. if
its 15th term is 1 less than twice its 8th term, find the
AP.
19. The 17th term of an AP is 5 more than twice its 8th
term. If 11th term of the AP is 43, find its 99th term
and nth term.
20. The sum of 5th and 9th terms of AP is 30. if its 25th
term is three times its 8th term, find the AP.
21.The 19th term of an AP is equal to 3 times its 6th
term. If its 9th term is 19, find AP
21. If 4 times the 4th term of an AP is equal to the 18
times its 18th term then find its 22nd term.
23. The 9th term of an AP is – 32 and the sum of its 11th
and 13th terms is – 94 . Find its common difference.
24. How many numbers are there between 101 and
999, which are divisible by both 2 and 5?

17. 25. Three numbers are in AP. If the sum is these numbers are 27
and sum of product 648, find the numbers.
Ans. Given
Let us consider three terms of AP: a – d, a, a + d
 Sum of three term = a – d + a + a + d = 27
 3a = 27
 a = 9
 Sum of product of terms = (a – d) (a) (a+d) = 648
 (a2 – d2 ) a = 648
(81 – d2)9 = 648
81– d2 = 648/9
d2 = 81 - 72
d2 = 9
d = 3
Therefore the AP : 9, 12, 15, …….
26. Three numbers are in AP. If the sum is these numbers are – 3 and
sum of product 8, find the numbers.

18. 27. Find four terms of AP whose sum is 28 and the sum of whose
square is 216.
Ans. Given,
Sum of four terms: (a-3d) + (a-d) + (a+d) + (a+3d) = 28
 a-3d + a-d+ a+d + a - 3d = 28
 4a= 28
 a = 7
The sum of whose squares are = 216
(a-3d)2 + (a-d)2 + (a+d)2 + (a+3d)2 = 216
[a2 + 9d2 – 6ad] + [a2 +d2 – 2ad] + [a2 + 9d2 +6ad] + [a2 +d2 + 2ad] = 216
4a2+ 20d2 = 216
4 *49 + 20 d2 = 216
196 + 20 d2 = 216
20 d2 = 20
d2 = 1 => d = 1
Therefore the 4 terms of AP is : 4, 6, 8, 10, ……..

19. 28. Find four terms of AP. whose sum is 20
and the sum of whose product is 120.
29. The sum of three numbers of AP is 3 and
the sum of their product is – 35. find the
numbers.
30. The sum of three consecutive terms of AP
is 21 and the sum of their product is 165.
Find the numbers.
31. Divide 24 in three parts such that they are
in AP and their product is 440.

20. 32. The sum first three terms of an AP is 48. if the product of
first and second terms exceeds 4 times the third term by 12.
Find the AP.
Ans. Given
First three terms of AP = a-d, a, a+d
=> a-d+a+a+d = 48
=> 3a = 48
=> a = 16
[a-d]a – 4 [a+d] = 12
=> a^2 – ad – 4a – 4d = 12
=> 256 – 16d – 64– 4d = 12
=> 192 – 20d = 12
=> - 20 d = 12 - 192
=> - 20 d = - 180
=> d= 9
Therefore the three term are = 7, 16, 25.

21. 33.The angles of a quadrilateral are in AP whose common
difference is 100. Find the angles.
34. The sum of three numbers of AP is 21 and the sum of their
product is 231. find the numbers.

22. Sum of n terms of an AP
Find the sum of n terms of an AP:
sn => sum of terms,
n => no. of terms
The other formula to use find the n
terms of AP is :

23. 3. Find the sum of first 24 terms of the list of
numbers whose nth term is given by an = 3 + 2n.
Ans. Given
an = 3 + 2n
a1 = 3 + 2 => a1 = 5
a2 = 3 + 2*2 => a1 = 7
a3 = 3 + 2*3 => a1 = 9
The AP : 5,7,9, ………… s24
a = 5, n = 24, d= 2
=> Sn = n/2 [2a + (n-1) d]
=> s24= 12 [ 2*5 + 23 *2]
=> = 12 [ 10 + 46]
= 12 * 56
= 672
Therefore the sum of 24 terms of AP is 672

24. 5. A manufacturer of TV sets produced 600 sets in the third
year and 700 sets in the seventh year. Assuming that the
production increases uniformly by a fixed number every
year, find :
(i) the production in the 1st year
(ii) the production in the 10th year
(iii) the total production in first 7 years.
6. How many terms of the AP : 9, 17, 25, . . . must be taken to
give a sum of 636?
7. 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.
8. Find the sum of first 51 terms of an AP whose second and
third terms are 14 and 18 respectively.

25. 9. If the sum of the first 14 terms of an AP is 1050 and its first
term is 10,find the 20th term.
Ans. Given,
s14= 1050, a = 10 a20 = ?
=> s14= n/2 [ 2a + (n-1)d]
=> 1050 = 7 [ 2*10 + 13 *d]
=> 1050 / 7 = 20 + 13d
=> 150 = 20 + 13d
=> 13d = 150 – 20
=> d = 130/13
=> d =10
Therefore a20 = 10 + 19 * 10
= 10 +190
= 200
Therefore the 20th term of AP is 200

26. Note: The nth term of an AP is the difference of the sum
to first n terms and the sum to first (n – 1) terms of it,
i.e., an = sn – sn-1.
i.e. a20 = s20 – s19.
10. 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.
11. A sum of ` 700 is to be used to give seven cash prizes
to students of a school for their overall academic
performance. If each prize is ` 20 less than its
preceding prize, find the value of each of the prizes.
12. The sum of 3rd term and 7th term of AP is 6 and their
product is 8. Find the sum of first 16 terms of the AP.
13. Find the sum of first 30 terms of AP in which 3rd term
is 7 and 7th term is two more than thrice the 3rd term.

27. • The sum of 3rd term and 7th term of AP is 6 and their
product is 8. Find the sum of first 16 terms of the AP.
Ans. Given
a3 + a7 = 6 => 2a + 8d = 6 => a + 4d = 3.
Therefore a = 3 – 4d --------- (1)
a3 * a7 = 8 => (a+ 2d) (a + 6d) = 8
(3 – 4d + 2d) (3 – 4d + 6d) = 8
(3 – 2d) (3 +2d) = 8
 9 – 4 d2 = 8
 4d2 = 9 –8
 d2 = 1
 d = ½ --------------------- (2)
By equation 1, we get
 a = 3 – 4 * ½
 a = 3 – 2 => a = 1

28. 14. Sum of the 4th and the 8th terms of an AP is 24 and 6th and
10th terms of AP is 44. find the sum of its first 29 terms of
AP.
15. Sum of the first 14 terms of an AP is 1505 and its first
term is 10. find its 25th term of AP.
16. An AP; 5,12,19,… has 50 terms. Find its last term. Hence,
find the sum of its last 15 terms.
17. An AP; 8,10,12,… has 60 terms. Find its last term. Hence,
find the sum of its last 9 terms

29. Higher Order thinking questions
i. In a given AP if pth term is q and qth term is p, then show that
nth term is (p + q – n).
Ans. Given
ap = a + ( p – 1) d = q
=> a+ pd – d = q --------------- (1)
aq = a + (q – 1) d = p
=> a + qd – d = p ---------------- (2)
By subtracting eqn 1 from eqn 2, we get
a + qd – d – a – pd + d = p – q
d ( q – p ) = p – q)
d = (p – q)/ - (p –q )
d = - 1
By substituting the value of d in eqn 1, we get
=> a + p(-1) – ( - 1) = q
=> a – p + 1 = q => a = (p + q – 1 )
Nth term of AP =>
an = a + ( n – 1) d = p + q – 1 - n + 1 = (p + q – n )

30. ii. If m times mth of AP is equals n times the nth term and m ≠
n, show that its ( m+ n )th term is 0.
Ans. Let a be the first term and d be the common difference
of the given AP.
Given m*am = n* an
=> m [a + ( m – 1 )d] = n [a + (n – 1)d]
=> m [a + ( m – 1 )d] - n [a + (n – 1)d] = 0
=> am + (m2 – m)d – an – ( n2 – n ) d = 0
=> a ( m – n) + [m2 - n2 – m + n] d = 0
=> a ( m – n) + [(m2 - n2) – (m – n)] d = 0
=> [m – n ] *{a + [( m + n ) – 1]d} = 0
=> [ m – n ] * a m+n = 0
=> a m+n = 0
Therefore a m+n term of AP is 0
iii. If the mth term of AP is 1/n and its nth term is 1/m, then
show that its (mn)th term is 1