Finding the next term:
An arithmetic progression is given below:-
3, 10, 17, ..................................................
The fourth term is 24 which is obtained by adding the common difference 7 to the third term which is 17.
Similarly,
Fifth term = 24 + 7 = 31
Sixth term = 31 + 7 = 38
Seventh term = 38 + 7 = 45
Eighth term = 45 + 7 = 52
So, the Arithmetic Progression is
3, 10, 17, 24, 31, 38, 45, 52
1. Find the next three terms of the arithmetic progressions given below:
(i) 5, 11, 17, 23, ........................
Ans.
The fifth term is 29 which is obtained by adding the common difference 6 to the fourth term which is 17.
Similarly,
Sixth term = 29 + 6 = 35
Seventh term = 35 + 6 = 41
(ii) -11, -8, -5, -2, ........................
Ans.
The fifth term is 1 which is obtained by adding the common difference 3 to the fourth term which is -2.
Similarly,
Sixth term = 1 + 3 = 4
Seventh term = 4 + 3 = 7
(iii) 4/9, 7/9, 10/9, 13/9 ...........
Ans.
The fifth term is (16/9) which is obtained by adding the common difference (1/3) to the fourth term which is (13/9).
Similarly,
Sixth term = (16/9) + (1/3) = 19/9
Seventh term = (19/9) + (1/3) = 22/3
(iv) 0, 9, 18, 27, ........................
Ans.
The fifth term is 36 which is obtained by adding the common difference 9 to the fourth term which is 27.
Similarly,
Sixth term = 36 + 9 = 45
Seventh term = 45 + 9 = 64
Generalized form of an arithmetic progression.
a, a + d, a + 2d, a + 3d, .........................
or
a, a + d, a + d + d, a + d + d + d, ...............
Finite arithmetic progression
When number of terms of the AP is finite it is said to be a finite arithmetic progression.
Example - 5,7, 9, 11, 13, 15, ......
Infinite arithmetic progression
when number of terms is infinite it is said to be an infinite arithmetic progression.
Examples- 1,3,5,7,9,11, .......
Note:- It is not necessary that the common difference should always be a natural number, it can be any real number.
Natural Number
A natural number is an integer greater than 0. Natural number begin at 1 and increment to infinity: 1, 2, 3, 4, 5, etc.
Natural number are also called "counting numbers" because they are used for counting.
1,2,3,4,5 ....... are natural number
Real Number
Positive or negative, large or small, whole numbers or decimal numbers are all Real Numbers.
1, 15.82, -0.1, 3/4 ........... are real numbers
1. Form an infinite arithmetic progression where the first term is 5 and common difference is 3.
Ans. 5, 8, 11, 14, 17
2. Form two finite arithmetic progressions each having 5 terms.
Ans.
5,7, 9, 11, 13
2, 4, 6, 8, 10
Example-
Write the first three terms of the arithmetic progression where the first term a = 10 and the common difference, d = -3.
Solution:
First term a = 10
Common difference d= -3
Second term = a+d
= 10 + (-3)
= 7
Third term = a+2d
= 10+2(-3)
= 10-6
= 4
Hence, the first three terms of the arithmetic progression are 10, 7, 4.
nth term of Arithmetic Progression
nth
term an = a + (n – 1)d
Example-
Find the 10th term of the arithmetic progression 4, 7, 10, 13, .....
Solution -
Here a = 4, d = 7 - 4 = 3, n = 10
a10 = a + (10 – 1)d
= 4 + 9 x 3
= 4 + 27
= 31
Arithmetic progression 2, 6, 10, ..... contains m terms. Find the last term.
Solution: Here the first term a = 2, the common difference d = 6 - 2 = 4 and the number of terms are m. Therefore the last term will be m. So n = m.
mth term am = a + (m – 1)d
= 2 + (m - 1) 4
= 2 + 4 m - 4
= 4m - 2
1. Arithmetic progression 3, 5, 7, ............ contains 15 terms. Find the last term.
Ans.
Last term = a + (n – 1)d
= 3 + (15 - 1) x 2
= 3 + 14 x 2
= 3 + 28
= 31
2. If the last term of the arithmetic progression -9, -5, -1, ............is 67, then how many terms are there in the progression?
Ans. Last term = a + (n – 1)d
67 = -9 + (n - 1) x 4
76 = (n - 1) x 4
(76/4) = n - 1
3. Find the mth and pth terms of arithmetic progression 10, 15, 20, ..........
Ans.
mth term am = a + (m – 1)d
= 10 + (m - 1) x 5
= 10 + 5 m - 5
= 5m + 5
pth term ap = a + (p – 1)d
= 10 + (p - 1) x 5
= 10 + 5p - 5
= 5p + 5
Example-6. Check whether 301 is a term in the AP 5, 11, 17, 23, .............Give reasons.
Solution:
Here, a = 5, d = 11-5 = 6
Let nth term be 301 i.e. an = 301
We have to find the value of n
an = a + (n – 1)d
301 = 5 + (n–1)6
301=5+6n–6
301 = 6n–1
6n = 302
n = 302/6
n = 151/3
Since number of terms is n, so the nth term should be a positive whole number, but
n is a fraction here. Therefore 301 is not a term of the given arithmetic progression.
1. An arithmetic progression whose third term is 12 and the last term is 106, contains 50 terms. Find the 21st term of this arithmetic progression.
Ans.
Third term a3 = a + (3 – 1)d
12 = a + 2d .... (i)
Last term = a + (n – 1)d
106 = a + 49d ......(ii)
equation (ii) - equation (i)
47d = 94
d = 2
Putting the value of d = 2 in equation (i), we get the value of a
a = 8
21st term a21 = a + (21 – 1)d
= 8 + (20) x 2
= 8 + 40
= 48
2. The first term of an arithmetic progression is 10 and common difference is -3. Find the 11th term.
Ans. a(first term) = 10, d(common difference) = -3
11th term a11 = a + (11 – 1)d
= 10 + 10 x (-3)
= 10 - 30
= -20
Example-8. How many two digit numbers are divisible by 5?
Solution: The list of two digit numbers divisible by 5 is :-
10, 15, 20, ............., 95
This is an arithmetic progression, whose first term is a = 10, the common difference
d = 5 and nth term an = 95
Since nth term
95 = 10 + (n-1) .5
95 = 10 + 5n -5
95 = 5+5n
5n = 95 - 5
n = 90/5
n = 18
Therefore, 18 two digit numbers are divisible by 5.
Example-9. Jyoti started to work in 1997 at a monthly salary of Rs.5000 and gets an annual increment of Rs.200 in her salary. In which year did her salary become Rs.7000/-?
Solution:
Monthly salaries (in Rs) for years 1997, 1998, 1999, 2000 ……..are
5000, 5200, 5400, 5600…..
This is an A.P., because difference of any two successive terms is 200, therefore the
common difference d = 200 and the first term a = 5000.
Suppose that Jyoti's salary becomes 7000 in n years.
Then,
an = 7000
a + (n – 1)d = 7000
5000 + (n–1) 200 = 7000
(n–1) 200 = 7000 – 5000
(n–1) 200 = 2000
n – 1 =
2000
200
n – 1 = 10
n = 11
Therefore in eleventh year i.e. in 2007 Jyoti's salary will become Rs.7000.
Till now you have solved examples in which series of numbers formed arithmetic progressions. Now we will solve examples where letter combinations (p, q, r etc.) form arithmetic progressions.
