Arithmetic Progression

Arithmetic Progression or arithmetic sequence is a sequence of numbers such

that the difference of any two successive terms is constant called common difference.
For example, the sequence 3, 6, 9, 12, …. Is an arithmetic progression with common
difference of 4.
Elements:

a1 = first term
an nth term
am = any term before an
d = common difference

d = a2 - a1 = a3 – a2 = a4 – a3

s = sum of all terms

Formula of finding nth term of an Arithmetic Progression

an = a1 + (n-1) d
or an = am + (n-m) d

Formula of finding the sum of n terms

n
S= (a1 + an) ; an = a1 + (n-1) d
2
n
S= [ 2a1 + (n – 1)d ]
2

Example:
1. Find the 25th term of Arithmetic Progression 3, 10, 17, ….
Given:

a1 = 3 a3 = 3
a1 = 10 n = 25

Solution:
The given A.P. 3, 10, 17, …. has a common difference of
d = 10 – 3 = 17 – 10 = 7

an = a1 + (n-1) d
an = 3 + (25-1) 7
an = 3 + 24(7)
an = 171 Answer

2. There are six arithmetic means between 3 and 45. Find the sum of all terms.
Given:

a1 = 3 n=8

an = 45
Solution:
The sequence is 3, , , , , , , 45. Thus, there are
8 terms with the first 3 and the eight term 45.
𝑛
S= (a1 + an)
2
8
S= (3 + 45)
2

S = 4 (48)
S = 192 Answer
3. How many terms of the sequence 4, 7, 10, …. Must be taken so that the sum is 329?
Given:

a1 = 4 a3 = 10
a2 = 7 S = 315

n=?
Solution:
The sequence 4, 7, 10, ….. forms an arithmetic progression with a common difference
of
 d = 10 – 7 = 7 – 4 = 3
𝑛
S= [2a1 + (n-1)d]
2
𝑛
329 = [ 2(4) + (n-1)3]
2

329(2) = n(8 + 3n -3)

658 = 8n +3𝑛2 -3n
3𝑛2 +5n -658 = 0
(3n + 47) (n-14) = 0
47
n = - 3 and n = 14

n = 14 Answer

4. The first term of an Arithmetic progression is 6 and the 10th term is 3 times the
second term. What is common difference?
Given:
a1 = 6
a10 = 3a2
d=?
Solution:

From an = a1 + (n-1)d
a10 = 6+ (10-1)d and a2 = 6 + (2-1)d
a10 = 6 + 9d a2 = 6 + d
then a10 = 3a2

6 + 9d = 3(6+d)
6 + 9d = 18 + 3d
9d – 3d = 18 – 6
6d = 12
d=2 Answer
5. Solve for x if x + 4x + 7x ……… 34x = 420.

Given: a1 = x

a2 = 4x
a3 = 7x
an = 35x
S = 420
Solution:

The given sequence is x, 4x,7x…..34x is an A.P. with d = a2- a1 = 4x – x = 3x

From an = a1 + (n-1) d
𝑛
34x = x + (n-1) 3x from S = (a1 + an)
2
12
34x = x + (n-1) 3x 420 = (x + 34x)
2
x
34 = 1 + (n-1)3 840 = 12 (x+34x)
34 = 1 + 3n -3 840 = 12 (35x)
3n = 34 -1 +3 70 = 35x
3n = 36 x=2 Answer
n = 12
Geometric Progression
Geometric Progression or geometric sequence is a sequence of numbers such
that the ratio of any two successive terms is constant called common ratio.
For example, the sequence 3, 9, 27, 81, ……. is an Geometric Progression with
common ratio of 3.
Elements:

a1 = first term
an = nth term
am = any term before an
r = common ratio

r = a2 ÷ a1 = a3 ÷ a2 = a4 ÷ a3………

s = sum of all terms

nth term of G.P.
an = a1rn-1 or an = amrn-m
Sum of n terms
𝑎1(𝑟 𝑛 −1)
s= when r>1
𝑟−1

𝑎1(𝑟 𝑛 −1)
s= when r<1
1−𝑟

Sum of Infinite Geometric Progression

For a geometric progression where -1 < r < 1 and n = ∞ “infinity”.
𝑎1
Sum of I.G.P. = 1−𝑟

Example:
1. What is the geometric mean of 2 and 32 ?
Given: a1 = 2
a3 = 32
a2 = ?
Solution:
from common ratio r = a2 / a1 = a3 / a2
𝑎2 32
= 𝑎2
2

(a2)2 = 64
(a2)2 = 64
a=8 Answer

2. Find the sum of the first 10 terms of geometric progression 3,6,12,24,….

Given : a1 = 3
a2 = 6
a3 = 12
a4 = 24
n = 10
s=?
Solution:
𝑎1(𝑟 𝑛 −1)
s= 𝑟−1
6
r= =2
3

3(210 −1)
s= 2−1

s = 3072 Answer

3. If the fourth term of a geometric progression is 162 and the sixth term is 1458, What
is the 8th term?
Given:
a4 = 162
a6 = 1458
a8 = ?
Solution: The 8th term
an = amrn-m a8 = a6r8-6
a6 = a4r6-4 a8 =1458r2
1458 = 162r2 a8 = 13, 122 Answer
1458
r2 = 162

r2 = 9
r=3

4. Determine x so that x, 2x + 5, 10x – 5 will form a geometric progression.

Given:
a1 = x
a2 = 2x + 5
a3 = 10x – 5
Solution:
r = a2 / a1 = a3 / a2
2𝑥+5 10𝑥−5
r= =
𝑥 2𝑥+5

(2x +5)2 = x(10x – 5)

4x2 + 20x + 25 = 10x2 - 5x
0 = 10x2 – 4x2 -5x -20x -25
0 = 6x2 – 25x -25
0 = (x – 5) (6x + 5)
x= 5 x = -5/6
x=5 Answer
5. What is the sum of the infinite geometric progression of 1, -1/5, 1/25,…?
Given: a1 = 1
a2 = -1/5
a3 = 1/25
Solution:
𝑎1
s= 1−𝑟

r = a2/a1
1
−
r= 5
1
1
r = −5

1
s= 1
1−(− )
5

1
s= 1
1+
5

1
s= 5 1
+
5 5

1
s= 6
5

𝟓
s= 𝟔
Answer
Harmonic Progression
Harmonic progression is a sequence of numbers formed by taking the reciprocal
of arithmetic progression.
Example:
1. the 10th term of the series 6,3,2 is?
Solution: their reciprocals are: 1/6, 1/3, ½, which forms an Arithmetic Progression with
a common difference of 1/3 – 1/6 = 2/6 – 1/6 = 1/6.
In A.P. the 10th term is:
a10 = a1 + (n-1)(d)
1
a10 = 6 + (10-1)(1/6)
1 1
a10 = + (9)(6)
6
1 9
a10 = +
6 6
10 5
a10 = =
6 3
𝟑
Therefore, in H.P. the 10th term is 𝟓
. Answer

2. Solve for x in the series: 1, 0.25, 1/7, x.

Solution:
Let x = 4th term of the series.
their reciprocals are 1, 4, 7, x which forms an Arithmetic Progression with a
common difference of 4 – 1 = 3.
In A.P. the 4th term is:
a4 = a1 + (n-1) (d)
x = 1 + (4-1) (3)
x = 1 + (3) (3)
x=1+9
x = 10
Therefore, in H.P. the 4th term which is equal to x is 1/10. Answer
 Exercises 2.1
Name: Date:
Section: Score:

1. Find the sum of all even numbers from 0 to 64

2. In geometric progression x, 2x + 3, 10x – 3, …. Find the common ratio.

3. What is the fourth term of the progression 0.5, 0.2, 0.125,….?