Sequence and Series - Arithmetic and Geometric Progressions Chapter Notes | Quantitative Aptitude for CA Foundation PDF Download

Chapter Overview

Sequence

Consider the following collections of numbers:

(1) 28, 2, 25, 27, …

(2) 2, 7, 11, 19, 31, 51, …

(3) 1, 2, 3, 4, 5, 6, …

(4) 20, 18, 16, 14, 12, 10, …

In (1) the numbers are not arranged by any definite rule. 

In (2) the numbers are in ascending order but there is no clear rule to obtain the next term from the previous; hence the next term cannot be determined uniquely. 

In (3) each term is obtained by adding 1 to the previous term; therefore the term after 6 is 6 + 1 = 7. 

In (4) each term is obtained by subtracting 2 from the preceding term; therefore the term after 10 is 10 − 2 = 8.

Collections (1) and (2) do not form sequences in the mathematical sense, whereas (3) and (4) do form sequences because there is a definite rule to generate terms.

Definition. An ordered collection of numbers a₁, a₂, a₃, a₄, …, a_n, … is called a sequence if there is a definite rule that assigns a unique value to a_n for each natural number n. The term a_n is the nth term or the general term of the sequence. The sequence is finite if it has a finite number of terms and infinite if the terms continue without end.

A finite sequence a₁, a₂, …, a_n may be denoted by appropriate notation and an infinite sequence by or simply {a_n}, where a_n denotes the nth element.

Examples of sequences:

• {1/n} : 1, 1/2, 1/3, 1/4, … (infinite)
• {(−1)ⁿ n} : −1, 2, −3, 4, −5, … (infinite)
• {n} : 1, 2, 3, 4, … (infinite)
• {n/(n+1)} : 1/2, 2/3, 3/4, 4/5, … (infinite)
• Sequence of even positive integers: 2, 4, 6, …
• Sequence of odd positive integers: 1, 3, 5, 7, …

All the above examples are infinite sequences unless explicitly restricted.

Finite examples:

• Even positive integers within 12: 2, 4, 6, 8, 10.
• Odd positive integers within 11: 1, 3, 5, 7, 9.

These examples are finite sequences.

Series

A series is the sum of the terms of a sequence. If {a_n} is a sequence, then an expression of the form a₁ + a₂ + a₃ + … + a_n + … is called a series. If the series contains a finite number of terms it is a finite series, otherwise an infinite series.

If S_n = u₁ + u₂ + u₃ + … + u_n, then S_n is called the sum to n terms (or the sum of the first n terms) of the series. The summation symbol ∑ is often used to denote sums.

Illustrations:

• 1 + 3 + 5 + 7 + … is a series whose terms are the odd numbers.
• 2 − 4 + 8 − 16 + … is a series with alternating signs and geometric growth.

Arithmetic Progression (A.P.)

A sequence a₁, a₂, a₃, …, a_n is called an arithmetic progression (A.P.) if the difference between any term and its preceding term is constant. That is, a₂ − a₁ = a₃ − a₂ = … = d, where d is the common difference.

If three numbers a, b, c are in A.P., then b − a = c − b or equivalently a + c = 2b. The number b is called the arithmetic mean between a and c.

General form of an A.P.

An A.P. with first term a and common difference d can be written as:

a, a + d, a + 2d, a + 3d, …

The nth term (t_n) of the A.P. is given by:

t_n = a + (n − 1)d

Example: 1) 2,5,8,11,14,17,…… is an A.P. in which d = 3 is the common diference.
2) 15,13,11,9,7,5,3,1,–1, is an A.P. in which –2 is the common difference.

Solution: In (1) 2^nd term = 5 , 1st term = 2, 3^rd term = 8,
so 2^nd term – 1st term = 5 – 2 = 3, 3rd term – 2^nd term = 8 – 5 = 3
Here the difference between a term and the preceding term is same that is always constant. This constant is called common difference.
Now in general an A.P. series can be written as
a, a + d, a + 2d, a + 3d, a + 4d, ……
where ‘a’ is the 1st term and ‘d’ is the common difference.
Thus 1^st term ( t₁ ) = a = a + ( 1 – 1 ) d
2^nd term ( t ₂ ) = a + d = a + ( 2 – 1 ) d 
3rd term (t3) = a + 2d = a + (3 – 1) d
4^th term (t₄) = a + 3d = a + (4 – 1) d
…………………………………………….
n^th term (t_n) = a + ( n – 1) d, where n is the position number of the term.
Using this formula we can get
50^th term (= t ₅₀) = a+ (50 – 1) d = a + 49d

Example 1: Find the 7th term of the A.P. 8, 5, 2, −1, −4, …

Solution:

a = 8

d = 5 − 8 = −3

t₇ = a + (7 − 1)d

t₇ = 8 + 6(−3)

t₇ = 8 − 18

t₇ = −10

Example 2: Which term of the A.P. 

Solution:

Assume the A.P. is of the form a, a + d, a + 2d, …

From the relation we obtain 17 = 3 + (n − 1)

Therefore n = 17 − 2 = 15

Hence, 17 is the 15th term of the A.P.

Example 3: If 5th and 12th terms of an A.P. are 14 and 35 respectively, find the A.P.

Solution:

Let a be the first term and d the common difference.

t₅ = a + 4d = 14

t₁₂ = a + 11d = 35

Subtracting, 7d = 21 ⇒ d = 3

Then a = 14 − 4×3 = 2

Hence the A.P. is 2, 5, 8, 11, 14, …

Example 4: Divide 69 into three parts which are in A.P. and the product of the first two parts is 483.

Solution:

Let the three parts be a − d, a, a + d.

(a − d) + a + (a + d) = 69

3a = 69 ⇒ a = 23

So the parts are 23 − d, 23, 23 + d.

Given (first part) × (second part) = 483 
⇒ 23(23 − d) = 483

23 − d = 483/23 = 21

d = 23 − 21 = 2

Therefore the parts are 21, 23, 25.

Example 5: Find the arithmetic mean between 4 and 10.

Solution: We know that the A.M. of a & b is = ( a + b ) /2
Hence, The A. M between 4 & 10 = ( 4 + 10 ) /2 = 7

Example 6: Insert 4 arithmetic means between 4 and 324.
4, –, –, –, –, 324

Solution: Here a= 4, d = ? n = 2 + 4 = 6, t_n = 324
Now t_n = a + ( n – 1 ) d
or 324= 4 + ( 6 – 1 ) d
or 320= 5d i.e., = i.e., d = 320 / 5 = 64 
So the 1st AM = 4 + 64 = 68
2^nd AM = 68 + 64 = 132
3^rd AM = 132 + 64 = 196
4^th AM = 196 + 64 = 260

Sum of the first n terms of an A.P.

Let S be the Sum, a be the 1st term and l the last term of an A.P. If the number of term is n, then t_n = l. Let d be the common difference of the A.P.
Now S = a + ( a + d ) + ( a + 2d ) + .. + (l– 2d ) + ( l– d ) + l
Again

On adding the above, we have

Note: The above formula may be used to determine the sum of n terms of an A.P. when the first term a and the last term is given.

Now l = t_n = a + ( n – 1 ) d

Note: The above formula may be used when the first term a, common difference d and the number of terms of an A.P. are given.

Sum of 1st n natural or counting numbers

On adding the above, we get

2S = ( n + 1 ) + ( n + 1 ) +....... to n terms

2S = n ( n + 1 )

S = n( n + 1 )/2

Then Sum of first n natural number is n( n + 1 ) / 2

Sum of 1st n odd number
S = 1 + 3 + 5 + …… + ( 2n – 1 )
Sum of first n odd number
S = 1 + 3 + 5 + …… + ( 2n – 1 )
Since S = n{ 2a + ( n –1 ) d } / 2, we find

S = n²
Then sum of first, n odd numbers is n, i.e. 1 + 3 + 5 + ..... + ( 2n – 1 ) = n²

Sum of the Squares of the first n natural nos.

Let S = 1² + 2² + 3² + …… + n²

We know m³ – ( m – 1 )³ = 3m² – 3m + 1
We put m = 1, 2, 3,……,n

Thus sum of the squares of the first n natural numbers is

 i.e.,

Similarly, sum of the cubes of first n natural numbers can be found out as by taking the identity
m⁴ – ( m – 1 )⁴ = 4m³ – 6m² + 4m – 1 and putting m = 1, 2, 3,…., n.
Thus

Geometric Progression (G.P.)

If in a sequence of terms each term is constant multiple of the proceeding term, then the sequence is called a Geometric Progression (G.P). The constant multiplier is called the common ratio
Examples: 1) In 5, 15, 45, 135,….. common ratio is 15/5 = 3
2) In 1, 1/2, 1/4, 1/9 … common ratio is (1/2) /1 = 1/2
3) In 2, –6, 18, –54, …. common ratio is (–6) / 2 = –3

Illustrations: Consider the following series :–

(i) 1 + 4 + 16 + 64 + …………….
Here second term / first term = 4/1 = 4; third term / second term = 16/4 = 4
fourth term/third term = 64/16 = 4 and so on.
Thus, we find that, in the entire series, the ratio of any term and the term preceding it, is a constant.

(ii) 1/3 – 1/9 + 1/27 – 1/81 + ………….
Here second term / 1st term = (–1/9) / ( 1/3) = –1/3
third term / second term = ( 1/27 ) / ( –1/9 ) = –1/3
fourth term / third term = ( –1/81 ) / (1/27 ) = –1/3 and so on.
Here also, in the entire series, the ratio of any term and the term preceding one is constant.
The above mentioned series are known as Geometric Series.
Let us consider the sequence a, ar, ar² , ar³, ….

1st term = a, 2nd term = a^r = ar ^2–1, 3rd term = ar² = ar^3–1, 4th term = ar³ = ar ^{4 –1} , …..

Similarly

Thus, common ratio =
ar ^n–1/ar ^n–2 = r
Thus, general term of a G.P is given by ar n–1 and the general form of G.P. is
a + ar + ar² + ar³ +……. ….
For example,

So

Example 1: If a, ar, ar² , ar³ , …. be in G.P. Find the common ratio.

Solution: 1st term = a, 2^nd term = ar
Ratio of any term to its preceding term = ar/a = r = common ratio.

Example 2: Which term of the progression 1, 2, 4, 8,… is 256?

Solution: a = 1, r = 2/1 = 2, n = ? t_n = 256 
t_n = ar^n–1
or 256 = 1 × 2^n–1 i.e., 2⁸ = 2 ^n–1 or, n – 1 = 8 i.e., n = 9 
Thus 9th term of the G. P. is 256

Geometric Mean

If a, b, c are in G.P., then b/a = c/b ⇒ b² = ac. The number b is called the geometric mean between a and c.

Example 1

Insert 3 geometric means between 1/9 and 9.

Solution:

1/9, –, –, –, 9

a = 1/9

Number of terms n = 2 + 3 = 5

t₅ = 9

Example 2

Find the G.P. where 4th term is 8 and 8th term is 128/625.

Solution:

Let a be the 1st term and r be the common ratio.
So
By the question t₄ = 8 and t₈ = 128/625
ar³ = 8 and ar⁷ = 128 / 625
Therefore

Now ar³ = 8 ⇒ a × (2/5) 3 = 8 ⇒ a = 125
Thus the G. P is
125, 50, 20, 8, 16/5, ………..
When r = –2/5 , a = –125 and the G.P is –125, 50, –20, 8, –16/5 ,………
Finally, the G.P. is 125, 50, 20, 8, 16/5, ………..
or, –125, 50, –20, 8, –16/5,………

Sum of first n terms of a G.P.

Let a be the first term and r the common ratio. The sum of the first n terms is:

Multiply S by r and subtract to obtain:

Sum of an infinite geometric series

Example 1: Find the sum of 1 + 2 + 4 + 8 + … to 8 terms.

Solution:

a = 1, r = 2, n = 8

S = a( rⁿ − 1 )/(r − 1) = 1(2⁸ − 1)/(2 − 1)

S = 2⁸ − 1 = 256 − 1 = 255

Example 2: Find the sum to n terms of 6 + 27 + 128 + 629 + ….

Solution: 

Example 3: Find the sum to n terms of 3 + 33 + 333 + ….

Solution: Let S = 3 + 33 + 333 + … (n terms) = 3(1 + 11 + 111 + … n terms)

Use decomposition and geometric-sum techniques to express the repeating-digit numbers in terms of geometric series.

Example 4: Find the sum of n terms of 0.7 + 0.77 + 0.777 + …

Solution: Let S = 0.7 + 0.77 + 0.777 + … (n terms)

S = 7(0.1 + 0.11 + 0.111 + … n terms)

Transform these repeating decimals into sums involving geometric series.

Example 5: Evaluate  using the sum of an infinite geometric series.

Solution:

= 21/100 + 1/132

= (693 + 25)/3300 = 718/3300 = 359/1650

Example 6: Find three numbers in G.P. whose sum is 19 and product is 216.

Solution:

Let the three numbers be a/r, a, ar.

Product a³ = 216 
⇒ a = 6.

So numbers are 6/r, 6, 6r and
 their sum is 6/r + 6 + 6r = 19.

Therefore 6/r + 6r = 13.

Multiply both sides by r: 6 + 6r² = 13r.

Rearrange: 6r² − 13r + 6 = 0.

Factor or solve the quadratic: (3r − 2)(2r − 3) = 0 ⇒ r = 2/3 or r = 3/2.

If r = 2/3: numbers = 9, 6, 4.

If r = 3/2: numbers = 4, 6, 9.

Illustrations - Applications

(I) A person is employed at ₹3000 per month and receives an increase of ₹100 per year. Find the total amount received in 25 years and the monthly salary in the last year.

SOLUTION:

He gets in the 1^st year at the Rate of 3000 per month; 
In the 2^nd year he gets at the rate of ₹ 3100 per month;
In the 3^rd year at the rate of ₹ 3200 per month so on.
In the last year the monthly salary will be
₹ {3000 + ( 25 – 1 ) × 100} = ₹ 5400
Total amount = ₹ 12 (3000 + 3100 + 3200 +… + 5400)

= ₹ 12 × 25/2 (3000 + 5400) 
= ₹ 150 × 8400 
= ₹ 12,60,000

(II) A person borrows ₹8,000 at 2.76% simple interest per annum. The principal and interest are to be paid in 10 monthly instalments, each instalment double the preceding one. Find the first and last instalment.

SOLUTION:

Interest for 10 months at 2.76% p.a. is calculated on simple interest basis.

Interest = 2.76 × (10/12) × 8000 / 100 = ₹184.

Total amount to be repaid = 8000 + 184 = ₹8184.

The instalments form a G.P. with ratio r = 2 and sum S = a(2¹⁰ − 1)/(2 − 1) = a(1024 − 1) = 1023a.

So 1023a = 8184 ⇒ a = 8184/1023 = ₹8.

First instalment = ₹8.

Last instalment = a × 2⁹ = 8 × 512 = ₹4096.