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

Chapter Overview

Sequence

Let us consider the following collection 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 nos. are not arranged in a particular order. In (2) the nos. are in ascending order but they do not obey any rule or law. It is, therefore, not possible to indicate the number next to 51.
In (3) we find that by adding 1 to any number, we get the next one. Here the number next to 6 is 6 + 1 = 7.
In (4) if we subtract 2 from any number we get the nos. that follows. Here the number next to 10 is 10 –2 = 8.
Under these circumstances, we say, the numbers in the collections (1) and (2) do not form sequences whereas the numbers in the collections (3) & (4) form sequences.
Thus a sequence may be defined as follows:—

An ordered collection of numbers a₁, a₂, a₃, a₄, ................., a_n, ................. is a sequence if according to some definite rule or law, there is a definite value of a n , called the term or element of the sequence, corresponding to any value of the natural number n.
Clearly, a₁ is the 1st term of the sequence , a₂ is the 2nd term, ................., a n is the nth term.
In the nth term a n , by putting n = 1, 2 ,3 ,......... successively , we get a₁, a₂ , a₃ , a₄, .........
Thus it is clear that the nth term of a sequence is a function of the positive integer n. The nth term is also called the general term of the sequence. To specify a sequence, nth term must be known, otherwise it may lead to confusion. A sequence may be finite or infinite.
If the number of elements in a sequence is finite, the sequence is called finite sequence; while if the number of elements is unending, the sequence is infinite.

A finite sequence a₁, a₂, a₃, a₄, ................., a_n is denoted by and an infinite sequence a₁, a₂, a₃, a ₄, ................., a _n , ................. is denoted by a or simply by { a_n } where a n is the nth element of the sequence.

Example :
1) The sequence { 1/n } is 1, 1/2, 1/3, 1/4……
2) The sequence { ( – 1 ) ⁿ n } is –1, 2, –3, 4, –5,…..
3) The sequence { n } is 1, 2, 3,…
4) The sequence { n / (n + 1) } is 1/2, 2/3, 3/4, 4/5 …….
5) A sequence of even positive integers is 2, 4, 6, .....................................
6) A sequence of odd positive integers is 1, 3, 5, 7, .....................................

All the above are infinite sequences.

Example:
1) A sequence of even positive integers within 12 i.e., is 2, 4, 6, 8, 10.
2) A sequence of odd positive integers within 11 i.e., is 1, 3, 5, 7, 9.

All the above are finite sequences.

Series

ILLUSTRATIONS:
(i) 1 + 3 + 5 + 7 + ............................ is a series in which 1st term = 1, 2nd term = 3 , and so on.
(ii) 2 – 4 + 8 –16 + ..................... is also a series in which 1st term = 2, 2nd term = –4 , and so on.

Arithmetic Progression (A.P.)

A sequence a₁, a₂ ,a₃, ……, a _n is called an Arithmetic Progression (A.P.) when a₂ – a₁ = a₃ – a₂ = ….. = a _n – a_n–1. That means A. P. is a sequence in which each term is obtained by adding a constant d to the preceding term. This constant ‘d’ is called the common difference of the A.P. If 3 numbers a, b, c are in A.P., we say
b – a = c – b or a + c = 2b; b is called the arithmetic mean between a and c.
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 generel 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: Here a = 8, d = 5 – 8 = –3

Now t₇ = 8 + (7 – 1) d 
= 8 + (7 – 1) (– 3) 
= 8 + 6 (– 3) 
= 8 – 18 
= – 10

Example 2: Which term of the AP

Solution:

We may write

or, 17 = 3 + ( n – 1)
or, n = 17 – 2 = 15
Hence, 15th term of the A.P. is . 17 7

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 & d be the common difference of A.P.

t₅ = a + 4d = 14 
t₁₂ = a + 11d = 35
On solving the above two equations,
7d = 21 = i.e., d = 3
and a = 14 – (4 × 3) = 14 – 12 = 2
Hence, the required A.P. is 2, 5, 8, 11, 14,……………

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

Solution: Given that the three parts are in A.P., let the three parts which are in A.P. be a – d, a, a + d.........
Thus a – d + a + a + d = 69
or 3a = 69
or a = 23
So the three parts are 23 – d, 23, 23 + d
Since the product of first two parts is 483, therefore, we have
23 ( 23 – d ) = 483
or 23 – d = 483 / 23 = 21
or d = 23 – 21 = 2
Hence, the three parts which are in A.P. are
23 – 2 = 21, 23, 23 + 2 = 25
Hence the three 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

Let S be the Sum, a be the 1st term and  the last term of an A.P. If the number of term is n, then t n = . 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 getor

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

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

Solution: 1/9, –, –, –, 9
a = 1/9, r = ?, n = 2 + 3 = 5, t_n = 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 be the common ratio. So the first n terms are a, ar, ar² , …... ar ^n–1 .
If S be the sum of n terms,

Subtracting (i) from (ii)

If r = 1 , then S_n = a + a + a+ ……….. to n terms = na

If the nth term of the G. P be l then  = ar^n–1

Therefore,

So, when the last term of the G. P is known, we use this formula.

Sum of infinite geometric series
S = a ( 1 – rⁿ ) / (1 – r) 
when r < 1 
= a (1 – 1/Rⁿ) / ( 1 – 1/R ) (since r < 1 , we take r = 1/R).

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

Solution: Here a = 1, r = 2/1 = 2 , n = 8
Let S = 1 + 2 + 4 + 8 + …… to 8 terms
= 1 ( 28 – 1 ) / ( 2 – 1 ) = 28 – 1 = 255

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

Solution: Required Sum= ( 5 + 1 ) + (5² + 2 ) + ( 5³ +3 ) + ( 5⁴ + 4 ) + … to n terms 
= ( 5 + 5² +5³ + …… + 5ⁿ ) + ( 1 + 2 + 3 + .. + n terms)
= {5 ( 5ⁿ – 1 ) / (5 – 1 )} + {n ( n + 1 ) / 2}
= {5 ( 5ⁿ – 1 ) /4} + {n ( n + 1 ) / 2}

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

Solution: Let S denote the required sum.
i.e. S = 3 + 33 + 333 + ………….. to n terms = 3 (1 + 11 + 111 + ……. to n terms)
= 3 / 9 (9 + 99 + 999 + …. to n terms)

=

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

Solution: Let S denote the required sum. i.e. S = 0.7 + 0.77 + 0.777 + ….. to n terms

= 7 (0.1 + 0.11 + 0.111 + …. to n terms) 
= 7  / 9 (0.9 + 0.99 + 0.999 + … to n terms ) 

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 3 numbers be a/r, a, ar.
According to the question or a³ = 6³ ⇒ a =6
a/r × a × ar = 216
So the numbers are 6/r, 6, 6r
Again 6/r + 6 + 6r = 19
or 6/r + 6r = 13
or 6 + 6r² = 13r
or 6r² – 13r + 6 = 0
or 6r² – 4r – 9r + 6 = 0
or 2r(3r –2) – 3 (3r – 2) = 2
or (3r – 2) (2r – 3) = 0 or, r = 2/3 , 3/2
So the numbers are
6/(2/3), 6, 6 × (2/3 ) = 9 , 6 , 4
or 6/(3/2), 6 , 6 × (3/2) = 4 , 6 , 9

ILLUSTRATIONS:

(I) A person is employed in a company at ₹ 3000 per month and he would get an increase of ₹ 100 per year. Find the total amount which he receives 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 the interest are to be paid in the 10 monthly instalments. If each instalment is double the preceding one, find the value of the first and the last instalment.

SOLUTION:
Interest to be paid = 2.76 × 10 × 8000 / 100 × 12 = ₹ 184
Total amount to be paid in 10 monthly instalment is ` (8000 + 184) = ₹ 8184
The instalments form a G P with common ratio 2 and so ₹ 8184 = a (210 – 1 ) / ( 2 – 1 ), a = 1st instalment
Here a = ₹8184 / 1023 = ₹ 8
The last instalment = ar 10—1 = 8 × 29 = 8 × 512 = ₹ 4096