Sequences and Series Class 11 Notes AP Inter 1st Year Maths Chapter 8

Students can go through AP Inter 1st Year Maths Notes 8th Lesson Sequences and Series will help students in revising the entire concepts quickly.

Sequences and Series Class 11 Notes AP Inter 1st Year Maths 8th Lesson

→ By a sequence, we mean an arrangement of numbers in a definite order according to some rule. Also, we define a sequence as a function whose domain is the set of natural numbers or some subsets of the type {1, 2, 3,… k}. A sequence containing a finite number of terms is called a finite sequence. A sequence is called infinite if it is not a finite sequence.

→ Let a₁, a₂, a₃,…. be the sequence, then the sum expressed as a₁ + a₂ + a₃ + …….. is called series. A series is called a finite series if it has got finite number of terms.

→ A sequence is said to be a geometric progression or G.P., if the ratio of any term to its preceding term is the same throughout. This constant factor is called the common ratio. Usually, we denote the first term of a G.P. by a and its common ratio by r. The general or the nth term of G.P. is given by a_n = ar^n-1. The sum S_n of the first n terms of G.P. is given by S_n = \(\frac{\mathrm{a}\left(\mathrm{r}^{\mathrm{n}}-1\right)}{\mathrm{r}-1}\) or \(\frac{a\left(1-r^n\right)}{1-r}\), r ≠ 1

→ The geometric mean (G.M) of any two positive numbers a and b is given by \(\sqrt{a b}\) i.e., the sequence a, G, b is G.P.

Example Problems

Question 1.
Write the first three terms in each of the following sequences defined by the following:
(i) a_n = 2n + 5
(ii) a_n = \(\frac{n-3}{4}\)
Solution:
(i) Here a_n = 2n + 5
Substituting n = 1, 2, 3, we get
a₁ = 2(1) + 5 = 7
a₂ = 9
a₃ = 11
Therefore, the required terms are 7, 9, and 11.

(ii) Here a_n = \(\frac{n-3}{4}\)
Thus a₁ = \(\frac{1-3}{4}=-\frac{1}{2}\)
a₂ = \(-\frac {1}{4}\)
a₃ = 0
Hence, the first three terms are \(-\frac {1}{2}\), \(-\frac {1}{4}\), and 0.

Question 2.
What is the 20th term of the sequence defined by a_n = (n – 1)(2 – n)(3 + n)?
Solution:
Putting n = 20, we obtain
a₂₀ = (20 – 1)(2 – 20)(3 + 20)
= 19 × (-18) × 23
= -7866

Question 3.
Let the sequence a_n be defined as follows:
a₁ = 1, a_n = a_n-1 + 2 for n ≥ 2.
Find the first five terms and write the corresponding series.
Solution:
We have a₁ = 1
a₂ = a₁ + 2 = 1 + 2 = 3
a₃ = a₂ + 2 = 3 + 2 = 5
a₄ = a₃ + 2 = 5 + 2 = 7
a₅ = a₄ + 2 = 7 + 2 = 9
Hence, the first five terms of the sequence are 1, 3, 5, 7, and 9.
The corresponding series is 1 + 3 + 5 + 7 + 9 +…

Question 4.
Find the 10th and nth terms of the G.P. 5, 25, 125,….
Solution:
Here a = 5 and r = 5
Thus, a₁₀ = 5(5)^10-1 = 5(5)⁹ = 5
and a_n = ar^n-1 = 5(5)^n-1 = 5ⁿ

Question 5.
Which term of the G.P., 2, 8, 32,… up to n terms, is 131072?
Solution:
Let 131072 be the n^th term of the given G.P.
Here, a = 2 and r = 4.
Therefore 131072 = a_n = 2(4)^n-1 or 65536 = 4^n-1
This gives 4⁸ = 4^n-1
So that n – 1 = 8, i.e., n = 9.
Hence, 131072 is the 9th term of the G.P.

Question 6.
In a G.P., the 3^rd term is 24, and the 6^th term is 192. Find the 10^th term.
Solution:
Here, a₃ = ar² = 24 ……….(1)
and a₆ = ar⁵ = 192 ……….(2)
Dividing (2) by (1), we get r = 2.
Substituting r = 2 in (1), we get a = 6.
Hence a₁₀ = 6(2)⁹ = 3072.

Question 7.
Find the sum of the first n terms and the sum of the first 5 terms of the geometric series \(1+\frac{2}{3}+\frac{4}{9}+\ldots \ldots\)
Solution:
Here a = 1 and r = \(\frac {2}{3}\)
Therefore

In particular,

Question 8.
How many terms of the G.P. 3, \(\frac {3}{2}\), \(\frac {3}{4}\),….. are needed to give the sum \(\frac {3069}{512}\)?
Solution:
Let n be the number of terms needed.
Given that a = 3, r = \(\frac {1}{2}\) and S_n = \(\frac {3069}{512}\)

Question 9.
The sum of the first three terms of a G.P. is \(\frac {13}{12}\) and their product is -1. Find the common ratio and the terms.
Solution:
Let \(\frac {a}{r}\), a, ar be the first three terms of the G.P.
Then \(\frac{a}{r}+a r+a=\frac{13}{12}\) ……(1)
and (\(\frac {a}{r}\))(a)(ar) = -1 ……..(2)
From (2), we get a³ = -1,
i.e., a = -1 (considering only real roots)
Substituting a = -1 in (1), we have
\(-\frac{1}{\mathrm{r}}-1-\mathrm{r}=\frac{13}{12}\)
or 12r² + 25r + 12 = 0
This is a quadratic in resolving, we get r = \(-\frac {3}{4}\) or \(-\frac {4}{3}\).
Thus, the three terms of G.P. are:
\(\frac {4}{3}\), -1, \(\frac {3}{4}\) for r = \(-\frac {3}{4}\)
and \(\frac {3}{4}\), -1, \(\frac {4}{3}\) for r = \(-\frac {4}{3}\)

Question 10.
Find the sum of the sequence 7, 77, 777, 7777, … to n terms.
Solution:
This is not a G.P.; however, we can relate it to a G.P. by writing the terms as
S_n = 7 + 77 + 777 + 7777 + … to n terms
= \(\frac {7}{9}\)[9 + 99 + 999 + 9999 + … to n term]
= \(\frac {1}{2}\)[(10 – 1) + (10² – 1) + (10³ – 1) + (10⁴ – 1) + … n terms]
= \(\frac {1}{2}\)[(10 + 10² + 10³ + … n terms) – (1 + 1 + 1 +… n terms)]
= \(\frac{7}{9}\left[\frac{10\left(10^n-1\right)}{10-1}-n\right]\)
= \(\frac{7}{9}\left[\frac{10\left(10^n-1\right)}{9}-n\right]\)

Question 11.
A person has 2 parents, 4 grandparents, 8 great-grandparents, and so on. Find the number of his ancestors during the ten generations preceding his own.
Solution:
Here a = 2, r = 2 and n = 10
Using the sum formula
S_n = \(\frac{a\left(\dot{r}^n-1\right)}{r-1}\)
We have S₁₀ = 2(2¹⁰ – 1) = 2046
Hence, the number of ancestors preceding the person is 2046.

Question 12.
Insert three numbers between 1 and 256 so that the resulting sequence is a G.P.
Solution:
Let G₁, G₂, G₃ be three numbers between 1 and 256 such that 1, G₁, G₂, G₃, 256 is a G.P.
Therefore 256 = r⁴ giving r = ±4 (Taking real roots only)
For r = 4, we have G₁ = ar = 4,
G₂ = ar² = 16,
G₃ = ar³ = 64
Similarly, for r = -4, numbers are -4, 16, and -64.
Hence, we can insert 4, 16, 64 between 1 and 256 so that the resulting sequences are in G.P.

Question 13.
If A.M. and G.M. of two positive numbers a and b are 10 and 8, respectively, find the numbers.
Solution:
Given that A.M = \(\frac{a+b}{2}\) = 10 ……(1)
and G.M = \(\sqrt{a b}\) = 8 …….(2)
From (1) and (2), we get
a + b = 20 ……(3)
ab = 64 …..(4)
Putting the values of a and b from (3), (4) in the identity
(a – b)² = (a + b)² – 4ab, we get
(a – b)² = 400 – 256 = 144
or a – b = ±12 …..(5)
Solving (3) and (5), we obtain
a = 4, b = 16 or a = 16, b = 4
Thus, the numbers a and b are 4, 16, or 16, 4, respectively.

Question 14.
If a, b, c, d, and p are different real numbers such that (a² + b² + c²)p² – 2(ab + bc + cd)p + (b² + c² + d²) ≤ 0, then show that a, b, c, and d are in G.P.
Solution:
Given that (a² + b² + c²)p² – 2(ab + bc + cd)p + (b² + c² + d²) ≤ 0 ……(1)
But L.H.S = (a²p² – 2abp + b²) + (b²p² – 2bcp + c²) + (c²p² – 2cdp + d²),
which gives (ap – b)² + (bp – c)² + (cp – d)² ≥ 0 …..(2)
Since the sum of squares of real numbers is non-negative,
Therefore, from (1) and (2), we have,
(ap – b)² + (bp – c)² + (cp – d)² = 0
or ap – b = 0, bp – c = 0, cp – d = 0
This implies that \(\frac{b}{a}=\frac{c}{b}=\frac{d}{c}\) = p
Hence, a, b, c, and d are in G.P.

Multiple Choice Questions

Question 1.
Which of the following is a finite sequence?
(1) 48, 24, 12,…….
(2) 1, 2, 3,…….
(3) 2, 4, 6, 8, 10,…….
(4) 2, 3, 5, 7, -11, 13,……
Answer:
(3) 2, 4, 6, 8, 10

Question 2.
Which of the following relations gives the Fibonacci sequence?
(1) a_n = a_n-1 + a_n-2, n > 2
(2) a_n-1 = a_n + a_n-2, n > 2
(3) a_n-2 = a_n + a_n-1, n > 2
(4) a_n = a_n+1 + a_n-2, n > 2
Answer:
(1) a_n = a_n-1 + a_n-2, n > 2

Question 3.
The first term of the Fibonacci sequence is
(1) 0
(2) 1
(3) 2
(4) 3
Answer:
(2) 1

Question 4.
If a_n = 4n + 6, then 15th term of the sequence is
(1) 6
(2) 10
(3) 60
(4) 66
Answer:
(4) 66

Question 5.
A series can also be denoted by symbol ________
(1) π a_n
(2) Σ a_n
(3) φ a_n
(4) θ a_n
Answer:
(2) Σ a_n

Question 6.
The sum of the first five terms of the series 2 + 4 + 6 +… is
(1) 14
(2) 16
(3) 20
(4) 30
Answer:
(4) 30

Question 7.
\(\sum_{n=1}^4 2 n+3\) = ________
(1) 5
(2) 12
(3) 21
(4) 32
Answer:
(4) 32

Question 8.
If a_n+1 = a_n.r then the sequence is called ________
(1) arithmetic progression
(2) geometric progression
(3) harmonic progression
(4) special progression
Answer:
(2) geometric progression

Question 9.
The nth term of G.P. is
(1) a_n = a + (n – 1)d
(2) a_n = a + nd
(3) a_n = a.r^n-1
(4) a_n = arⁿ
Answer:
(3) a_n = a.r^n-1

Question 10.
If r = 1 in G.P., then the sum of the first n terms is
(1) na
(2) \(\frac {a}{n}\)
(3) (n – 1)a
(4) (n + 1)a
Answer:
(1) na

Question 11.
If 4, 8, 16, 32,……. are in G.P., then the sum upto the 5th term is
(1) 16
(2) 64
(3) 128
(4) 124
Answer:
(4) 124

Question 12.
The sum of series \(1+\frac{1}{2}+\frac{1}{4}+\ldots \ldots \ldots .\) upto 6 terms is
(1) \(\frac {63}{32}\)
(2) \(\frac {32}{63}\)
(3) \(\frac {26}{53}\)
(4) \(\frac {53}{26}\)
Answer:
(1) \(\frac {63}{32}\)

Question 13.
The geometric mean of 3 and 12 is ________
(1) 4
(2) 6
(3) 9
(4) 12
Answer:
(2) 6

Question 14.
If A.M. of two numbers is \(\frac {15}{2}\) and their G.M. is 6 then the two numbers are
(1) 6 and 8
(2) 12 and 3
(3) 24 and 6
(4) 27 and 3
Answer:
(2) 12 and 3

Question 15.
If A means arithmetic mean and G means geometric mean, then which of the following is true?
(1) A > G
(2) A ≥ G
(3) G > A
(4) G ≥ A
Answer:
(2) A ≥ G

Question 16.
The ratio of A.M and G.M. of two positive numbers a and b is 5 : 3. Then the ratio of a to b is
(1) 9 : 1
(2) 3 : 5
(3) 1 : 9
(4) 3 : 1
Answer:
(1) 9 : 1