 Introduction
 Recursive definition
 Semantic Definition
 Arithmetic Sequence
 Arithmetic mean
 Arithmetic Series
 Geometric sequence
 Geometric mean
 Geometric Series
 Infinite Sum
 Harmonic sequence
 Harmonic Mean
 Harmonic Series
 A.M, G.M, H.M and their relations
 Visualization of the proof
 Application of AM, GM, HM
 Exercise
 Exercise: Miscellaneous Exercise [NCERT, pg 147]
 EXERCISE 8.2 [NCERT, page 145]
In mathematics, the word, “sequence” we usually mean a ordered collection with an identified first member, second member, third member and so on. is also called progression (AP).
Introduction
Sequence is a pattern of ordered numbers. Since the numbers follow a pattern, we can relate each number to its numerical position with a rule. Such a rule is called sequence. Each number in a sequence is a term of the sequence.
Presumably, a sequence continues by following the pattern that first few “terms”
suggest. To understand the pattern more explicitely, it is also useful to think of a sequence
as a function.
Thus,
So, a realvalued function defined on N = {1, 2, 3, . . .} is a
sequence.
The range of the function is
real numbers, the term of sequences
Mathematically, it is written as
\(f : N \to R\) .
A sequence can be defined with two different ways
 Recursive definition (Syntactic definition), Implicit definition
 Formal definition (Semantic definition), Explicit definition
Recursive definition
A sequence can be described by comparing each term to the one that comes before it, i.e., by defining the later term in relationto previous terms. Such rule of describing sequence is called recursive definition. The recursive definition contains two parts. twwo parts, the initial condition and definition. For example, in a sequence
133,130,127,124,...,
each term after the first term is equal to three less than the previous term.
Therefore, a recursive definition for this sequence is as follows
 an initial condition (the value of the first term): a_{1}=133
 a recursive definition (relates each term after the first term to the one before it): a_{n}=a_{n1}3 for n>1
Example 1
Writing a recursive definition for a sequenceThe number of blocks in two dimensional pyramid is a sequence that follows a recursive formula. What is the recursive definition of the sequence?
The solution is as follows.
 Let us count the number of blocks in each pyramid: it is 1,3,6,10,....
 Now, subtract consecutive terms to find out what happens from one term to the next
a_{2}a_{1}=31=2
a _{3}a_{2}=63=3
a _{4}a_{3}=106=4
 Now, use n to express the relationship between successive terms
a _{n}a_{n1}=n  To write the recursive definition, state the initial condition and the recursive formula
a _{1}=1 and a_{n}=a_{n1}+n
Example 2
What is the 100th term of the pyramid sequence in the example given below?
Solution
To find the explicit formula, expand the first few terms of the pyramid sequence. which is as below.
a_{1}  a_{2}  a_{3}  a_{4}  ...  a_{n} 
1  3  6  10  ...  a_{n} 
1  1+2  1+2+3  1+2+3+4  ...  1+=2+3+4+...+n 
Therefore,
a
_{n}=1+2+3+4++(n2)+(n1)+n (1)
Writing in reverse order, we get
a _{n}=n+(n1)+(n2)+...+4+3+2+1 (2)
Adding (1) and (2), we get
a_{n}  =1  +2  3  ++  (n2)  +(n1)  +n 
a_{n}  =n  +(n1)  +(n2)  ++  +3  +2  +1 
2a_{n}  =(n+1)  +(n+1)  +(n+1)  ++  +(n+1)  +(n+1)  +(n+1) 
2a_{n}  =n(n+1)  
a_{n}  =\(\frac{n(n+1)}{2} \) 
Therefore, the explicit formula, for this sequence is
a
_{n}=\( \frac{n(n+1)}{2} \)
Now, we substitute n by 100 to find the 100th term. Which is
a
_{n}=\( \frac{n(n+1)}{2} \)
or
a
_{100}=\( \frac{100(100+1)}{2}\)
or
a
_{100}=5050
Now, we define sequence mathematically.
Semantic Definition
A sequence is a function defined on the set \( \mathbb{N}\). For example, f(a_{n})=3a_{n}. There are different types of such sequences. Among them we discuss three basic types of sequence in the following section.
 Arithmetic sequence
 Geometric sequence
 Harmonic sequence
Arithmetic Sequence
An arithmetic sequence is a list of numbers with a definite pattern based on addition. If you take any number in the sequence then subtract it by the previous one, and the result is always the same or constant then it is an arithmetic sequence.
In the arithemetic sequence
The constant difference in all pairs of consecutive or successive numbers is called the common difference. It is denoted by the letter d.
Please note that
Difference here means the second minus the first. We add the common difference to go from one term to
another.
Therefore,
An arithmetic sequence is a sequence of numbers such that the difference between the consecutive terms is constant.
For example,
the sequence 5, 7, 9, 11, 13, 15, . . . is an arithmetic progression with common
difference of 2.
General term of arithmetic sequence
An arithmetic sequence with a starting value \(a\) and common difference \(d\) is a sequence of the form
\(a,a+d,a+2d,a+3d,\cdots\)
A recursive definition for this sequence has two parts
\(t_1=a\): initial condition
\(t_n=t_{n1}+d\) for n>1:recursive formula
Therefore, an semantic definition for this sequence is a single formula as general terms given following calulations
\(t_n=a+(n1)d\) for n > 1
\(t_1=a\)
\(t_2=a+(d)\)
or \(t_2=a+(21)d)\)
\(t_3=a+(21)d+d\)
or \(t_3=a+(31)d)\)
\(t_4=a+(31)d+d\)
or \(t_4=a+(41)d)\)
Similarly,
\(t_n=a+(n1)d\) for n > 1
Arithmetic mean
Let a, b, and c are three terms in an arithmetic sequence, then b is called Arithemetic mean of a and c.
In this case
, first two terms a and b will have the difference which will be equal to the next two terms b and c.
So we can
Thus,
\(ba = cb\).
Rearranging the terms, we get
or
\(2b = a + c \)
or
\(b = \frac{a+c}{2} \)
Arithmetic Series
An arithmetic series is the sum of the terms of an arithmetic sequence, denoted by \(S_n\)
To find the explicit formula for the S_{n}: Sum of arithmetic sequence up to n th term, we write
\(S_n\)  =a  +(a+d)  +(a+2d)  +...  +[a+(n2)d]  +[a+(n1)d] 
\(S_n\)  =[a+(n1)d]  +[a+(n2)d]  +...  +(a+2d)  +(a+d)  +a 
\(2S_n\)  =[2a+(n1)d]  +[2a+(n1)d]  +...  +...  +[2a+(n1)d]  +[2a+(n1)d] 
\(2S_n=n[2a+(n1)d]) \)
or
\(S_n=\frac{n}{2}[2a+(n1)d] \)
Example 3
If the first term of an AP is 67 and the common difference is 13, find the sum of the first 20 terms.
Here, a = 67 and d= 13, thus
S_{n} = \(\frac{n}{2}[2a+(n1)d]\)
or
S_{n} = \(\frac{20}{2}[2 \times 67+(201)(13)]=1130\)
So, the sum of first 20 terms is 1130.
Example 4
[Modeling]:Look at the figure below. The length of the side of each cube is 1cm. Copy the figure in isometric dot paper Based on the pattern
 Draw Figure no. 4 of this pattern on the dot paper
 Find the volume of the four Figures
 What would be the volume of Figure no. 12 of this pattern
 Write an equation to represent the volume of Figure n
Example 5
Show that sum of first n odd natural number is S_{n}=\(n^2\)
Here, a=1,d=2, thus the sum is
S_{n}=\( \frac{n}{2} [2a+(n1)d]\)
or
S_{n}=\( \frac{n}{2} [2.1+(n1)2]\)
or
S_{n}=\(\frac{n}{2} [2+2n2]\)
or
S_{n}=\(n^2\)
Example 6
Show that sum of first n even natural number is S_{n}=n(n+1)
Here, a=2,d=2, thus the sum is
S_{n}=\(\frac{n}{2} [2a+(n1)d]\)
or
S_{n}=\(\frac{n}{2} [2.2+(n1)2]\)
or
S_{n}=\(\frac{n}{2} [4+2n2]\)
or
S_{n}=n(n+1)
Geometric sequence
Euclid’s book The Elements (300 BC, Book VIII) introduces a “geometric progression” as a progression in which the ratio of any element to the previous element is a constant.
The Greeks, over two thousand years ago, considered sequences such as
\(\frac{1}{2},\frac{1}{4},\frac{1}{8},\frac{1}{16},....,\frac{1}{2^k},...\)
and their sums, such as
\(\displaystyle \sum_{k=1}^{k=\infty} \frac{1}{2^k}\)
The sum, above, at any finite step, is always less than the number 1.
Since the sum is less than 1 at any finite step , we can conclude that the series converges in the limit to 1
Does a series have a sum?
\( \sum_{n=1}^\infty \frac{1}{2^n}\)
Answer
\( \sum_{n=1}^\infty \frac{1}{2^n}=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+...=1\)
The sum of the series is 1.
The visualization is as follows
Imagine that we paint a blank canvas in steps. At each step, we paint half of the unpainted area. The total area painted after
"n" steps is therefore the "n"th partial sum
\( \frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+...+ \frac{1}{2^n}\)
The total area remaining unpainted is
\( \frac{1}{2^n}\)
After an infinite number of steps we will have painted all of the canvas, of which the area is 1.
Geometric sequence
An geometric sequence is a list of numbers with a definite pattern based on multiplication. If you take any number in the sequence then divide it by the previous one, and the result is always the same or constant then it is an geometric sequence.
In the geometric sequence,
The constant ratio in all pairs of consecutive or successive numbers in a sequence is called the common ratio. It is denoted by a letter r.
Please note that
Ratio here means the second divide the first. We use the common ratio to go from one term to another.
Therefore, a geometric sequence is a sequence of numbers in which the ratio between the consecutive terms is always constant. For instance,
the sequence 5, 15, 45,135, . . . is an geometric progression with common ratio of 3.
General term of geometric sequence
A geometric sequence with a starting value a and common ratio r is a sequence of the form
\(a,ar,ar^2,ar^3,...\)
A recursive definition for this sequence has two parts
\(t_1=a:\) initial condition
\(t_n=t_{n1}r\) for n>1 recursive formula
Therefore, an explicit definition for this sequence is a single formula as general terms given by
\(t_n=ar^{n1} \)for n > 1
Example 1
When a ball bounces as shown below, the height of consecutive bounces become a geometric sequence. If \(a_1=100,a_3=49\), thn what is the height of 5th bounce?
The height of the first and third bounces are given in the figure above.
Solution
\(a_1=100,a_3=49\)
Use the explicit formula to relate a_{1} to a_{3} and to find r.
or
a_{n}=\(a_1 r^{n1}\)
or
\(a_3=a_1 r^{31}\)
or
\(49=100 r^2\)
or
\(r=\frac{7}{10}\)
Find the fifth using explicit formula, thus we have
\( a_5=a_1 r^4\)
or
\(a_5=100 \times \left ( \frac{7}{10} \right ) ^4 \approx 24\)
Geometric mean
In the geometric sequence, if a, b and c are three consecutive terms, then b is called geometric mean of a and b.
In this sequence, the first two terms a and b will have the ratio which will be equal to the next two terms b and c.
So we can say,
\(\frac{b}{a}=\frac{c}{b}\)
Rearranging the terms, we get
\( b^2 = a \times c \)
or
\( b = \sqrt{ac}\)
Geometric Series
A geometric series is the sum of the terms of an geometric sequence, denoted by \(S_n\)
Let a geometric sequence is given by
\(a,ar,ar^2,ar^3,...\)
If we are adding up the terms of the geometric sequence given above, then we have a geometric sum or geometric series given by
\(\displaystyle \sum_{k=0}^{k=\infty} ar^k\)
Method 1
The explicit formula for the sum of finite terms is given as below
Sum of geometric sequence up to n th term: \(a,ar,ar^2,ar^3,...ar^{n1}\) is written as
S_{n}=\(a+(a r)+(a r^2)+(a r^3)++[a r^{n2}]+[a r^{n1}]\)
Taking a common, we get
S_{n}=\(a[1+r+r^2+r^3++r^{n2}+r^{n1}] \)
Multiplying by \(\frac{1r}{1r}\) , we get
S_{n}=\(a \frac{1r}{1r} [1+r+r^2+r^3++r^{n2}+r^{n1}] \)
or\(S_n=\frac{a(r^n1)}{r1}\)
Method 2
Sum of geometric sequence up to n th term: \(a,ar,ar^2,ar^3,...ar^{n1}\) is written as
S_{n}=\(a+(a r)+(a r^2)+(a r^3)++[a r^{n2}]+[a r^{n1}]\)
Multiplying by r, we get
rS_{n}=\( (a r)+(a r^2)+(a r^3)++[a r^{n1}]+[a r^n]\)
Substracting , we get
rS_{n}S_{n}=\(ar^na \)
or\(S_n=\frac{a(r^n1)}{r1}\)
Infinite Sum
We know that, sum of the first n terms of a geometric series with first term a and
common ratio r is
\(S_n=\frac{a(r^n1)}{r1} ; r \ne 1\)
In the case when r has magnitude less than 1, the term \(r^n\) approaches 0 as n becomes
very large. So, in this case, the sequence of partial sums S1,S2,S3,... has a limit:
So,
\(S_\infty= \displaystyle \lim_{n \to \infty} \frac{a(r^n1)}{r1} = \frac{a}{1r} \)
The limiting sum is usually referred to as the sum to infinity of the series and denoted
by S∞. Thus, for a geometric series with common ratio r such that r  < 1, we have
\(S_\infty= \frac{a}{1r} \)
Example 2
A person saves NRs100 in a bank account at the beginning of each month. The bank offers a return of 12% compounded monthly.
(a) Determine the total amount saved after 12 months.
Solution
Let
S_{n}: Sum of geometric sequence up to n th term.
It is written as
S_{n}=\( \frac{a(1r^n)}{1r} \)
Based on the formula, the solution of above problem with
\(a = 100, r=1.12, n = 12\)
The solution is
\(S_{12}=\frac{a(r^n1)}{r1}= 1268.25\)
Example 3
What is the sum of the finite geometric series 3+6+12+24+...+3072
Given that, the first term is 3, the common ratio is 2, and the nth term is 3072, therefore, we use explicit formula to find the n
or \(3072=3 \times 2^{n1} \)
or \(1024=2^{n1} \)
or \(n=11 \)
Therefore, the sum up to 11th term is
\(S_n= \frac{a(1r^n)}{1r} \)
or \(S_{11}= \frac{3(12^{11})}{12} \)
or \(S_{11}= 6141\)
Harmonic sequence
In mathematics, a harmonic sequence formed by taking the reciprocals of an arithmetic progression.
The sequence 1,2,3,4,5,6,... is an arithmetic progression, so its reciprocals
\( \frac{1}{1}, \frac{1}{2},
\frac{1}{3}, \frac{1}{4}, ... \)
is a harmonic sequence.
If each term of an harmonic sequence is multiplied or divide by a constant, the sequence of the resulting number are also harmonic sequence. For example,
if \(a,b,c,d, \cdots \) is harmonic sequence, then then \(ka,kb,kc,kd, \cdots \) is harmonic sequence where k is constant and \(k \ne 0\)
 \(\frac{a}{k},\frac{b}{k},\frac{c}{k},\frac{d}{k},\cdots \) is a harmonic sequence where \(k \ne 0\)
General term of Harmonic sequence
Let \(a,a+d,a+2d,a+3d, \cdots\) are the terms of arithemetic sequence
Then
terms of Harmonic sequence are given by
\( \frac{1}{a},\frac{1}{a+d},\frac{1}{a+2d}+\frac{1}{a+3d},\cdots\)
Now,
The nth term of the sequence is given by
\( a_n=\frac{1}{a+(n1)d}\)
Harmonic Mean
Harmonic mean is finding taking the reciprocal of the arithmetic mean of the reciprocals. The formula to calculate the harmonic mean is given by:
Harmonic Mean =\( \frac{2}{(\frac{1}{a})+(\frac{1}{b})}=\frac{2ab}{a+b} \)
Where a,b are the values of arithmetic sequence.
Harmonic Series
If \( \frac{1}{a},\frac{1}{a+d},\frac{1}{a+2d},…,\frac{1}{a+(n1)d}\) is given harmonic progression, the formula to find the sum of n terms in the harmonic progression is given by the formula:
Sum of nth terms, \( S_n= \frac{1}{d} \ln \left [ \frac{2a+(2n–1)d}{2a–d} \right ]\)
where,
“a” is the first term of A.P
“d” is the common difference of A.P
“ln” is the natural logarithm
Proof
Given \( \frac{1}{a},\frac{1}{a+d},\frac{1}{a+2d},…,\frac{1}{a+(n1)d}\) are in harmonic progression,
We set the values as follows,
a=1,d=1
Then, \( \frac{1}{1},\frac{1}{1+1},\frac{1}{1+2.1},…,\frac{1}{1+(n1)1}\) are the terms
\( 1,\frac{1}{2},\frac{1}{3}, ...,\)are the terms
Now, the Riemann sum of the function \( f(x)=\frac{1}{x}\) approximates the sum of the harmonic series given above
Therefore,
We set the values as follows,
\( S_n= \int_{a+(0\frac{1}{2})d}^{a+(n\frac{1}{2})d} \frac{1}{x} dx \)
For any common difference,the formula becoms
\( S_n=\frac{1}{d} \int_{a+(0\frac{1}{2})d}^{a+(n\frac{1}{2})d} \frac{1}{x} dx \)
or \( S_n= \frac{1}{d} \int_{a\frac{d}{2}}^{a+(n\frac{1}{2})d} \frac{1}{x} dx \)
or \( S_n= \frac{1}{d} \left [ \ln x \right]_{a\frac{d}{2}}^{a+(n\frac{1}{2})d} \)
or \( S_n= \frac{1}{d} \ln \left [ \frac{2a+(2n–1)d}{2a–d} \right ]\)
Therefore
Sum of nth terms, \( S_n= \frac{1}{d} \ln \left [ \frac{2a+(2n–1)d}{2a–d} \right ]\)
A.M, G.M, H.M and their relations
A finite sequence consisting more than two terms has one or more terms in between the first and last terms. Theses between terms are called means of the sequence. Precisely
 if a,b,c,d are in arithmetic sequence then b,c are arithmetic means (AM)
So, given n terms, the AM is
\(AM=\frac{a+b+ \cdots}{n} \)
 item if a,b,c,d are in geometric sequence then b,c are geometric means (GM)
So, given n terms, the GM is
\(GM=\sqrt[n]{a \times \times \cdots } \)
 if a,b,c,d are in harmonic sequence then b,c are harmonic means (HM)
So, given n terms, the HM is
\(HM=\frac{n}{(\frac{1}{a})+(\frac{1}{b})+ \cdots}\)
Theorem 1
Given any two numbers a and b the AM, GM, and HM are as follows.
 AM=\(\frac{a+b}{2}\)
 GM=\(\sqrt{ab}\)
 HM=\(\frac{2ab}{a+b}\)
The proof are as follows:
 Let AM is the single mean between a and b, then
AMa=bAM
or 2AM=a+b
or AM=\(\frac{a+b}{2}\)
 Let GM is the single mean between a and b, then
\( \frac{GM}{a}=\frac{b}{GM}\)
or \(GM^2=ab \)
or \( GM=\sqrt{ab}\)
 Let HM is the single mean between a and b, then
\( \frac{1}{HM}\frac{1}{a}=\frac{1}{b}\frac{1}{HM}\)
or \( \frac{2}{HM}=\frac{1}{a}+\frac{1}{b}\)
or \( HM=\frac{2ab}{a+b}\)
Theorem 2
Let a and b are two nonnegative numbers, then
 \( GM^2=AM \times HM \)
 \( AM \ge GM \ge HM \) Arithmetic mean is greater than geometric mean and harmonic mean, and geometric mean is greater than harmonic mean.
Let a and b are two nonnegative numbers then,
\( AM=\frac{a+b}{2}, GM=\sqrt{ab}, HM=\frac{2ab}{a+b}\)
The proof are as follows:

Now, we have
\( GM^2=ab\)
or \( GM^2=\frac{a+b}{2} \times \frac{2ab}{a+b} \)
or \( GM^2=AM \times HM \) 
Now, we have
\(AMGM=\frac{a+b}{2}\sqrt{ab}\)
or \(AMGM=\frac{a+b2\sqrt{ab}}{2}\)
or \(AMGM=\frac{{{\sqrt{a}}^{2}}+{{\sqrt{b}}^{2}}2\sqrt{a}\sqrt{b}}{2}\)
or \(AMGM=\frac{{{( \sqrt{a}\sqrt{b} )}^{2}}}{2}\)
or \(AM\ge GM\) (1)
Similarly,
\(GMHM=\sqrt{ab}\frac{2ab}{a+b}\)
or \(GMHM=\frac{\sqrt{ab}( a+b )2ab}{a+b}\)
or \(GMHM=\frac{\sqrt{ab}( a+b )2\sqrt{ab}\sqrt{ab}}{a+b}\)
or \(GMHM=\frac{\sqrt{ab}}{a+b}( a+b2\sqrt{ab} ) \)
or \(GMHM=\frac{\sqrt{ab}}{a+b}{{( \sqrt{a}\sqrt{b} )}^{2}}\)
or \(GM\ge HM\)(2)
Combining (1) and (2), we get
\(AM\ge GM\ge HM\)
Visualization of the proof
Let us suppose that a and b are two given numbers. Now, draw a semi circle with diameter a+b.

Visualization of AM
By the property of radius and diameter, we get that
\( AM =\frac{a+b}{2} \) 
Visualization of GM
By the mean proportionality property (squaring a rectangle), we can obtain by using the property of similarity that, DQ is the geometric mean given by
\( GM =\sqrt{ab} \) 
Visualization of HM
By using proportionality, we get
Triangle ADQ and QDB are similar with AD=a, DB=b, so we have
\( \frac{GM}{a}=\frac{b}{GM} \)
or \( GM= \sqrt{ab} \)
Again, by using the property of similarity on OCDE, we get that, QR is the harmonic mean given by
\( HM =\frac{2ab}{a+b} \)
By using proportionality, we get
Triangle DRQ and ODQ are similar with QR=GM,QD=\(\sqrt{ab}\), OD=\(\frac{ab}{2}\), so we have
\( \frac{HM}{\sqrt{ab}}=\frac{\sqrt{ab}}{\frac{a+b}{2}} \)
or \( HM= \frac{2ab}{a+b} \)
Application of AM, GM, HM

Let Ram has Rs 1oo and Shyam has Rs 120, then the avarage amount is AM, which is answered by
\( AM =\frac{a+b}{2} =\frac{100+120}{2} =Rs 110\) 
Let Ram deposited Rs 100 in a bank, on which 4% profit in first year and 16% profit in second year, then the avarage profit is GM, which is answered by
\( GM =\sqrt{4 \times 16}=8\%\)
Please note that, the situation can NOT be explained by \(\frac{4+16}{2} =10\%\). 
Let Ram travelled 100km with fuel efficiency 25KM per liter and next 100km with fuel efficiency 16KM per liter, then the avarage fuel efficiency is HM, which is answered by
\( HM =\frac{2*25*16}{25+16}=19.51\)
Please note that, the situation can NOT be explained by AM or GM
Because
The full efficiency for first 100 km is \( \frac{100}{25}=4\) liter
The full efficiency for second 100 km is \( \frac{100}{16}=6.25\) liter
The total fuel efficiency is
\(\frac{200}{10.25}=19.51\)
Solved Example
Compare the sum of n terms of the series: \(1+2a+3a^2+4a^3+........ .\) and \(a+2a+3a+ 4a ...\) up to n term
Solution
Given that
\(S_n=1+2a+3a^2+4a^3+\cdots+na^{n1} \)
\(T_n=a+2a+3a+ 4a +\cdots+na \)
Now, substracting, we get
\(S_nT_n= (1+2a+3a^2+4a^3+\cdots+na^{n1})(a+2a+3a+ 4a +\cdots+na) \)
or \(S_nT_n= (1+2a+3a^2a2a3a)+4a(a^21)+5a(a^31)+\cdots + na(a^{n2}1) \)
or \(S_nT_n= (1+3a^2a3a)+4a(a^21)+5a(a^31)+\cdots + na(a^{n2}1) \)
or \(S_nT_n= (a1)(3a1)+4a(a^21)+5a(a^31)+\cdots + na(a^{n2}1) \)
Case 1: If a=1, then
\(S_nT_n= (a1)(3a1)+4a(a^21)+5a(a^31)+\cdots + na(a^{n2}1)\)
\(S_nT_n= (0)+(0)+(0)+\cdots+(0)\)
or \(S_nT_n= 0\)
or \(S_n =T_n\) (1)
Case 1: If a >1, then
\(S_nT_n= (a1)(3a1)+4a(a^21)+5a(a^31)+\cdots + na(a^{n2}1)\)
\(S_nT_n= (+ve)+(+ve)+(+ve)+\cdots+(+ve)\)
or \(S_nT_n= Positive\)
or \(S_n T_n>0\)
or \(S_n >T_n= 0\) (2)
Thus, the comparision is
\(S_n = T_n \)if a =1
\(S_n > T_n \)if a > 1
Exercise
 Integration to Investment and Growth: Determine the monthly repayments needed to repay a 100000 loan which is paid back over 25 years when the interest rate is 8% compounded annually[Ans: 780.66]
 Show that if three quantities form any two of the three sequences AS,GS,and HS then they also form the remaining third sequence
 The sum of three numbers in AP is 36. When the numbers are increased by 1,4,43 respectively, the resulting numbers are in GP, find the numbers.
 Divide 69 in three parts in AP and the product of first two parts is 483
 If a be the AM between b and c, b be the GM between c and a, then prove that c will be HM between a and b.
 Show that b^2 is greater than, equal to, or less than according as a,b,c are in AP, GP or HP
 If a and b are two positive numbers then prove that AM,GM,HM form GP.
 If the \(4^{th}\) and \(15^{th}\) terms of an arithmetic series are 11 and 44 respectively then, find the sum of its first 20 terms. (Ans:610)
 Puja gets an employment of Rs.25000 in a month with an annual increment of Rs. 1500. How much does she earn in six years? Find her salary at 12^{th} year. (Ans: Rs.190500, Rs.41500)
 In a geometric series, if the sixth term is 16 times the second term and the sum of first seven terms is \(\frac{127}{4}\) , then find the sum of the first 15 term. (Ans: \(\frac{32767}{4}\))
 There are 8 bags full of books. The numbers of books in each bag form a GP. If the \(4^{th}\) and \(6^{th}\) bags contain 24 and 96 books respectively, find the number of the books in the \(1^{th}\) and last bags. ( Ans: 3, 384)
 One side fo a square is 10 cm . The mid point of its sides are joined to form another square, whose mid point are again joined t form one more square. The process is continued indefinitely. Find the sum of the areas of all the squares so formed. (Ans:200 square cm)
Exercise: Miscellaneous Exercise [NCERT, pg 147]
 If f is a function satisfying f (x +y) = f(x) f(y) for all x, y ∈ N such that f(1) = 3 and \( \displaystyle \sum_{x=1}^{n} f(x)=120\),find the value of n.
 The sum of some terms of G.P. is 315 whose first term and the common ratio are 5 and 2, respectively. Find the last term and the number of terms.
 The first term of a G.P. is 1. The sum of the third term and fifth term is 90. Find the common ratio of G.P.
 The sum of three numbers in G.P. is 56. If we subtract 1, 7, 21 from these numbers in that order, we obtain an arithmetic progression. Find the numbers.
 A G.P. consists of an even number of terms. If the sum of all the terms is 5 times the sum of terms occupying odd places, then find its common ratio.
 If \( \frac{a+bx}{abx}=\frac{b+cx}{bcx}=\frac{c+dx}{cdx}\)(x≠ 0), then show that a, b, c and d are in G.P.
 Let S be the sum, P the product and R the sum of reciprocals of n terms in a G.P. Prove that \(P^2R^n = S^n\).
 If a, b, c, d are in GP, prove that \((a^n + b^n), (b^n + c^n), (c^n + d^n)\) are in GP
 If a and b are the roots of \(x^2 – 3x + p = 0\) and c, d are roots of \(x^2 – 12x + q = 0\), where a, b, c, d form a GP. Prove that (q + p) : (q – p) = 17:15.
 The ratio of the A.M. and G.M. of two positive numbers a and b, is m : n. Show that \(a:b=\left ( m+\sqrt{m^2n^2} \right ) : \left ( m\sqrt{m^2n^2} \right ) \)
 Find the sum of the following series up to n terms:
(i) 5 + 55 +555 + …
(ii) 0.6 +0.66 +0.666+…  Find the 20th term of the series 2 × 4 + 4 × 6 + 6 × 8 + ... + n terms.
 A farmer buys a used tractor for Rs 12000. He pays Rs 6000 cash and agrees to pay the balance in annual instalments of Rs 500 plus 12% interest on the unpaid amount. How much will the tractor cost him?
 Bidhan buys a scooter for Rs 22000. He pays Rs 4000 cash and agrees to pay the balance in annual instalment of Rs 1000 plus 10% interest on the unpaid amount. How much will the scooter cost him?
 A person writes a letter to four of his friends. He asks each one of them to copy the letter and mail to four different persons with instruction that they move the chain similarly. Assuming that the chain is not broken and that it costs 50 paise to mail one letter. Find the amount spent on the postage when 8th set of letter is mailed.
 A man deposited Rs 10000 in a bank at the rate of 5% simple interest annually. Find the amount in 15th year since he deposited the amount and also calculate the total amount after 20 years.
 A manufacturer reckons that the value of a machine, which costs him Rs. 15625, will depreciate each year by 20%. Find the estimated value at the end of 5 years.
 150 workers were engaged to finish a job in a certain number of days. 4 workers dropped out on second day, 4 more workers dropped out on third day and so on. It took 8 more days to finish the work. Find the number of days in which the work was completed.
EXERCISE 8.2 [NCERT, page 145]
 Find the 20th and nth terms of the G.P. \(\frac{5}{2},\frac{5}{4},\frac{5}{8},...\)
 Find the 12th term of a G.P. whose 8th term is 192 and the common ratio is 2.
 The 5th, 8th and 11th terms of a G.P. are p, q and s, respectively. Show that \(q^2 = ps\).
 The 4th term of a G.P. is square of its second term, and the first term is – 3. Determine its 7th term.
 Which term of the following sequences:
 \(2,2\sqrt{2},4,... \) is 128 ?
 \(\sqrt{3},3,3\sqrt{3},... \) is 729 ?
 \(\frac{1}{3},\frac{1}{9},\frac{1}{27},...\) is \(\frac{1}{19683}\)
 For what values of x, the numbers \( \frac{2}{7},x,\frac{7}{2}\) are in G.P.?
 0.15, 0.015, 0.0015, ... 20 terms
 \(\sqrt{7},\sqrt{21},3\sqrt{7},...\) n terms
 \(1,a,a^2,a^3,...\) n terms (if a≠1)
 \(x^3,x^5,x^7,...\) n terms (if \(n \ne \pm\) 1)
 Evaluate \(\displaystyle \sum_{k=1}^{11} (2+3^k)\)
 The sum of first three terms of a G.P. is \(\frac{39}{10}\) and their product is 1. Find the common ratio and the terms.
 How many terms of G.P. \(3, 3^2, 3^3\), … are needed to give the sum 120?
 The sum of first three terms of a G.P is 16 and the sum of the next three terms is 128. Determine the first term, the common ratio and the sum to n terms of the G.P.
 Given a G.P. with a = 729 and 7th term 64, determine \(S_7\).
 Find a G.P. for which sum of the first two terms is – 4 and the fifth term is 4 times the third term.
 If the 4th, 10th and 16th terms of a G.P. are x, y and z, respectively. Prove that x, y, z are in G.P.
 Find the sum to n terms of the sequence, 8, 88, 888, 8888… .
 Find the sum of the products of the corresponding terms of the sequences 2, 4, 8,16, 32 and \(128, 32, 8, 2,\frac{1}{2}\)
 Show that the products of the corresponding terms of the sequences \(a, ar, ar^2,…ar^{n – 1}\) and \(A, AR, AR^2, … AR^{n – 1}\) form a G.P, and find the common ratio.
 Find four numbers forming a geometric progression in which the third term is greater than the first term by 9, and the second term is greater than the 4th by 18.
 If the pth, qth and rth terms of a G.P. are a, b and c, respectively. Prove that \(a^{q – r} b^{r – p}c^{p – q} = 1\).
 If the first and the nth term of a G.P. are a and b, respectively, and if P is the product of n terms, prove that \(p^2 = (ab)^n\).
 Show that the ratio of the sum of first n terms of a G.P. to the sum of terms from \((n+1)^{th}\) to \((2n)^{th}\) term is \(\frac{1}{r^n}\)
 If a, b, c and d are in G.P. show that \((a^2 + b^2 + c^2) (b^2 + c^2 + d^2) = (ab + bc + cd)^2\) .
 Insert two numbers between 3 and 81 so that the resulting sequence is G.P.
 Find the value of n so that \(\frac{a^{n+1}+b^{n+1}}{a^n+b^n} \) may be the geometric mean between a and b.
 The sum of two numbers is 6 times their geometric mean, show that numbers are in the ratio \( \left ( 3+2\sqrt{2} \right ): \left ( 32\sqrt{2}\right )\)
 If A and G be A.M. and G.M., respectively between two positive numbers, prove that the numbers are \( A \pm \sqrt{(A+G)(AG)}\)
 The number of bacteria in a certain culture doubles every hour. If there were 30 bacteria present in the culture originally, how many bacteria will be present at the end of 2nd hour, 4th hour and nth hour ?
 What will Rs 500 amounts to in 10 years after its deposit in a bank which pays annual interest rate of 10% compounded annually?
 If A.M. and G.M. of roots of a quadratic equation are 8 and 5, respectively, then obtain the quadratic equation.
Find the sum to indicated number of terms in each of the geometric progressions in Exercises 7 to 10:
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