Introduction to geometric progressions - class-X

The sequence is also called as Progression.

A progression of the form $a, ar, ar^2$, ..... is a geometric progression.
Geometric Progression refers to a sequence in which successor term of each term is obtained by multiplying a constant term.

The geometric progression which have infinite terms is called infinite geometric progression.
$1 + 0.5 + 0.25 + 0.125....$ is an example of infinite geometric progression.

$b^{2} = ac$ satisfies (ii).

$Given\>series\>is\>an\>infinite\>GP\>with\>first\>term\>=1\>and\>common\>ratio=1/2\\\therefore\>sum=(\frac{a}{1-r})\\=(\frac{a}{1-1/2})=2$

$\begin{array}{l}S = {3^{10}} + {3^9} + \cfrac{{{3^9}}}{4} + \cfrac{{{3^7}}}{2} + \cfrac{{{{5.3}^6}}}{{16}} + \cfrac{{{3^2}}}{{16}} + \cfrac{{{{7.3}^4}}}{{64}} + .........\infty \S = \cfrac{{1 \times {3^{10}}}}{{{2^0}}} + \cfrac{{2 \times {3^9}}}{{{2^1}}} + \cfrac{{3 \times {3^8}}}{{{2^2}}} + \cfrac{{4 \times {3^7}}}{{{2^3}}} + \cfrac{{5 \times {3^6}}}{{{2^4}}} + \cfrac{{6 \times {3^5}}}{{{2^5}}} + \cfrac{{7 \times {3^4}}}{{{2^6}}} + .....\infty \\cfrac{S}{6} = \cfrac{{{3^9}}}{2} + \cfrac{{2 \times {3^8}}}{{{2^2}}} + .......\infty \S - \cfrac{S}{6} = \cfrac{{{3^{10}}}}{2^0} + \cfrac{{{3^9}}}{{{2^1}}} + \cfrac{{{3^8}}}{{{2^2}}} + ........\infty \\cfrac{{6S - S}}{6} = \cfrac{{{3^{10}}}}{{1 - \cfrac{1}{6}}} = \cfrac{{{3^{10}}}}{5}\left( 6 \right) \end{array}$

$4,64,p$ are in GP 

G.P series with common ratio $\dfrac 13$

Given series is $1+ \dfrac {1}{3}+ \dfrac {1}{9 }+ \dfrac {1}{27}+......$

$\displaystyle { ar }^{ 4 }=48$
$\displaystyle { ar }^{ 8 }=768$
$\displaystyle \therefore \quad { r }^{ 4 }=16$
$\displaystyle \therefore \quad r=2$
$\displaystyle a.{ 2 }^{ 4 }=48$
or, $\displaystyle a=\frac { 48 }{ 16 } =3$
The GP is 3,6, 12,24,.....

A geometric series is a series for which the ratio of each two consecutive terms is a constant function of the summation index .

A geometric progression is a sequence of numbers where each term in the sequence is found by multiplying the previous term with a with a unchanging number called the common ratio.
Example: $2, 6, 18, 54, 108....$
This geometric sequence has a common ratio $3$.

$10, 20, 40, 80$ is an example of geometric sequence.
In geometric sequence, the ratio of succeeding term to the preceeding term is always equal.

A geometric series is the sum of the numbers in a geometric progression.

Geometric series is of the following form:

The sequence $6, 12, 24, 48....$ is a geometric progression as the ratio here is common.
The common ratio in the given series is $2$.

$1, 3, 9, 27, 81$ is a geometric sequence.
A geometric progression is a sequence of numbers such that the quotient of any two successive members of the sequence is a constant called the common ratio of the sequence.

A geometric sequence is a sequence of numbers such that the quotient of any two successive members of the sequence is a constant called the common ratio of the sequence.
So, $1, 3, 9, 27, 81...$ is a geometric progression.
Here the common ratio is $3$.

A sequence of number, ${ a } _{ 1 }+{ a } _{ 2 }+......{ a } _{ n }$ quotient of any two successive number is a constant,

$4, -4, 4, -4, 4$ is a geometric progression.
Here the common ratio is $-1$.

The general form of geometric progression is $2, 4, 8, 16$.

The common ratio is used in geometric progression.

{$2, -2, 2, -2, 2$} is a general form of geometric sequence.

The common ratio is calculated in G.P.
For example: $2,4,8,16,....$

$a, ar, ar^2, ar^3, ar^4....$ is an infinite geometric progression.

Given sequence is $a,ar,ar^2, ar^3, ar^4$.

$1 + 0.5 + 0.25 + 0.125....$ is an example of infinite geometric progression.
Here the common ratio is $0.5$.
An infinite geometric series is the sum of an infinite geometric progression.

An geometric sequence containing infinite number of terms. It has a common ratio which is same throughout.
Example: $1 + 0.5 + 0.25 + 0.125....$ is an infinite geometric sequence.
Here the common ratio is $0.5$.

If a sequence is a finite geometric sequence, then :

$1 + 0.5 + 0.25 + 0.125 $$ is a finite geometric progression.
Here the common ratio is $0.5$.
An finite geometric series is the sum of an finite geometric sequence.

An infinite geometric series is the sum of an infinite geometric sequence.
So, $1 + 2 + 4 + 8 +....$ is an infinite geometric sequence.
Here the common ratio is $2$ and it is never ending.

Given:

This is G.P in which  $a=2,r=\cfrac{{2}^{2}}{2}=2$ and $n=9$
$\therefore$ ${S} _{n}=\cfrac{a({r}^{n}-1)}{(r-1)}=\cfrac{2\times ({2}^{9}-1)}{(2-1)}=2\times (512-1)=2\times 511=1022$.

Here $a=3$ and $r=\cfrac{6}{3}=2$. Let the number of terms be $n$$.
Then, ${t}_{n}=384$ $\Rightarrow$ $a{r}^{n-1}=384$
$\Rightarrow$ $3\times {2}^{n-1}=384$
$\Rightarrow$ ${2}^{n-1}=128={2}^{7}$
$\Rightarrow$ $n-1=7$
$\Rightarrow$ $n=8$
$\therefore$ Number of terms $=8$.

1. Consider the two numbers $1$ and $4$, geometric mean of $1$ and $4$ is $\sqrt {1 \times 4} = 2$.
When each number is reduced by $2$, the numbers become $-1$ and $2$ whose geometric mean does not exist.
2. Now consider two numbers $2$ and $7$. Their geometric mean is $\sqrt {14}$. The new numbers are $4$ and $9$ whose geometric mean is $\sqrt {4\times 9} = 6$ which is not equal to $2\sqrt {14}$.
Thus only statement $1$ is true.

We are given $a,b,c$ are in G.P.

The common ratio between two numbers has to be non-zero for it to form a G.P.

Difference between consecutive terms is $(4-2), (8-4), (16-8)$ and so on, i.e. 2,4,8 and so on.
Since the difference is not constant, it is not an AP.

The series can be split into two geometric series: S1 = 1/7 + 1/7^3 + 1/7^5... and S2 = 2/7^2 + 2/7^4 + 2/7^6... For S1, a=1/7, r=1/49, sum = (1/7)/(1-1/49) = 7/48. For S2, a=2/49, r=1/49, sum = (2/49)/(1-1/49) = 2/48. Total sum = 9/48 = 3/16.

If the original series is a, ar, ar^2, ..., the sum is a/(1-r). The series of squares is a^2, a^2r^2, a^2r^4, ..., which is a geometric series with first term a^2 and common ratio r^2. The sum is a^2/(1-r^2).

We have,

$1+n+{ n }^{ 2 }+...{ n }^{ 127 }=\cfrac { { n }^{ 128 }-1 }{ n-1 } \ =\cfrac { \left( { n }^{ 64 }-1 \right) \left( { n }^{ 64 }+1 \right)  }{ (n-1) } \ ({ a }^{ n }-{ b }^{ m })$ is divisible by $\quad (a-b)\bigvee  m\in { z }^{ + }$(positive Integers)

For the curve y = x^3, the tangent at x1 meets the curve at x2 = -2x1. This forms a geometric progression with common ratio -2. The area of the triangle formed by points on the curve scales with the coordinates, leading to a ratio of 1/16 for consecutive triangles.

Given:  a, b, c are in G.P.

Given:-

Last $3$ of the $4$ numbers are in AP.

$\Rightarrow \frac { a }{ r } ,a,ar=8\quad \quad \Rightarrow { a }^{ 3 }=8\quad \quad \Rightarrow a=2$
$a=2$ is a root of the given equation 
$\Rightarrow 8-4k+28-8=0\quad \quad \Rightarrow k=7$

Given 

$\\Let\>the\>numbers\>an\>a,\>ar,\>ar^2\\where\>a=first\>term\>and\>r=common\>ratio\\\therefore\>a+ar+ar^2=14\\and\\a+1,\>ar+1,\>ar^2-1\>are\>in\>AP\\\therefore\>2(ar+1)=(a+1)+(ar^2-1)\\or\>2ar+2=a+ar^2\\or\>3ar+2=a+ar+ar^2\\or\>3ar+2=14\\\therefore\>ar=4\\or\>a=(\frac{4}{r})\\\therefore(\frac{4}{r})+(\frac{4}{r})\times\>r+(\frac{4}{r})\times\>r^2=14\\or\>(\frac{4}{r})+4+4r=14\\or\>4+4r+4r^2=14r\\or\>4r^2-10r+4=0\\or\>4r^2-8r-2r+4=0\\or\>4r(r-2)-2(r-2)=0\\or\>(4r-2)(r-2)=0\\\therefore\>r=(\frac{1}{2})\>or\>2\\ifr=(\frac{1}{2}),then\>a=(\frac{4}{(\frac{1}{2})})=8\\\therefore\>sequence\>8,4,2\>\\\>and\>if\>r=2,\>then\>a=(\frac{4}{2})=2\\\therefore\>sequence\>2,4,8$

A series is said to be geometric when ratio between one term and its previous term is constant/same throughout the series.

Consider given the series, $1+\left( 1+x \right)+{{\left( 1+x \right)}^{2}}+{{\left( 1+x \right)}^{3}}+...............{{\left( 1+x \right)}^{n}}$

Given series is G.P. which have first term $a=1$,comman ratio $r=1+x$ and number of term $=n$

So sum is,

${{s} _{n}}=\dfrac{a\left( {{r}^{n}}-1 \right)}{r-1}=\dfrac{[{{\left( 1+x \right)}^{n}}-1]}{1+x}={{\left( 1+x \right)}^{n-1}}-{{\left( 1+x \right)}^{-n}}$

Now rth term of ${{\left( 1+x \right)}^{n-1}}-{{\left( 1+x \right)}^{-n}}$ is $={}^{n-1}{{C} _{r}}{{x}^{r}}-{}^{-n}{{C} _{r}}.{{x}^{r}}=\left( {}^{n-1}{{C} _{r}}-{}^{-n}{{C} _{r}} \right){{x}^{r}}$

Hence, coefficient of ${{x}^{r}}=\left( {}^{n-1}{{C} _{r}}-{}^{-n}{{C} _{r}} \right)$

Given $3,p,12$ are in GP

Given series is $4,8,16,32,....$

Let the common ratio of G.P is $r$ Therefore $\quad \beta =\alpha t,\alpha { t }^{ 2 }$
Equation $\alpha { x }^{ 2 }+2\alpha rx+\alpha { t }=0\quad 
\Rightarrow { x }^{ 2 }+2rx+{ t }^{ 2 }=0....(i)$
Given equation (i) and ${ x }^{ 2 }+x-1=0....(ii)$ has a common root
$(i)-(ii)\Rightarrow (2e-1)x+({ r }^{ 2 }+1)=0\Rightarrow x=\cfrac { -\left( { r }^{ 2 }+1 \right)  }{ 2r-1 } ....(iii)\quad $
Putting (iii) in equation (ii) $\Rightarrow { \left( { r }^{ 2 }+1 \right)  }^{ 2 }-\left( { r }^{ 2 }+1 \right) (2r-1)-{ \left( { 2r }^{ 2 }-1 \right)  }^{ 2 }=0\Rightarrow { r }^{ 4 }-2{ r }^{ 3 }-{ r }^{ 2 }+2r+1=0....(iv)$
dividing equation (iv) by ${r}^{2}$ $\Rightarrow { \left( r-\cfrac { 1 }{ r }  \right)  }^{ 2 }-2{ \left( r-\cfrac { 1 }{ r }  \right)  }+1=0\Rightarrow { \left( r-\cfrac { 1 }{ r } -1 \right)  }^{ 2 }=0\Rightarrow \cfrac { r-1 }{ r } =1....(v)\quad $
$\left( \gamma -\alpha  \right) \beta =\left( \alpha { r }^{ 2 }-\alpha  \right) \times \alpha r={ \alpha  }^{ 2 }\left( { \alpha  }^{ 2 }-1 \right) r={ \alpha  }^{ 2 }(r-1)={ \alpha  }^{ 2 }{ r }^{ 2 }$
(using $(v)=\alpha \times \alpha { t }^{ 2 }\quad $

Let a and d be the first term and common ratio of the GP respectively.
Given a=1 and d=4.
Now, $a _n=ar^{n-1}$
$\therefore a _1=a=1$
$a _2=ar=1\times4=4$
$a _3=ar^2=1\times(4)^2=16$
$a _4=ar^3=1\times(4)^3=64$
$a _5=ar^4=1\times(4)^4=256$

Given series

It is fundamental concept that if $y^2 = xz$, then $x,y,z\in$  G.P

We know, $S _{\infty }=\dfrac{a}{1-r}$
From the given series, $a=\dfrac{2}{3} ,r=-\dfrac{2}{3}$
$\therefore S _{\infty }=\dfrac{\frac{2}{3}}{1-\left ( -\frac{2}{3} \right )}$

We know, $S _{\infty }=\dfrac{a}{1-r}$
From the given series, $a=1 ,r=-\dfrac{1}{3}$
$\therefore S _{\infty }=\dfrac{1}{1-\left ( -\dfrac{1}{3} \right )}$
$= \dfrac{1}{1+\dfrac{1}{3}}$

Given $ a,b,c $ are in GP.
So, the common ratio between the first and second term ; second and third will be the same.
$ => \dfrac {b}{a} = \dfrac {c}{b} $

The sequence $-6 + 42 - 294 + 2058$ is a finite geometric sequence.
Here the common ratio is $-7$.

Given series is $10-5+2.5-1.25.....$

Given that ${(16-x)}^{2}=(8-x)(40-x)$
$\Rightarrow 256-32x+{x}^{2} = 320-48x+{x}^{2}$
$\Rightarrow 16x = 64$ 

Ratio of second term to first term is $ \frac {150}{50} = 3$

Ratio of third term to second term is  $ \frac {200}{150} = 1.33$

Thus, the ratio is not matching. 

Hence, it is not a GP.

In series $2,4,6,8,....$ difference is same i.e. $2$

Combine the terms in threes, to get the geometric series
$\cfrac { 1 }{ 4 } +\cfrac { 1 }{ 32 } +\cfrac { 1 }{ 256 } +....;\quad \quad S=\cfrac { \cfrac { 1 }{ 4 }  }{ 1-\cfrac { 1 }{ 8 }  } =\cfrac { 2 }{ 7 } $ or
rearrange the terms into three series:
$1+\cfrac { 1 }{ 8 } +\cfrac { 1 }{ 64 } +...\quad -\cfrac { 1 }{ 2 } -\cfrac { 1 }{ 16 } -\cfrac { 1 }{ 128 } -....,\quad -\cfrac { 1 }{ 4 } -\cfrac { 1 }{ 32 } -\cfrac { 1 }{ 256 } -....\quad $
${ S } _{ 1 }=\cfrac { 1 }{ 1-\cfrac { 1 }{ 8 }  } =\cfrac { 8 }{ 7 } ;{ S } _{ 2 }=\cfrac { -\cfrac { 1 }{ 2 }  }{ 1-\cfrac { 1 }{ 8 }  } =-\cfrac { 4 }{ 7 } ;{ S } _{ 3}=\cfrac { -\cfrac { 1 }{ 4 }  }{ 1-\cfrac { 1 }{ 8 }  } =-\cfrac { 2 }{ 7 } ;\quad \therefore S=\cfrac { 2 }{ 7 } $

${ a }^{ x }=b\ { b }^{ y }=c=>{ a }^{ xy }=c\ { c }^{ z }=a\ { c }^{ z }=a=>{ b }^{ yz }=a\ \left( { a }^{ x } \right) ^{ yz }=a\ { a }^{ xyz }=a\ xyz=1$