Tag: introduction to geometric progressions

Questions Related to introduction to geometric progressions

The geometric sequence is also called as

maths geometric sequences introduction to geometric progression understanding geometric progressions introduction to geometric progressions

  1. geometric progression

  2. arithmetic sequence

  3. harmonic sequence

  4. geometric series

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Correct Option: A
Explanation:

The sequence is also called as Progression.

So, the geometric sequence can be called as the geometric progression.

A progression of the form $a, ar, ar^2$, ..... is a

maths geometric sequences introduction to geometric progression understanding geometric progressions introduction to geometric progressions

  1. geometric series

  2. harmonic series

  3. arithmetic progression

  4. geometric progression

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Correct Option: D
Explanation:

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

maths geometric sequences introduction to geometric progression understanding geometric progressions introduction to geometric progressions

  1. finite geometric progression

  2. finite arithmetic progression

  3. infinite geometric progression

  4. finite harmonic progression

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Correct Option: C
Explanation:

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.

If $a, b, c$ are in G.P., then

maths geometric sequences introduction to geometric progression understanding geometric progressions introduction to geometric progressions

  1. $a(b^{2} + a^{2}) = c(b^{2} + c^{2})$

  2. $a(a^{2} + c^{2}) = c(a^{2} + b^{2})$

  3. $a^{2}(b + c) = c^{2}(a + b)$

  4. None of these

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Correct Option: B
Explanation:

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

The sum $1+\dfrac { 2 }{ x } +\dfrac { 4 }{ { x }^{ 2 } } +\dfrac { 8 }{ { x }^{ 3 } } +....\left( up\ to\ \infty  \right) ,x\neq 0,$ is finite if

maths geometric sequences introduction to geometric progression understanding geometric progressions introduction to geometric progressions

  1. $\left| x \right| < 2$

  2. $\left| x \right| > 2$

  3. $\left| x \right| < 1$

  4. $2\left| x \right| < 1$

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Correct Option: B
Explanation:
$1 + \dfrac{2}{x} + \dfrac{4}{{{x^3}}} + \dfrac{8}{{{x^3}}}.......\,upto\,\,\infty $
It is infinite $G.P$ with Common ratio $r=\dfrac{2}{x}$
It's sum is infinite if $\left| r \right| < 1$
i-e    
 $\left| {\dfrac{2}{x}} \right| < 1$
 $ \Rightarrow \dfrac{2}{{\left| x \right|}} < 1$
 $ \Rightarrow 2 < \left| x \right|$
$ \Rightarrow \left| x \right| > 2$       

The sum of the infinite series $1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+......$

maths geometric sequences introduction to geometric progression understanding geometric progressions introduction to geometric progressions

  1. Cannot be determined.

  2. Equals $\dfrac{15}{8}$

  3. Equals $2$

  4. Will be higher than $2$.

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Correct Option: C
Explanation:

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

$S = {3^{10}} + {3^9} + \frac{{{3^9}}}{4} + \frac{{{3^7}}}{2} + \frac{{{{5.3}^6}}}{{16}} + \frac{{{3^2}}}{{16}} + \frac{{{{7.3}^4}}}{{64}} + .........$ upto infinite terms, then $\left( {\frac{{25}}{{36}}} \right)S$ equal to 

maths geometric sequences introduction to geometric progression understanding geometric progressions introduction to geometric progressions

  1. ${6^9}$

  2. ${3^{10}}$

  3. ${3^{11}}$

  4. ${2.3^{10}}$

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Correct Option: B
Explanation:

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

$\dfrac56S=\dfrac{3^{10}\times 6 }{5}$
$\therefore \dfrac {25}{36}S=3^{10}$

If $4,64,p$ re in GP find p

maths geometric sequences introduction to geometric progression understanding geometric progressions introduction to geometric progressions

  1. 1024

  2. 2944

  3. 512

  4. 256

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Correct Option: A
Explanation:

$4,64,p$ are in GP 

Condition to be in GP is 
$b^2=ac\64^2=4p\p=\dfrac{64^2}{4}=1024$

In each of the following questions, a series of number is given which follow certain rules. One of the number is missing. Choose the missing number from the alternatives given below and mark it on your answer-sheet as directed. $1, \dfrac {1}{3}, \dfrac {1}{9}, \dfrac {1}{27}, \dfrac {1}{81}, \dfrac {1}{243}, $?

maths geometric sequences introduction to geometric progression understanding geometric progressions introduction to geometric progressions

  1. $\dfrac {1}{729}$

  2. $\dfrac {1}{829}$

  3. $\dfrac {1}{749}$

  4. $\dfrac {1}{769}$

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Correct Option: A
Explanation:

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

$\therefore$, next term is $\dfrac {1}{243} \times \dfrac {1}{3}=\dfrac {1}{729}$

Find the sum of an infinite G.P : $\displaystyle 1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+.......$

maths geometric sequences introduction to geometric progression understanding geometric progressions introduction to geometric progressions

  1. $\displaystyle \frac{3}{5}$

  2. $\displaystyle \frac{3}{2}$

  3. $\displaystyle \frac{49}{27}$

  4. $\displaystyle \frac{8}{5}$

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Correct Option: B
Explanation:

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

$a=1, r= \dfrac {1}{3}$

$\therefore S _{\infty}=\dfrac{a}{1-r}$

$S _{\infty}=\dfrac{1}{1-\dfrac{1}{3}}$

$S _{\infty}=\dfrac{1}{\dfrac{2}{3}}$

$\therefore S _{\infty}=\dfrac{3}{2}$