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Questions Related to maths

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

Show answer
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 $2$.

Show answer
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. ${2.3^{10}}$

Show answer
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

Show answer
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. $\dfrac {1}{769}$

Show answer
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. $\displaystyle \frac{8}{5}$

Show answer
Correct Option: B
Explanation:

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

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

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

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

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

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

Find the GP whose $5^{th}$ term is $48$ and $9^{th}$ term is$ 768$.

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

  1. $3,6,12,24$

  2. $2,4,8,16$

  3. $6,12,24,48$

  4. $12,24,36,48$

Show answer
Correct Option: A
Explanation:

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

The reciprocals of all the terms of a geometric progression form a ________ progression.

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

  1. AP

  2. HP

  3. GP

  4. AGP

Show answer
Correct Option: C
Explanation:
Let  $ a $ be the first term  and $ r $ be the common ratio of the GP. 

So, the series is $ a, ar, ar^2... $

Their reciprocals are $ \dfrac {1}{a}, \dfrac {1}{ar}, \dfrac {1}{ar^2} .. $

It is also a GP, with first term $ \dfrac {1}{a} $ and common ratio $ \dfrac {1}{r} $

In a _______ each term is found by multiplying the previous term by a constant.

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

  1. arithmetic sequence

  2. geometric series

  3. arithmetic series

  4. harmonic progression

Show answer
Correct Option: B
Explanation:

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

Or,
In a Geometric series each term is found by multiplying the previous term by a constant.
$(Ans \to B)$

A _________ is a sequence of numbers where each term in the sequence is found by multiplying the previous term with a unchanging number called the common ratio.

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

  1. geometric progression

  2. arithmetic series

  3. arithmetic progression

  4. harmonic progression

Show answer
Correct Option: A
Explanation:

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 $3$.