Tag: geometric sequences

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Questions Related to geometric sequences

Multiple choice maths geometric sequences sum of terms of g.p sum of n terms of an gp summing geometric series

The sum of series $\displaystyle \frac{3}{4} + \frac{15}{16} + \frac{63}{64}+ ..... $ up to $n$ terms is

  1. $\displaystyle n - \frac{4^n}{3} - \frac{1}{3}$
  2. $\displaystyle n + \frac{4^{-n}}{3} - \frac{1}{3}$
  3. $\displaystyle n + \frac{4^n}{3} - \frac{1}{3}$
  4. $\displaystyle n - \frac{4^{-n}}{3} - \frac{1}{3}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For $n=1$, we have
$\displaystyle n - \dfrac{4^n }{3} - \dfrac{1}{3} = 1  - \dfrac{4}{3} - \dfrac{1}{3} = - \dfrac{2}{3}$
$\displaystyle n + \dfrac{4^n}{3} - \dfrac{1}{3} = 1 + \dfrac{4}{3} - \dfrac{1}{3} = 2$
$n - \displaystyle \dfrac{4^{-n}}{3} + \dfrac{1}{3} = 1 - \dfrac{4^{-1}}{3} + \dfrac{1}{3}= \dfrac{5}{4}$
Also, for $n = 2$, we have
$ \displaystyle n + \dfrac{4^{-n}}{3} - \dfrac{1}{3} = 2 + \dfrac{1}{48} - \dfrac{1}{3} = \dfrac{27}{16}$ and $\displaystyle \dfrac{3}{4} + \dfrac{15}{16} = \dfrac{27}{16}$
Hence, option (b) is correct.
ALTER We have,
$\displaystyle \dfrac{3}{4} + \dfrac{15}{16} + \dfrac{63}{64}+ ..... $ to n terms
$= \displaystyle \dfrac{2^2 - 1}{2^2} + \dfrac{2^4 - 1}{2^4} + \dfrac{2^6 - 1}{2^6}+ .... $ to n terms.
$= \displaystyle \left ( 1 - \dfrac{1}{2^2} \right ) + \left ( 1 - \dfrac{1}{2^4} \right ) + \left( 1 - \dfrac{1}{2^6} \right ) + ..... $ to n terms
$= n - \left \{ \dfrac{1}{2^2} + \dfrac{1}{2^4} + \dfrac{1}{2^6} + .... \text{to n terms} \right \}$
$= n \displaystyle - \dfrac{1}{2^2} \left \{ \dfrac{1 - \left (\dfrac{1}{2^2} \right )^n }{1 - \dfrac{1}{2^2}} \right \}$
$= \displaystyle n - \dfrac{1}{3} (1 - 4^{-n})$
$= n + \displaystyle \dfrac{4^{-n}}{3} - \dfrac{1}{3}$

Multiple choice maths geometric sequences sum of terms of g.p sum of n terms of an gp summing geometric series

If the sum of $n$ terms of a GP (with common ratio $r$) beginning with the $\displaystyle p^{th}$ term is $k$ times the sum of an equal number of the same series beginning with the $\displaystyle q^{th}$ term, then the value of $k$ is

  1. $\displaystyle r^{p/q}$
  2. $\displaystyle r^{q/p}$
  3. $\displaystyle r^{p-q}$
  4. $\displaystyle r^{p+q}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$p^{th}$ term of the series  $=ar^{p-1}$     ($a$ is first term)

$q^{th }$ term of th series $= ar^{q-1}$
Sum of $n$ term beginning with $p^{th} $ term 
$=\dfrac{ar^{p-1}(r^n - 1)}{r-1}$
Sum of $n$ term beginning with $q^{th} $ term 
$=\dfrac{ar^{q-1}(r^n - 1)}{r-1}$
Sum of $n$ term beginning with $p^{th} $ term $= k$ (sum of $n$ term beginning with $q^{th} $ term )
Thus $\dfrac{ar^{p-1}(r^n - 1)}{r-1}$$=k\dfrac{ar^{q-1}(r^n - 1)}{r-1}$
$\Rightarrow k = \dfrac{r^{p-1}}{r^{q-1}}$
$\Rightarrow k= r^{p-q}$

Multiple choice maths geometric sequences sum of terms of g.p sum of n terms of an gp summing geometric series

The sum of $1 + \dfrac {2}{5} + \dfrac {3}{5^{2}} + \dfrac {4}{5^{3}} + ....$ up to $n$ terms is

  1. $\dfrac {25}{16} - \dfrac {4n + 5}{16\times 5^{n - 1}}$
  2. $\dfrac {3}{4} - \dfrac {2n + 5}{16\times 5^{n + 1}}$
  3. $\dfrac {3}{7} - \dfrac {3n + 5}{16\times 5^{n - 1}}$
  4. $\dfrac {1}{2} - \dfrac {5n + 1}{3\times 5^{n + 2}}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Let $S=1+\cfrac { 2 }{ 5 } +\cfrac { 3 }{ { 5 }^{ 2 } } +\cfrac { 4 }{ { 5 }^{ 3 } } +...\quad \quad (1)$
Multiplying $S$ with $\cfrac{1}{5}$ we get
$\cfrac { 1 }{ 5 } S=\cfrac { 1 }{ 5 } +\cfrac { 2 }{ { 5 }^{ 2 } } +\cfrac { 3 }{ { 5 }^{ 3 } } +\cfrac { 4 }{ { 5 }^{ 4 } } +...\quad \quad (2)$
Subtracting $(2)$ from $(1)$
$\quad \quad \quad S\;\;=1+\cfrac { 2 }{ 5 } +\cfrac { 3 }{ { 5 }^{ 2 } } +\cfrac { 4 }{ { 5 }^{ 3 } } +....{ T } _{ n }\\ \underline { \quad \quad -\cfrac { 1 }{ 5 } S=-\left[ \cfrac { 1 }{ 5 } +\cfrac { 2 }{ { 5 }^{ 2 } } +\cfrac { 3 }{ { 5 }^{ 3 } } +...{ T } _{ n-1 } \right] -{ T } _{ n } } \\ \left( 1-\cfrac { 1 }{ 5 }  \right) S=1+\cfrac { 1 }{ 5 } +\cfrac { 1 }{ { 5 }^{ 2 } } +\cfrac { 1 }{ { 5 }^{ 3 } } +.....-{ T } _{ n-1 }\\ \left( 1-\cfrac { 1 }{ 5 }  \right) S=\cfrac { (1)\left( 1-\cfrac { 1 }{ { 5 }^{ n } }  \right)  }{ \left( 1-\cfrac { 1 }{ 5 }  \right)  } -\cfrac { n }{ { 5 }^{ n-1 } } \\ \cfrac { 4 }{ 5 } S=\cfrac { 5 }{ 4 } -\cfrac { (4n+5) }{ 4\times { 5 }^{ n } } \\ S=\cfrac { 25 }{ 16 } -\cfrac { (4n+5) }{ 16\times { 5 }^{ n-1 } } $
Multiple choice maths geometric sequences sum of terms of g.p sum of n terms of an gp summing geometric series

In a $G.P$. the ratio of the sum of the first eleven terms to the sum of last eleven terms is $\displaystyle \frac{1}{8}$ and the ratio of the sum of all terms without the first nine to the sum of all the terms without the last nine is $2$. Then the number of terms of the $G.P$ is

  1. $15$
  2. $43$
  3. $38$
  4. $56$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

We have: 
$\dfrac { \frac { a({ r }^{ 11 }-1) }{ r-1 }  }{ \frac { a{ r }^{ n-11 }({ r }^{ 11 }-1) }{ r-1 }  } =\dfrac { 1 }{ 8 }$
$\Rightarrow { r }^{ n-11 }=8$ ...(i)
Also:
$\dfrac { \frac { a{ r }^{ 9 }({ r }^{ 11 }-1) }{ (r-1) }  }{ \frac { a({ r }^{ n-9 }-1) }{ (r-1) }  } =2$
$\Rightarrow { r }^{ 9 }=2 $
$\Rightarrow r={ 2 }^{ \frac { 1 }{ 9 }  }$ ...(ii)
Substituting (ii) in (i):
${ 2 }^{ \frac { n-11 }{ 9 }  }={ 2 }^{ 3 }$
$\Rightarrow \dfrac { n-11 }{ 9 } =3$
$\Rightarrow n=38$
Hence, (c) is correct.

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

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

  1. geometric series

  2. harmonic series

  3. arithmetic progression

  4. geometric progression
Reveal answer Fill a bubble to check yourself
D Correct answer
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.

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

The geometric progression which have infinite terms is called

  1. finite geometric progression

  2. finite arithmetic progression

  3. infinite geometric progression

  4. finite harmonic progression
Reveal answer Fill a bubble to check yourself
C Correct answer
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.

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

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

  1. $\left| x \right| < 2$
  2. $\left| x \right| > 2$
  3. $\left| x \right| < 1$
  4. $2\left| x \right| < 1$
Reveal answer Fill a bubble to check yourself
B Correct answer
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$       
Multiple choice maths geometric sequences introduction to geometric progression understanding geometric progressions introduction to geometric progressions

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

  1. Cannot be determined.
  2. Equals $\dfrac{15}{8}$
  3. Equals $2$
  4. Will be higher than $2$.
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

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