Tag: geometric sequences

Questions Related to geometric sequences

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

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

  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. $\displaystyle n - \frac{4^{-n}}{3} - \frac{1}{3}$

Show answer
Correct Option: B
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
$= n - \left \{ \dfrac{1}{2^2} + \dfrac{1}{2^4} + \dfrac{1}{2^6} + .... \text{to n terms} \right \}$
$= n \displaystyle - \dfrac{1}{2^2} \left \{ \dfrac{1 - \left (\dfrac{1}{2^2} \right )^n }{1 - \dfrac{1}{2^2}} \right \}$
$= \displaystyle n - \dfrac{1}{3} (1 - 4^{-n})$
$= n + \displaystyle \dfrac{4^{-n}}{3} - \dfrac{1}{3}$

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

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

  1. $\displaystyle r^{p/q}$

  2. $\displaystyle r^{q/p}$

  3. $\displaystyle r^{p-q}$

  4. $\displaystyle r^{p+q}$

Show answer
Correct Option: C
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}$
$\Rightarrow k = \dfrac{r^{p-1}}{r^{q-1}}$
$\Rightarrow k= r^{p-q}$

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

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

  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. $\dfrac {1}{2} - \dfrac {5n + 1}{3\times 5^{n + 2}}$

Show answer
Correct Option: A
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 } } $

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

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

  1. $15$

  2. $43$

  3. $38$

  4. $56$

Show answer
Correct Option: C
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.

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

Show answer
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

Show answer
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

Show answer
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

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

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:

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