Tag: summing geometric series

Questions Related to summing geometric series

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.