Sum of n Terms of a G.P. - Class XI

Sum of infinite series of a GP $=\dfrac {a} {1-r}$, where $a$ is the first term and $r$ is the common ratio

Given that

We have,

The three numbers be $a,ar\,and\,a{{r}^{2}}$.

Given that,

$ a+ar+a{{r}^{2}}=70 $

$ \Rightarrow a\left( 1+r+{{r}^{2}} \right)=70\,\,......\,\,\left( 1 \right) $

Also given that,

$4a,\,5ar\,and\,4a{{r}^{2}}$ in an A.P.

$\begin{align}

$ \Rightarrow 2\left( 5ar \right)=4a+4a{{r}^{2}} $

$ \Rightarrow 5r=2+2{{r}^{2}} $

$ \Rightarrow 2{{r}^{2}}-5r+2=0 $

$ \Rightarrow 2{{r}^{2}}-\left( 4+1 \right)r+2=0 $

$ \Rightarrow 2{{r}^{2}}-4r-1r+2=0 $

$ \Rightarrow 2r\left( r-2 \right)-1\left( r-2 \right)=0 $

$ \Rightarrow \left( r-2 \right)\left( 2r-1 \right)=0 $

$ \Rightarrow r-2=0,\,\,2r-1=0 $

$ \Rightarrow r=2,\,\,r=\dfrac{1}{2} $

From (1) we get,

$a=10\,\,\,at\,\,\,r=2$

And $a=40\,\,\,at\,\,\,r=\dfrac{1}{2}$

Sum of this series

$ =\dfrac{a}{1-r} $

$ =\dfrac{40}{1-2} $

$ =-40 $

This is the answer.

The series form a GP with $a=1,r=\dfrac 14$

$2+{ 2 }^{ 2 }+{ 2 }^{ 3 }+......+{ 2 }^{ 9 }$

$\displaystyle { s } _{ 5 }=\frac { 1\times \left[ { 1-\left( { 2 }/{ 3 } \right)  }^{ 5 } \right]  }{ 1-\left( { 2 }/{ 3 } \right)  } =\frac { 211 }{ 81 } $

B is in G.P. with $a=2^0, r=2, n=65$
$\therefore S _n=\frac {a(r^n-1)}{r-1}=\frac {2^0(2^{65}-1)}{2-1}$
$\therefore B=2^{65}-1$
$\Rightarrow B=A-1$
$\therefore$ A is larger than B by $1$.

The first term in given Geometric series, $a _1=x$
The common ratio $r$ $=\dfrac{4x}{2x}=2$
No. of terms, $n$ $=5$
Applying sum of GP formula,
$S _n=\dfrac{a(1-r^n)}{1-r}$
      $=\dfrac{x(1-2^5)}{1-2}$
      $=\dfrac{x(1-32)}{-1}=31x$
Hence,option D is correct.

The series is 0.5 + 0.55 + 0.555... = (5/9) * (0.9 + 0.99 + 0.999...) = (5/9) * [(1-0.1) + (1-0.01) + (1-0.001)...]. This simplifies to (5/9) * [n - (0.1 + 0.01 + ... + 0.1^n)]. The sum of the geometric part is (1/10)(1-(1/10)^n)/(1-1/10) = (1/9)(1-1/10^n). Multiplying through leads to the correct expression.

We have,

Let the G.P be $\dfrac{a}{r}, a, ar$.

The series is a geometric progression with first term i and common ratio i. The sum of 4 consecutive powers of i (i, i^2, i^3, i^4) is i - 1 - i + 1 = 0. Since 1000 is divisible by 4, the sum of 1000 terms is 0.

Since the sequence in Geometric progression 

The sum is S = sum from k=0 to n-1 of (1+x+...+x^k). This is sum from k=0 to n-1 of (1-x^(k+1))/(1-x). Expanding this gives (n - (x + x^2 + ... + x^n)) / (1-x). Using the geometric sum formula for the numerator yields the result in option C.

$\underset{x \rightarrow 1}{lim} \dfrac{x + x^2 + x^3  + .... + x^n - x}{x - 1} \left(\dfrac{0}{0} \right)$ form

The series is 1, 3, 7, 15... which can be written as (2^1-1), (2^2-1), (2^3-1), (2^4-1)... The sum of n terms is sum(2^k - 1) for k=1 to n. This equals (2^(n+1) - 2) - n.

Given, 

$a=2,r=3$

Let the numbers be $a-d,a,a+d$

For an infinite G.P., S = a / (1-r), so r = 1 - a/S. The sum of the first n terms is S_n = a(1-r^n) / (1-r). Substituting r = 1 - a/S and 1-r = a/S, we get S_n = a(1 - (1-a/S)^n) / (a/S) = S(1 - (1-a/S)^n).

Let ${ S } _{ n }=2+6+18+...+486$

$\displaystyle x(x+y)+x^{2}(x^{2}+y^{2})+x^{3}(x^{3}+y^{3})+$.......to n terms.

Given  $\delta _{rs}\, =\, 0   if \quad r\neq s\quad \delta  _{rs}=1\quad if \quad r=s$

$\therefore \displaystyle \sum _{r=1}^{n}\, \sum _{s=1}^{n}\, \delta _{rs}\, 2^r\, 3^s$

$=\displaystyle

\sum _{r=1}^{n} 2^r3^r\, =\, \displaystyle \sum _{r=1}^{n} 6^r\, \,

=6+6^2\, +\, 6^3+..6^n\,=\, \displaystyle \frac {6(6^n-1)}{5}$

$\displaystyle A.M.=\frac { 1+2+4+8+...+{ 2 }^{ n } }{ n+1 } =\frac { { 2 }^{ n+1 }-1 }{ \left( n+1 \right) \left( 2-1 \right)  } =\frac { { 2 }^{ n+1 }-1 }{ n+1 } $

Ans: B

Given expression = $\displaystyle 1+\frac{1}{12}+\frac{1}{36}+\frac{1}{108}=\frac{108+9+3+1}{108}=\frac{121}{108}=1.120\left ( approx \right )$

Given $6^{\frac{1}{2}}.6^{\frac{1}{4}}.6^{\frac{1}{8}}....\infty$
Here power of 6 are in G.P
Sum of $\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8} ...\infty$

$S=1+\cfrac { 1 }{ 2 } +\cfrac { 1 }{ 4 } +\cfrac { 1 }{ 8 } .......$ first $20$ terms

The given series is a Geometric Progression, with first terms $ a = \dfrac {1}{100} $ and common ratio $ r = \dfrac {T _2}{T _1} = \dfrac {\dfrac {1}{10000}}{\dfrac {1}{100}} = \dfrac {1}{100} $

For a GP, the $ nth $ term is given by $ T _n = ar^{n-1} =\dfrac {1}{100}  \times (\dfrac {1}{100})^{n-1} =(\dfrac {1}{100})^{n} = 10^{-2n}$

Given, first term, $a = 120$, common ratio, $r = -2$ and $a _5=1$
We know $S _n=\dfrac{a _1-a _nr}{1-r}$
$S _n=\dfrac{120-(-2)}{1-(-2)}$
$S _n=\dfrac{120-(-2)}{1-(-2)}$
$S _n=\dfrac{122}{3}$
$S _n=40.66$

First term, $a$ is $80$
Common ratio, $r =$ $\dfrac{-20}{80}=\dfrac{-1}{4}$
$S _n=\dfrac{a(1-r^2)}{1-r}$
$S _n=\dfrac{80(1-(\frac{-1}{4})^2)}{1-\frac{-1}{4}}$
$S _n = \dfrac{79.98}{1.25}$
$S _n = 63.98$

Given, $a = 1, r = 3$
$S _n = \dfrac{a _1(1-r^n)}{1-r}$
$S _7 = \dfrac{1(1-3^7)}{1-3}$
$ = \dfrac{1(1-3^7)}{-2}$
$ = 1093$

A GP can be written as:

The given series is a G.P
Sum of n terms of a G.P is $ a* {(\frac{(r^n;-;1)}{(r-1)})} $
Here a=1 and r=3
Substituting it in the equation and finding n we get n=5

Given series is a sum of terms of GP with common ratio $-\dfrac{1}{2}$
Sum of $n$ terms of GP with common ratio $r$ and first term $a$ is $\dfrac { a(1-r^{ n }) }{ 1-r } $
Putting $a=1,n=8$ and $r=-\dfrac { 1 }{ 2 } $ in above equation, we have 

Given : $1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+......+\dfrac{1}{2^{n−1}} < 2−\dfrac{1}{1000}$

$\displaystyle \sum _{ n=1 }^{ 13 }{ ({ i }^{ n }+{ i }^{ n+1 }) }$ 

Let $S=1+2a+3a^{2}+4a^{3}+$   ..........$+na^{n-1}$
Multiply both sides by $a$, we get
$Sa=0+a+2a^{2}+3a^{3}$   .............$(n-1)a^{n-1}+na^{n}$
Subtract both equations,
$S(1-a)=1+a+a^{2}+a^{3}$    .............$a^{n-1}-na^{n}$
Clearly above series is G.P
Common ratio $= a$
$S(1-a)=\dfrac{1(a^{n}-1)}{a-1}-na^{n}$
$S=\dfrac{1-(a^{n})}{(a-1)^{2}}-\dfrac{na^{n}}{1-a}$

$1, 2, 4, ....., 2^n$

Let the first two terms be $a$ and $ar$ where $-1 < r < 1$.
$\therefore a(1 + r) = 4\dfrac {1}{2}$
Since $s = a/(1 - r) = 6, a = 6(1 - r)$
$\therefore 6(1 - r)(1 + r) = 4\dfrac {1}{2}; \therefore r = \pm \dfrac {1}{2}; \therefore a = 3$ or $9$.

$\begin{array}{l} { T _{ 1 } }=1 \ { T _{ 2 } }=a \ { T _{ 3 } }=Ca \ { T _{ 4 } }=C{ a^{ 2 } } \ { T _{ 2n } }=a\frac { { 2n } }{ 2 } \cdot C\frac { { 2n } }{ 2 } -1={ a^{ n } }{ C^{ n-1 } } \ { 5 _{ 2n } }=1+a+ca.....{ a^{ n } }{ C^{ n-1 } } \ =1+\left[ { a+c{ a^{ 2 } }+{ c^{ 2 } }{ a^{ 3 } }...{ a^{ n } }{ c^{ n-1 } } } \right]  \ +\left[ { ca+{ c^{ 2 } }{ a^{ 2 } }+{ c^{ 3 } }{ a^{ 3 } }.....{ c^{ n-1 } }{ a^{ n-1 } } } \right]  \ =1+\frac { { a\left( { { a^{ n } }{ c^{ n-1 } } } \right)  } }{ { ac-1 } } +\frac { { ac\left( { { a^{ n-1 } }{ c^{ n-1 } }-1 } \right)  } }{ { ac-1 } }  \ =\frac { { \left( { { a^{ n } }{ c^{ n } }-1 } \right) \left( { a+1 } \right)  } }{ { ac-1 } }  \end{array}$

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

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

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.