Tag: maths

Questions Related to maths

The sum of sequence $0.15,0.015,0.0015,.....$ upto 20 term is ?

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

  1. $\dfrac{1}{6}[1-(0.1)^{20}]$

  2. $\dfrac{1}{6}[1+(0.1)^{20}]$

  3. $\dfrac{1}{3}[1-(0.1)^{20}]$

  4. None of these

Show answer
Correct Option: A
Explanation:

Since the sequence in Geometric progression 

where, $a=0.15;\ r=\dfrac { 0.015 }{ 0.15 } =0.1;\ n=20\ \therefore { S } _{ n }=\dfrac { a({ r }^{ n }-1) }{ (r-1) } \ =\dfrac { 0.15({ \left( 0.1 \right)  }^{ 20 }-1) }{ 0.1-1 } \ =\dfrac { 1 }{ 6 } \left( 1-{ (0.1) }^{ 20 } \right) $

The sum of $10$ terms of GP $\frac { 1 } { 2 } + \frac { 1 } { 4 } + \frac { 1 } { 8 } + \ldots$ is-

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

  1. $\frac { 2 ^ { 10 } - 1 } { 2 ^ { 10 } }$

  2. $\frac { 2 ^ { 9 } - 1 } { 2 ^ { 9 } }$

  3. $\frac { 2 ^ { 10 } - 1 } { 2 ^ { 9 } }$

  4. $\frac { 2 ^ { 9 } - 1 } { 2 ^ { 10 } }$

Show answer
Correct Option: A
Explanation:
$S=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...$ is in G.P
where $a=\dfrac{1}{2}$ and $r=\dfrac{\dfrac{1}{4}}{\dfrac{1}{2}}=\dfrac{1}{2}$
${S} _{n}=\dfrac{a\left({r}^{n}-1\right)}{r-1}$
${S} _{10}=\dfrac{\dfrac{1}{2}\left({\left(\dfrac{1}{2}\right)}^{10}-1\right)}{\dfrac{1}{2}-1}$
${S} _{10}=\dfrac{\dfrac{1}{2}\left({\left(\dfrac{1}{2}\right)}^{10}-1\right)}{\dfrac{-1}{2}}$

$=1-\dfrac{1}{{2}^{10}}$
$=\dfrac{{2}^{10}-1}{{2}^{10}}$


The geometric series $a+ar+ar^{2}+ar^{3}+......\infty$ has sum $7$ and the terms involving odd powders of $r$ has sum $'3'$, then the value of $(a^{2}-r^{2})$ is-

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

  1. $\dfrac{5}{4}$

  2. $\dfrac{5}{2}$

  3. $\dfrac{25}{4}$

  4. $5$

Show answer
Correct Option: B
Explanation:
Sum of geometric series = $\dfrac{a(r^{n}-1)}{r-1}=7$
Also odd powers of r terms :-
$ar,ar^{3},ar^{5},ar^{7}$
$\therefore sum=\dfrac{ar(r^{2n}-1)}{r^{2}-1}$
$\therefore $ sum of infinite G.P = $\dfrac{a}{1-r}=7$ _________(1)
$\therefore $ sum of infinite second = $\dfrac{ar}{1-r^{2}}=3$ _____(2)
$\therefore $ From (1) & (2)
$\dfrac{1+r}{r}=\dfrac{7}{3} ; r=\dfrac{3}{4}$
$\therefore a=\dfrac{7}{4}$
$\therefore a^{2}-r^{2}=\dfrac{5}{2}$

The sum 
$1 + \left( {1 + x} \right) + \left( {1 + x + {x^2}} \right) + \left( {1 + x + {x^2} + {x^3}} \right) +  \ldots n$  terms equals 

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

  1. $\frac{{1 - {x^n}}}{{1 - x}}$

  2. $\frac{{x\left( {1 - {x^n}} \right)}}{{1 - x}}$

  3. $\frac{{n\left( {1 - x} \right) - x\left( {1 - {x^n}} \right)}}{{{{\left( {1 - x} \right)}^2}}}$

  4. None of these

Show answer
Correct Option: C

$\lim _{ x\leftarrow 1 }{ \cfrac { x+{ x }^{ 2 }+{ x }^{ 3 }+....+{ x }^{ n }-n }{ x-1 }  } =$

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

  1. $\cfrac{n(n+1)}{2}$

  2. $\cfrac{n+1}{2}$

  3. $\cfrac{2}{n}$

  4. $n$

Show answer
Correct Option: A
Explanation:

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

By L Hospital rule
$= \underset{x \rightarrow 1}{lim} \dfrac{1 + 2x + 3x^2 + ... + nx^{n - 1}}{1}$
$= 1 + 2 + 3 + ... + n$
$= \dfrac{n (n + 1)}{2}$

The sum of first $10$ terms of the series $\sqrt{2}+\sqrt{6}+\sqrt{18}+...$ is

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

  1. $121(\sqrt{6}+\sqrt{2})$

  2. $243(\sqrt{3}+1)$

  3. $\cfrac{121}{\sqrt{3}-1}$

  4. $242(\sqrt{3}-1)$

Show answer
Correct Option: A
Explanation:
$s=\sqrt{2}+\sqrt{6}+\sqrt{18}+....$

$\Rightarrow s=\sqrt{2}+\sqrt{2}\times \sqrt{3}+\sqrt{2}(\sqrt{3})^2+.....$

This is a G.P with $1$st term $(a)=\sqrt{2}$ and common ratio$(r)=\sqrt{3}$

Sum of $10$ terms of this G.P., $S=\dfrac{a(r^{10}-1)}{r-1}$

$=\dfrac{\sqrt{2}((\sqrt{3})^{10}-1)}{\sqrt{3}-1}$

$=\dfrac{\sqrt{2}(242)}{\sqrt{3}-1}\times \dfrac{\sqrt{3}+1}{\sqrt{3}+1}$

$=121\times \sqrt{2}(\sqrt{3}+1)$

$=121(\sqrt{6}+\sqrt{2})$

$\Rightarrow (A)$ Option. 

If ${S} _{n}=\sum _{ r=1 }^{ n }{ \cfrac { 1+2+{ 2 }^{ 2 }+..Sum\quad to\quad r\quad terms }{ { 2 }^{ r } }  } $, then ${S} _{n}$ is equal to 

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

  1. ${2}^{n}-n-1$

  2. $1-\cfrac{1}{{2}^{n}}$

  3. $n-1+\cfrac{1}{{2}^{n}}$

  4. ${2}^{n}-1$

Show answer
Correct Option: C
Explanation:
$\displaystyle  \rightarrow S _{n} = \sum _{r=1}^{n}\left(\dfrac{\frac{2^{r}-1}{2-1}}{2^{r}}\right)$ sum of G.P

$\displaystyle  \Rightarrow S _{n} = \sum _{r=1}^{n}(1-2^{-r})$

$ \displaystyle \Rightarrow S _{n} = n- \sum _{r=1}^{n}2^{-r}$

$ \displaystyle \Rightarrow S _{n} = n -\left(2^{-1}(\frac{2^{-n}-1}{2^{-1}-1})\right)$

$\displaystyle  \Rightarrow S _{n} = n-\left(\frac{1}{2}(\frac{1-2^{n}}{2^{n}(\frac{-1}{2})})\right)$

$\displaystyle  \Rightarrow S _{n} = n+\left(\frac{1-2^{n}}{2^{n}}\right) = n-1+\frac{1}{2^{n}}$

$ \Rightarrow (C)$

Find the sum of 8 terms of the G.P: 3+6+12+24.........

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

  1. 381

  2. 384

  3. 128

  4. None of these

Show answer
Correct Option: A

If $S _{1}=\left{2\right},\ S _{2}=\left{3,6\right},\ S _{3}=\left{4,8,16\right},\ S _{4}=\left{5,10,20,40\right},...$ then the sum of numbers in the set $S _{15}$ is

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

  1. $5\left(2^{15}\right)$

  2. $16\left(2^{15}-1\right)$

  3. $16\left(2^{16}-1\right)$

  4. $15\left(2^{15}-1\right)$

Show answer
Correct Option: B
Explanation:

$S _1=\left{ 2 \right} $

$S _2=\left{ 3,6 \right} $
$S _3=\left{ 4,8,16 \right} $
$S _4=\left{ 5,10,20,40 \right} $
looking at the trend we can write a general set $S _n$ as :
$\Rightarrow S _n=\left{ 2^0\left( n+1 \right) ,2\left( n+1 \right) ,2^{ 2 }\left( n+1 \right) ,.....2^{ n-1 }\left( n+1 \right)  \right} $
the elements in $S _{15}$ set can be obtained by a substituting $n=15$ in the above equation.
$\Rightarrow S _{15}=\left{ \left( 16 \right) ,2\left( 16 \right) ,2^{ 2 }\left( 16 \right) ,.....2^{ 14}\left( 16 \right)  \right} $
Sum of the elements :
$=2^0 \left( 16 \right) ,2^1 \left( 16 \right) ,2^{ 2 }\left( 16 \right) ,.....2^{ 14 }\left( 16 \right)  $
$=16\left{ 2^0+2^1+2^2+.....+2^{14}\right}$
$=16\left{ 1+2+2^2+......2^{14}\right}$
we know that the sum of an GP containing n terms is given by :
$\Rightarrow S _n=\dfrac{a\left( r^n-1\right)}{r-1};\;\left| r\right| >1$ , where a is the first term and 'r' is the common ratio. 
Using this,
$=16\left{ 1+2+2^2+.......2^{14}\right}$
$=16\dfrac{\left( 2^{15}-1\right)}{2-1}$
$=16\left( 2^{15}-1\right)$
Hence, the answer is $16\left( 2^{15}-1\right).$
 

If $a _{0},a _{1},a _{3},....$ and $b _{0},b _{1},b _{2},b _{3},...$ are two geometric progressions with $a _{1}=2\surd 3$ and $b _{1}=\dfrac {52}{9}\sqrt {3}$ if $3a _{99}b _{99}=104$ then $\displaystyle \sum^{101} _{i=0}a _{1}b _{1}$ is

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

  1. $102$

  2. $3536$

  3. $2040$

  4. $3120$

Show answer
Correct Option: D