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Questions Related to maths

$n$ is an integer. The largest integer $m$, such that ${n^m} + 1$ divides $1 + n + {n^2} + .....{n^{127}},$ is

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

  1. $127$

  2. $63$

  3. $64$

  4. $32$

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Correct Option: C
Explanation:

$1+n+{ n }^{ 2 }+...{ n }^{ 127 }=\cfrac { { n }^{ 128 }-1 }{ n-1 } \ =\cfrac { \left( { n }^{ 64 }-1 \right) \left( { n }^{ 64 }+1 \right)  }{ (n-1) } \ ({ a }^{ n }-{ b }^{ m })$ is divisible by $\quad (a-b)\bigvee  m\in { z }^{ + }$(positive Integers)

$\left( { n }^{ 64 }-1 \right) $ is divisible by $(n-1)$
$\cfrac { \left( { n }^{ 64 }-1 \right)  }{ (n-1) } $ is Integer value
$1+n+{ n }^{ 2 }+...{ n }^{ 127 }$ is divisible by $\left( { n }^{ 64 }+1 \right) $
Given $1+n+{ n }^{ 2 }+...{ n }^{ 127 }$ is divisible by $\quad { n }^{ m }+1$
Maximum possible value of m $m=64$

Tangent at a point ${P _1}$ (other than (0, 0) on the curve $y = {x^3}$ meets the curve again at ${P _2}$. The tangent at ${P _2}$ meets the curve again at ${P _3}$ and so on. Show that the abscissae of ${P _1},{P _2},..........,{P _n}$ form a G.P. Also find the ratio $\left[ {area\,\left( {\Delta {P _1}.{P _2}.{P _3}} \right)/area\,\left( {\Delta {P _2}{P _3}{P _4}} \right)} \right].$

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

  1. $\dfrac{1}{2}$

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

  3. $\dfrac{1}{8}$

  4. $\dfrac{1}{16}$

Show answer
Correct Option: D

If $a, b, c$ are in G.P., then

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

  1. $a^2, b^2, c^2$ are in G.P.

  2. $a^2(b+c), c^2 (a+b), b^2 (a+c)$ are in G.P.

  3. $\displaystyle \frac{a}{b+c}, \frac{b}{c+a}, \frac{c}{a+b}$ are in G.P.

  4. None of the above.

Show answer
Correct Option: A
Explanation:

Given:  a, b, c are in G.P.

$\therefore b^{2}=ac$                                    ...eq ( 1 )
Squaring both sides:
$\Rightarrow b^{4}=a^{2}c^{2}$
$\Rightarrow (b^{2})^{2}=a^{2}c^{2}$
$\therefore a^{2},b^{2},c^{2}$ are in G.P.                    

Consider an infinite $G.P$. with first term $a $ and common ratio $r$, its sum is $4$ and the second term is $\dfrac {3}{4}$, then?

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

  1. $a=\dfrac{4}{7}, r=\dfrac{3}{7}$

  2. $a=\dfrac{3}{2}, r=\dfrac{1}{2}$

  3. $a=1, r=\dfrac{3}{4}$

  4. $a=3, r=\dfrac{1}{4}$

Show answer
Correct Option: C,D
Explanation:

Given:-

${S} _{\infty} = 4$
${a} _{2} = \cfrac{3}{4}$
$\Rightarrow ar = \cfrac{3}{4}$
$\Rightarrow 4ar = 3 ..... \left( 1 \right)$
As we know that,
${S} _{\infty} = \cfrac{a}{1 - r}$
$\therefore \cfrac{a}{1 - r} = 4$
$\Rightarrow 4r = 4 - a ..... \left( 2 \right)$
From equation $\left( 1 \right) & \left( 2 \right)$, we have
$a \left( 4 - a \right) = 3$
$\Rightarrow {a}^{2} - 4a + 3 = 0$
$\Rightarrow \left( a - 3 \right) \left( a - 1 \right) = 0$
$\Rightarrow a = 1$ or $a = 3$
Substituting the value of $a$ in equation $\left( 1 \right)$, we get
For $a = 3$
$\Rightarrow r = \cfrac{1}{4}$
For  $a = 1$
$\Rightarrow r = \cfrac{3}{4}$

The first term of an infinite geometric progression is x and its sum is $5$. then 

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

  1. $x < -10$

  2. $0 < x < 10$

  3. $-10 < x < 10$

  4. $x > 10$

Show answer
Correct Option: B
Explanation:
Given:-
${S} _{\infty} = 5$
$a = x$
As we know that,
${S} _{\infty} = \cfrac{a}{1 - r}$
$\therefore \cfrac{x}{1 - r} = 5$
$\Rightarrow \cfrac{x}{5} = 1 - r$
$\Rightarrow r = 1 - \cfrac{x}{5}$
Now,
$\left| r \right| < 1$
$\left| 1 - \cfrac{x}{5} \right| < 1$
$\Rightarrow -1 < 1 - \cfrac{x}{5} < 1$
$\Rightarrow -2 < \cfrac{-x}{5} < 0$
$\Rightarrow 0 < \cfrac{x}{5} < 2$
$\Rightarrow 0 < x < 10$

The first three of four given numbers are in G.P. and last three are in A.P. whose common difference is $6$. If the first and last numbers are same, then first will be?

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

  1. $2$

  2. $4$

  3. $6$

  4. $8$

Show answer
Correct Option: D
Explanation:

Last $3$ of the $4$ numbers are in AP.


Let they are, $a-d, a, a+d$. Also the first number is same as $4th, a+d$.

Therefore the $4$ numbers are $a+d, a-d, a, a+d$

The first $3$ of these are in G.P.


$\therefore (a-d)^2=a(a+d)$ But $d=6$

$\therefore (a-6)^2=a(a+6)$

$\therefore a^2-2a+36=0$

Solving the above quadratic equation, we get,

$a=2$

Therefore the series is:

$8,-4,2,8$


The sum of $1 + \left( {1 + a} \right)x + \left( {1 + a + {a^2}} \right){x^2} + ....\infty ,\,0 < a,\,x < 1$ equals

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

  1. $\dfrac{1}{{\left( {1 - x} \right)\left( {1 - a} \right)}}$

  2. $\dfrac{1}{{\left( {1 - a} \right)\left( {1 - ax} \right)}}$

  3. $\dfrac{1}{{\left( {1 - x} \right)\left( {1 - ax} \right)}}$

  4. $\dfrac{1}{{\left( {1 - x} \right)\left( {1 + a} \right)}}$

Show answer
Correct Option: C
Explanation:
$1+\left( 1+a \right) x+\left( 1+a+{ a }^{ 2 } \right) { x }^{ 2 }+\dots \infty $

$ General\quad term\quad of\quad series={ T } _{ n }=\frac { \left( 1-{ a }^{ r+1 } \right)  }{ \left( 1-a \right)  } { x }^{ r }$  

$ { S } _{ n }=\sum _{ r=1 }^{ \infty  }{ \dfrac { \left( 1-{ a }^{ r+1 } \right) { x }^{ r } }{ \left( 1-a \right)  }  } $  

$ \left( 1-a \right) { S } _{ n }=\sum _{ r=1 }^{ \infty  }{ { x }^{ r } } -a\sum _{ r=1 }^{ \infty  }{ \left( ax \right) ^{ r } } $ 

$\left( 1-a \right) { S } _{ n }=\dfrac { 1 }{ 1-x } -\dfrac { a }{ 1-ax } $     ...........  $\because G.P.$ series 

$ \left( 1-a \right) { S } _{ n }=\dfrac { 1-ax-a+ax }{ \left( 1-x \right) \left( 1-ax \right)  } $  

$ \left( 1-a \right) { S } _{ n }=\dfrac { 1-a }{ \left( 1-x \right) \left( 1-ax \right)  } $  

$ { S } _{ n }=\dfrac { 1 }{ \left( 1-x \right) \left( 1-ax \right)  } $
B is correct

If roots of the equations $(b-c)x^2+(c-a)x+a-b=0$, where $b\neq c$, are equal, then a, b, c are in?

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

  1. G.P.

  2. H.P.

  3. A.P.

  4. A.G.P.

Show answer
Correct Option: C
Explanation:
$(b-c)x^{2}+(c-a)x+a-b=0$
Root are equal, so $D=0$
$\Rightarrow (c-a)^{2}-4(b-c)(a-b)=0$
$\Rightarrow c^2+a^2-2ac-4ab+4b^2+4ac-4bc=0$
$\Rightarrow c^2+a^2+4b^2+2ac-4ab-4bc=0$
$\Rightarrow (a+c-2b)^{2}=0$
$a+c=2b$

























If the roots of ${ x }^{ 2 }-k{ x }^{ 2 }+14x-8=0$ are in geometric progression, then $k=$

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

  1. $-3$

  2. $7$

  3. $4$

  4. $0$

Show answer
Correct Option: B
Explanation:

$\Rightarrow \frac { a }{ r } ,a,ar=8\quad \quad \Rightarrow { a }^{ 3 }=8\quad \quad \Rightarrow a=2$
$a=2$ is a root of the given equation 
$\Rightarrow 8-4k+28-8=0\quad \quad \Rightarrow k=7$

The third term of a geometric progression is $4$. The product of the first five terms is 

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

  1. ${4}^{3}$

  2. ${4}^{4}$

  3. ${4}^{5}$

  4. ${4}^{6}$

Show answer
Correct Option: C
Explanation:
Let $a$ and $r$ the first term and common ratio, respectively.
Given:Third term$=a{r}^{2}=4$
Product of first $5$ terms is
$=a.ar.a{r}^{2}.a{r}^{3}.a{r}^{4}$
$={a}^{5}{r}^{10}$
$={\left(a{r}^{2}\right)}^{5}$
$={4}^{5}$