Tag: understanding geometric progressions

Questions Related to understanding geometric progressions

The sum of infinity of $\frac{1}{7} + \frac{2}{7^2} + \frac{1}{7^3} + \frac{2}{7^4} + ......$ is:

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

  1. $\frac{1}{5}$

  2. $\frac{1}{24}$

  3. $\frac{5}{48}$

  4. $\frac{3}{16}$

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Correct Option: D

The limit of the sum of an infinite number of terms in a geometric progression is $a/(1 - r)$ where a denotes the first term and $-1 <r<1$ denotes the common ratio. The limit of the sum of their squares is:

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

  1. $\dfrac{a^2}{(1 - r)^2}$

  2. $\dfrac{a^2}{1 + r^2}$

  3. $\dfrac{a^2}{1 - r^2}$

  4. $\dfrac{4a^2}{1 + r^2}$

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Correct Option: C

If $S=1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+....\infty$.
then, the sum of the given series is $2$.

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

  1. True

  2. False

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Correct Option: A
Explanation:

We have,

$S=1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+....\infty$

Then,
$a=1$, $r=\dfrac{1}{2}$

We know that
$S=\dfrac{a}{1-r}$

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

$S=\dfrac{1}{\dfrac{1}{2}}$

$S=2$

Hence, this is the answer.

Given a sequence of $4$ members, first three of which are in G.P. and the last three are in A.P. with common difference six. If first and last terms of this sequence are equal, then the last term is:

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

  1. $8$

  2. $16$

  3. $2$

  4. $4$

Show answer
Correct Option: A
Explanation:
Let for terms be $a,ar,ar^{2},a$

$\because ar,ar^{2}$ and a are in A.P

$\therefore ar^{2}-ar=6$

$\Rightarrow ar(r-1)=6$

And $a-ar=2\times 6$

$\Rightarrow a(r-1)=-12$

$\Rightarrow \dfrac{ar(r-1)}{a(r-1)}=\dfrac{-6}{12}$

$\Rightarrow r=-\dfrac{1}{2}(\because r\neq 1)$

$\therefore a(1-r)=12$

$\Rightarrow a\left(1+\dfrac{1}{2}\right)=12$

$\Rightarrow \dfrac{39}{2}=12$

$\Rightarrow a=8$

$\therefore $ Last term = $8$

$n$ 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. $32$

Show answer
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 $\left( 1 \right) &amp; \left( 2 \right)$, 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:

$8,-4,2,8$