Tag: introduction to geometric progression

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

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

  1. $\frac{1}{5}$
  2. $\frac{1}{24}$
  3. $\frac{5}{48}$
  4. $\frac{3}{16}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The series can be split into two geometric series: S1 = 1/7 + 1/7^3 + 1/7^5... and S2 = 2/7^2 + 2/7^4 + 2/7^6... For S1, a=1/7, r=1/49, sum = (1/7)/(1-1/49) = 7/48. For S2, a=2/49, r=1/49, sum = (2/49)/(1-1/49) = 2/48. Total sum = 9/48 = 3/16.

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

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:

  1. $\dfrac{a^2}{(1 - r)^2}$
  2. $\dfrac{a^2}{1 + r^2}$
  3. $\dfrac{a^2}{1 - r^2}$
  4. $\dfrac{4a^2}{1 + r^2}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

If the original series is a, ar, ar^2, ..., the sum is a/(1-r). The series of squares is a^2, a^2r^2, a^2r^4, ..., which is a geometric series with first term a^2 and common ratio r^2. The sum is a^2/(1-r^2).

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

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 $2$.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
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.

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

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:

  1. $8$
  2. $16$
  3. $2$
  4. $4$
Reveal answer Fill a bubble to check yourself
A Correct answer
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 = $8$
Multiple choice maths geometric sequences introduction to geometric progression understanding geometric progressions introduction to geometric progressions

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

  1. $127$
  2. $63$
  3. $64$
  4. $32$
Reveal answer Fill a bubble to check yourself
C Correct answer
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 $m=64$

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

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 $\left[ {area\,\left( {\Delta {P _1}.{P _2}.{P _3}} \right)/area\,\left( {\Delta {P _2}{P _3}{P _4}} \right)} \right].$

  1. $\dfrac{1}{2}$
  2. $\dfrac{1}{4}$
  3. $\dfrac{1}{8}$
  4. $\dfrac{1}{16}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

For the curve y = x^3, the tangent at x1 meets the curve at x2 = -2x1. This forms a geometric progression with common ratio -2. The area of the triangle formed by points on the curve scales with the coordinates, leading to a ratio of 1/16 for consecutive triangles.

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

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

  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.
Reveal answer Fill a bubble to check yourself
A Correct answer
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.                    

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

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?

  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. $a=3, r=\dfrac{1}{4}$
Reveal answer Fill a bubble to check yourself
C,D Correct answer
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  $a = 1$
$\Rightarrow r = \cfrac{3}{4}$

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

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

  1. $x < -10$
  2. $0 < x < 10$
  3. $-10 < x < 10$
  4. $x > 10$
Reveal answer Fill a bubble to check yourself
B Correct answer
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,
$\left| r \right| < 1$
$\left| 1 - \cfrac{x}{5} \right| < 1$
$\Rightarrow -1 < 1 - \cfrac{x}{5} < 1$
$\Rightarrow -2 < \cfrac{-x}{5} < 0$
$\Rightarrow 0 < \cfrac{x}{5} < 2$
$\Rightarrow 0 < x < 10$
Multiple choice maths geometric sequences introduction to geometric progression understanding geometric progressions introduction to geometric progressions

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?

  1. $2$
  2. $4$
  3. $6$
  4. $8$
Reveal answer Fill a bubble to check yourself
D Correct answer
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:

$8,-4,2,8$