Tag: geometric sequences

Questions Related to geometric sequences

For a set of positive numbers, consider the following statements:
1. If each number is reduced by $2$, then the geometric mean of the set may not always exists.
2. If each number is increased by $2$, then the geometric mean of the set is increased by $2$.
Which of the above statements is/are correct?

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

  1. $1$ only

  2. $2$ only

  3. Both $1$ and $2$

  4. Neither $1$ nor $2$

Show answer
Correct Option: A
Explanation:

1. Consider the two numbers $1$ and $4$, geometric mean of $1$ and $4$ is $\sqrt {1 \times 4} = 2$.
When each number is reduced by $2$, the numbers become $-1$ and $2$ whose geometric mean does not exist.
2. Now consider two numbers $2$ and $7$. Their geometric mean is $\sqrt {14}$. The new numbers are $4$ and $9$ whose geometric mean is $\sqrt {4\times 9} = 6$ which is not equal to $2\sqrt {14}$.
Thus only statement $1$ is true.

If $a, b, c$ are in G.P., then $\dfrac {a - b}{b - c}$ is equal to

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

  1. $\dfrac {a}{b}$

  2. $\dfrac {b}{a}$

  3. $\dfrac {a}{c}$

  4. $\dfrac {c}{b}$

Show answer
Correct Option: A
Explanation:

We are given $a,b,c$ are in G.P.


Hence, ${b}^{2}=a\times c$


$\dfrac { a-b }{ b-c }=\dfrac { a-b }{ b-\dfrac { { b }^{ 2 } }{ a }  } $

$=\dfrac { a\left( a-b \right)  }{ b\left( a-b \right)  } $

$=\dfrac { a }{ b } $

Hence, option A is correct.

Say true or false.

Zero can be the common ratio of a G.P.

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

  1. True

  2. False

Show answer
Correct Option: B
Explanation:

The common ratio between two numbers has to be non-zero for it to form a G.P.

Say true or false.
$2, 4, 8, 16, .....$ is not an $A.P.$

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

  1. True

  2. False

Show answer
Correct Option: A
Explanation:

Difference between consecutive terms is $(4-2), (8-4), (16-8)$ and so on, i.e. 2,4,8 and so on.
Since the difference is not constant, it is not an AP.

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. $\frac{3}{16}$

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

Show answer
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

Show answer
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