Tag: maths

Questions Related to maths

Find the sum the infinite G.P.: $\displaystyle {\frac{2}{3}\, -\, \frac{4}{9}\, +\, \frac{8}{27}\, -\, \frac{16}{21}\, +\, ........}$ 

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

  1. $\displaystyle \frac{2}{5}$

  2. $\displaystyle \frac{3}{5}$

  3. $\displaystyle \frac{19}{27}$

  4. $\displaystyle \frac{8}{5}$

Show answer
Correct Option: A
Explanation:

We know, $S _{\infty }=\dfrac{a}{1-r}$
From the given series, $a=\dfrac{2}{3} ,r=-\dfrac{2}{3}$
$\therefore S _{\infty }=\dfrac{\frac{2}{3}}{1-\left ( -\frac{2}{3} \right )}$


$\Rightarrow \dfrac{\frac{2}{3}}{1+\frac{2}{3}}$

$\Rightarrow \dfrac{\frac{2}{3}}{\frac{5}{3}}$

$\Rightarrow \dfrac{2}{5}$

Sum the series: $\displaystyle {1\, -\, \frac{1}{3}\, +\, \frac{1}{3^2}\, -\, \frac{1}{3^3}\, +\, \frac{1}{3^4}.......\infty}$

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

  1. $\displaystyle \frac{3}{4}$

  2. $\displaystyle \frac{4}{3}$

  3. $\displaystyle \frac{2}{3}$

  4. $\displaystyle \frac{1}{3}$

Show answer
Correct Option: A
Explanation:

We know, $S _{\infty }=\dfrac{a}{1-r}$
From the given series, $a=1 ,r=-\dfrac{1}{3}$
$\therefore S _{\infty }=\dfrac{1}{1-\left ( -\dfrac{1}{3} \right )}$
$= \dfrac{1}{1+\dfrac{1}{3}}$


$ =\dfrac{1}{\dfrac{4}{3}}$

$= \dfrac{3}{4}$

If a, b and c are in geometric progression, then $a^2$, $b^2$ and $c^2$ are in _____ progression.

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

  1. AP

  2. GP

  3. HP

  4. AGP

Show answer
Correct Option: B
Explanation:

Given $ a,b,c $ are in GP.
So, the common ratio between the first and second term ; second and third will be the same.
$ => \dfrac {b}{a} = \dfrac {c}{b} $


$ => b^2 = ac $

If we square both sides, we see that
$ (b^2)^2 = a^2 \times c^2 $

This means, even, $ a^2, b^2, c^2 $ are also in GP.

The sequence $-6 + 42 - 294 + 2058$ is a

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

  1. finite geometric sequence

  2. finite arithmetic sequence

  3. infinite geometric sequence

  4. infinite harmonic sequence

Show answer
Correct Option: A
Explanation:

The sequence $-6 + 42 - 294 + 2058$ is a finite geometric sequence.
Here the common ratio is $-7$.

As it has finite terms, therefore the series is finite sequence.

The sum of the series $10 - 5 + 2.5 - 1.25.....$ is called

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

  1. finite geometric sequence

  2. finite arithmetic sequence

  3. infinite geometric sequence

  4. infinite harmonic sequence

Show answer
Correct Option: C
Explanation:

Given series is $10-5+2.5-1.25.....$

Here the common ratio is $\dfrac {-5}{10}=-\dfrac {1}{2}$.
It is also never ending and continued.
Hence, the given series is infinite geometric series.

When a number $x$ is subtracted from each of the numbers $8, 16$, and $40$, the resulting three numbers form a geometric progression. Find the value of $x$.

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

  1. $3$

  2. $4$

  3. $6$

  4. $12$

  5. $18$

Show answer
Correct Option: B
Explanation:

Given that ${(16-x)}^{2}=(8-x)(40-x)$
$\Rightarrow 256-32x+{x}^{2} = 320-48x+{x}^{2}$
$\Rightarrow 16x = 64$ 

$\Rightarrow x = 4$

Say true or false.
The total savings (in $Rs.$) after every month for $10$ months when $Rs. 50$ are saved each month are $50, 150, 200, 250, 300, 350, 400, 450, 500$ represent 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:

Ratio of second term to first term is $ \frac {150}{50} = 3$

Ratio of third term to second term is  $ \frac {200}{150} = 1.33$

Thus, the ratio is not matching. 

Hence, it is not a GP.

Say true or false.
Given series:
$15, 30, 60, 120, 240$ is in G.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:

Given series is $15,30,60,120,240$
Ratio between first two terms $=$ $\dfrac{30}{15}$ $=2$
Ratio between second and third terms $=$ $\dfrac{60}{30}$ $=2$
Ratio between third and fourth terms $=$ $\dfrac{120}{60}$ $=2$
Ratio between fourth and fifth terms $=$ $\dfrac{240}{120}$ $=2$
Since, the ratio between the terms is the same. The series forms a G.P.

Which of the following is not a G.P.?

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

  1. $2, 4, 6, 8....$

  2. $5, 25, 125, 625....$

  3. $1.5, 3.0, 6.0, 12.0....$

  4. $8, 16, 24, 32, ....$

Show answer
Correct Option: A,D
Explanation:

In series $2,4,6,8,....$ difference is same i.e. $2$

In $8,16,24,32,......$ difference again is same $8$
$\therefore$ both the series (a) and (b) are in AP as the difference between their consecutive terms is the same.

For the infinite series $1-\cfrac { 1 }{ 2 } -\cfrac { 1 }{ 4 } +\cfrac { 1 }{ 8 } -\cfrac { 1 }{ 16 } -\cfrac { 1 }{ 32 } +\cfrac { 1 }{ 54 } -\cfrac { 1 }{ 128 } -....\quad $ let $S$ be the (limiting) sum. Then $S$ equals

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

  1. $0$

  2. $\cfrac { 2 }{ 7 } $

  3. $\cfrac { 6 }{ 7 } $

  4. $\cfrac { 9 }{ 32 } $

  5. $\cfrac { 27 }{ 32 } $

Show answer
Correct Option: B
Explanation:

Combine the terms in threes, to get the geometric series
$\cfrac { 1 }{ 4 } +\cfrac { 1 }{ 32 } +\cfrac { 1 }{ 256 } +....;\quad \quad S=\cfrac { \cfrac { 1 }{ 4 }  }{ 1-\cfrac { 1 }{ 8 }  } =\cfrac { 2 }{ 7 } $ or
rearrange the terms into three series:
$1+\cfrac { 1 }{ 8 } +\cfrac { 1 }{ 64 } +...\quad -\cfrac { 1 }{ 2 } -\cfrac { 1 }{ 16 } -\cfrac { 1 }{ 128 } -....,\quad -\cfrac { 1 }{ 4 } -\cfrac { 1 }{ 32 } -\cfrac { 1 }{ 256 } -....\quad $
${ S } _{ 1 }=\cfrac { 1 }{ 1-\cfrac { 1 }{ 8 }  } =\cfrac { 8 }{ 7 } ;{ S } _{ 2 }=\cfrac { -\cfrac { 1 }{ 2 }  }{ 1-\cfrac { 1 }{ 8 }  } =-\cfrac { 4 }{ 7 } ;{ S } _{ 3}=\cfrac { -\cfrac { 1 }{ 4 }  }{ 1-\cfrac { 1 }{ 8 }  } =-\cfrac { 2 }{ 7 } ;\quad \therefore S=\cfrac { 2 }{ 7 } $