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

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.

  1. True

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

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

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

  1. True

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

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

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

  1. $2, 4, 6, 8....$
  2. $5, 25, 125, 625....$
  3. $1.5, 3.0, 6.0, 12.0....$
  4. $8, 16, 24, 32, ....$
Reveal answer Fill a bubble to check yourself
A,D Correct answer
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.

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

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

  1. $0$
  2. $\cfrac { 2 }{ 7 } $
  3. $\cfrac { 6 }{ 7 } $
  4. $\cfrac { 9 }{ 32 } $
  5. $\cfrac { 27 }{ 32 } $
Reveal answer Fill a bubble to check yourself
B Correct answer
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 } $

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

If there exists a geometric progression containing 27, 8 and 12 as three of its terms (not necessarily consecutive) then no. of progressions possible are

  1. $1$
  2. $2$

  3. infinite

  4. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
$8, 12, 29$

Let the first term be $8$

$\therefore a = 8$    ...(i)

Let the $p^{th}$ ther be $12$

$\therefore ar^{p-1} = 12$     ...(ii)

and $q^{th}$ term be $27$

$\therefore ar^{q-1} = 27$    ...(iii)

(i) / (ii)

$\dfrac{a}{ar^{p-1}} = \dfrac{8}{12}$

$\Rightarrow r^{p-1} = \dfrac{3}{2} = 1.5$      ...(iv)

(iii) / (i)

$r^{q-1} = \dfrac{27}{8} = \left(\dfrac{3}{2}\right)^3$

$r^{q-1} = (1.5)^3$    ...(v)

(v) / (iv)

$\Rightarrow \dfrac{r^{q-1}}{r^{p-1}} = \dfrac{(1.5)^3}{1.5}$

$\Rightarrow r^{q-1-p+1} = (1.5)^2$

$\Rightarrow r^{q-p} = (1.5)^2$

              $= (r^{p-1})^2$     ...from (iv)

$\Rightarrow r^{q-p} = r^{2p-2}$

$\therefore q-p = 2p-2$

$\Rightarrow \boxed{q=3p-2}$

for every distinct value of '$p$' there will be a district integer value of '$q$'

$\therefore $ Infinite no. of progression are possible.

option $C$