Tag: introduction to geometric progressions

Questions Related to introduction to geometric progressions

How would you find the sequence is finite geometric sequence?

  1. An arithmetic sequence containing finite number of terms

  2. A geometric sequence containing finite number of terms

  3. An arithmetic sequence containing infinite number of terms

  4. A geometric sequence containing infinite number of terms


Correct Option: B
Explanation:

If a sequence is a finite geometric sequence, then :

It will have the finite number of terms.
it will be a geometric sequence i.e. its ratio will be constant throughout.
Option B is the correct answer.

Identify the finite geometric progression.

  1. $3, 6, 12, 24...$

  2. $81, 27, 9, 3..$

  3. $10 - 5 + 2.5 - 1.25.....$

  4. $1 + 0.5 + 0.25 + 0.125$


Correct Option: D
Explanation:

$1 + 0.5 + 0.25 + 0.125 $$ is a finite geometric progression.
Here the common ratio is $0.5$.
An finite geometric series is the sum of an finite geometric sequence.

Identify the correct sequence represents a infinite geometric sequence.

  1. $3, 6, 12, 24, 48$

  2. $1 + 2 + 4 + 8 +....$

  3. $1, -1, 1, -1, 1$

  4. $1, 3, 4, 5, 6....$


Correct Option: B
Explanation:

An infinite geometric series is the sum of an infinite geometric sequence.
So, $1 + 2 + 4 + 8 +....$ is an infinite geometric sequence.
Here the common ratio is $2$ and it is never ending.

If $\dfrac{a-b}{b-c}=\dfrac{a}{b}$, then $a, b, c $ are in

  1. GP

  2. HP

  3. AP

  4. SP


Correct Option: A
Explanation:

Given:

$\dfrac{a-b}{b-c}=\dfrac{a}{b}$
$\Rightarrow ab-b^{2}=ab-ac$
$\therefore b^{2}=ac  \rightarrow G.P.$
$(Ans \to A)$

$2+{2}^{2}+{2}^{3}+.......+{2}^{9}=$?

  1. $2044$

  2. $1022$

  3. $1056$

  4. None of these


Correct Option: B
Explanation:

This is G.P in which  $a=2,r=\cfrac{{2}^{2}}{2}=2$ and $n=9$
$\therefore$ ${S} _{n}=\cfrac{a({r}^{n}-1)}{(r-1)}=\cfrac{2\times ({2}^{9}-1)}{(2-1)}=2\times (512-1)=2\times 511=1022$.

How many terms are there in the G.P $3,6,12,24,.........,384$?

  1. $8$

  2. $9$

  3. $10$

  4. $11$

  5. $7$


Correct Option: A
Explanation:

Here $a=3$ and $r=\cfrac{6}{3}=2$. Let the number of terms be $n$$.
Then, ${t}_{n}=384$ $\Rightarrow$ $a{r}^{n-1}=384$
$\Rightarrow$ $3\times {2}^{n-1}=384$
$\Rightarrow$ ${2}^{n-1}=128={2}^{7}$
$\Rightarrow$ $n-1=7$
$\Rightarrow$ $n=8$
$\therefore$ Number of terms $=8$.

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?

  1. $1$ only

  2. $2$ only

  3. Both $1$ and $2$

  4. Neither $1$ nor $2$


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

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

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

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

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


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.

  1. True

  2. False


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 $A.P.$

  1. True

  2. False


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.