
If 8 GM’s are inserted between 2 and 3, then the product of 8 GM’s is:
A.6
B.36
C.216
D.1296
Answer
591.3k+ views
Hint: If 8 GM’s are inserted between 2 and 3 (GM= geometric progression), then the first value is 2 and the last value is 3, then in geometric progression first value is denoted as $a$ and second value is denoted as \[ar\], third value as\[a{r^2}\], and so on, compare the values and find the value of $a\,and\,r$, which will help in finding out the value of product of 8GM’s.
Complete step-by-step answer:
If 8 GM’s are inserted between 2 and 3, then it looks like
$2,\,G{M_1},\,G{M_2},\,G{M_3},\,G{M_4},\,G{M_5},\,G{M_6},\,G{M_7},\,G{M_8},\,3$, and we know that geometric progression series is $a,\,ar,\,a{r^2},\,a{r^3},\,a{r^4},............$ compare the values with the series to figure out the values of a and r.
So, the value of $a = 2,\,ar = G{M_1},\,a{r^2} = G{M_2},............,a{r^9} = 3$
Now, we have $a{r^9} = 3$ and $a = 2........\left( 1 \right)$,
$
\Rightarrow a{r^9} = 3 \\
\Rightarrow {r^9} = \dfrac{3}{a} \\
\Rightarrow {r^9} = \dfrac{3}{2}...........\left( 2 \right) \\
$
To find the value of product of 8GM’s, we have
\[
\Rightarrow G{M_1} \times G{M_2} \times ........ \times G{M_8} = ar \times a{r^2} \times a{r^3} \times ........ \times a{r^8} \\
\Rightarrow G{M_1} \times G{M_2} \times ........ \times G{M_8} = {a^8}{r^{\dfrac{{8\left( {8 + 1} \right)}}{2}}} \\
\Rightarrow G{M_1} \times G{M_2} \times ........ \times G{M_8} = {a^8}{r^{36}} \\
\Rightarrow G{M_1} \times G{M_2} \times ........ \times G{M_8} = {a^8}{\left( {{r^9}} \right)^4} \\
\Rightarrow G{M_1} \times G{M_2} \times ........ \times G{M_8} = {\left( 2 \right)^8} \times {\left( {\dfrac{3}{2}} \right)^4} \\
\Rightarrow G{M_1} \times G{M_2} \times ........ \times G{M_8} = 4 \times 4 \times 4 \times 4 \times \dfrac{3}{2} \times \dfrac{3}{2} \times \dfrac{3}{2} \times \dfrac{3}{2} \\
\Rightarrow G{M_1} \times G{M_2} \times ........ \times G{M_8} = 1296 \\
\]
So, the product of 8GM’s is 1296.
Option D is correct.
Note: In geometric progression series, a is the first term and r is the common ratio between the numbers of the series. Don’t complicate the question by finding the values of each GM.
Complete step-by-step answer:
If 8 GM’s are inserted between 2 and 3, then it looks like
$2,\,G{M_1},\,G{M_2},\,G{M_3},\,G{M_4},\,G{M_5},\,G{M_6},\,G{M_7},\,G{M_8},\,3$, and we know that geometric progression series is $a,\,ar,\,a{r^2},\,a{r^3},\,a{r^4},............$ compare the values with the series to figure out the values of a and r.
So, the value of $a = 2,\,ar = G{M_1},\,a{r^2} = G{M_2},............,a{r^9} = 3$
Now, we have $a{r^9} = 3$ and $a = 2........\left( 1 \right)$,
$
\Rightarrow a{r^9} = 3 \\
\Rightarrow {r^9} = \dfrac{3}{a} \\
\Rightarrow {r^9} = \dfrac{3}{2}...........\left( 2 \right) \\
$
To find the value of product of 8GM’s, we have
\[
\Rightarrow G{M_1} \times G{M_2} \times ........ \times G{M_8} = ar \times a{r^2} \times a{r^3} \times ........ \times a{r^8} \\
\Rightarrow G{M_1} \times G{M_2} \times ........ \times G{M_8} = {a^8}{r^{\dfrac{{8\left( {8 + 1} \right)}}{2}}} \\
\Rightarrow G{M_1} \times G{M_2} \times ........ \times G{M_8} = {a^8}{r^{36}} \\
\Rightarrow G{M_1} \times G{M_2} \times ........ \times G{M_8} = {a^8}{\left( {{r^9}} \right)^4} \\
\Rightarrow G{M_1} \times G{M_2} \times ........ \times G{M_8} = {\left( 2 \right)^8} \times {\left( {\dfrac{3}{2}} \right)^4} \\
\Rightarrow G{M_1} \times G{M_2} \times ........ \times G{M_8} = 4 \times 4 \times 4 \times 4 \times \dfrac{3}{2} \times \dfrac{3}{2} \times \dfrac{3}{2} \times \dfrac{3}{2} \\
\Rightarrow G{M_1} \times G{M_2} \times ........ \times G{M_8} = 1296 \\
\]
So, the product of 8GM’s is 1296.
Option D is correct.
Note: In geometric progression series, a is the first term and r is the common ratio between the numbers of the series. Don’t complicate the question by finding the values of each GM.
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