Question

If 1, x, y, z, 2 are in Geometric progression, the xyz=
A) $2\sqrt 2 $
B) 4
C) 8
D) None of this

Answer 1

Hint:
Geometric progression has the sequence $a,\,ar,\,a{r^2},..........,a{r^n}$ compare the values with the sequence and find the value of xyz. Then on solving the equation for a constant value, we’ll get our required result.

Complete step by step solution:
Given that 1, x, y, z, 2 are in Geometric progression, and the sequence of geometric progression is $a,\,ar,\,a{r^2},..........,a{r^n}$.
Where, a= first term and r= common ratio.
Now compare the values with the series, where
$
   \Rightarrow a = 1......\left( 1 \right) \\
  \, \Rightarrow x = ar........\left( 2 \right) \\
   \Rightarrow y = a{r^2}.........\left( 3 \right) \\
   \Rightarrow z = a{r^3}..........\left( 4 \right) \\
   \Rightarrow 2 = a{r^4}...........\left( 5 \right) \\
$
Substitute the value of a=1 in equation (5)
$
   \Rightarrow 2 = a{r^4} \\
   \Rightarrow a{r^4} = 2 \\
   \Rightarrow \left( 1 \right){r^4} = 2 \\
   \Rightarrow {r^4} = 2..........\left( 6 \right) \\
   \Rightarrow {r^2} = \sqrt 2 .........\left( 7 \right) \\
 $
For finding the value of xyz, multiply (2),(3) and (4)
$
   \Rightarrow xyz = \left( {ar} \right)\left( {a{r^2}} \right)\left( {a{r^3}} \right) \\
   \Rightarrow xyz = {a^3}{r^6} \\
   \Rightarrow xyz = {a^3}\left( {{r^2} \times {r^4}} \right) \\
 $
Where, ${r^4} = 2$, a=1, and ${r^2} = \sqrt 2 $
Substitute the values in the above equation
$
   \Rightarrow xyz = {a^3}\left( {{r^2} \times {r^4}} \right) \\
   \Rightarrow xyz = 1\left( {2\sqrt 2 } \right) \\
   \Rightarrow xyz = 2\sqrt 2 \\
 $

So, the value of $xyz = 2\sqrt 2 $.

Note:
Terms are in Geometric progression only when the ratio of any two adjacent values in the sequence is the same throughout the series. Whenever we need to choose three terms in GP we’ll always choose $\dfrac{a}{r}, a, ar$.