Question

If $x,1,z$ are in A.P and $x,2,z$ are in G.P the $x,4,z$ will be in:

• A. A.P
• B. G.P
• C. H.P
• D. None of these

Answer 1

The correct option is C

H.P

Explanation for the correct option.

As $x,1,z$ are in A.P, so

$\begin{array}{rcl}1-x& =& z-1\\ & \Rightarrow & z+x=2\end{array}$

As $x,2,z$ are in G.P, so

$\begin{array}{rcl}\frac{2}{x}& =& \frac{z}{2}\\ & \Rightarrow & 4=xz\end{array}$

If $x,4,z$ are in H.P, then

$\begin{array}{rcl}\frac{1}{4}-\frac{1}{x}& =& \frac{1}{z}-\frac{1}{4}\\ \Rightarrow \frac{2}{4}& =& \frac{z+x}{xz}\\ \Rightarrow \frac{2}{4}& =& \frac{2}{4}\end{array}$

So, $x,4,z$ are in H.P.

Hence, option C is correct.

Explanation for the incorrect options.

Option A

If $x,4,z$ are in A.P, then

$\begin{array}{rcl}4-x& =& z-4\\ \Rightarrow 8& =& z+x\\ \Rightarrow 8& \ne & 2\end{array}$

Hence, option A is incorrect.

Option B

If $x,4,z$ are in G.P, then

$\begin{array}{rcl}\frac{4}{x}& =& \frac{z}{4}\\ \Rightarrow 16& =& zx\\ \Rightarrow 16& \ne & 4\end{array}$

Hence, option B is incorrect.

Hence, option C is the correct answer.