Question

If $In(a+c),In(c-a),In(a-2b+c)$are in A.P., then

• A. $a,b,c$ are in A.P
• B. $a^2,b^2,c^2$ are in A.P
• C. $a,b,c$ are in G.P
• D. $a,b,c$ are in H.P

Answer 1

The correct option is D

$a,b,c$ are in H.P

Explanation for correct option:

Find the relation:

$\text{In(a+c), In(c–a), In(a–2b+c)}$ are in A.P.

So,

$$\begin{gathered} \text{2In(c–a)=In(a+c)+In(a–2b+c)}\mathbf{[}\mathbf{\because }\mathbf{2}\mathbf{b}\mathbf{=}\mathbf{a}\mathbf{+}\mathbf{c}\mathbf{]} \\ \Rightarrow {(c–a)}^2=(a+c)(a–2b+c)\mathbf{[}\mathbf{\because }\mathbf{}\mathbf{n}\mathbf{l}\mathbf{o}\mathbf{g}\mathbf{a}\mathbf{=}\mathbf{l}\mathbf{o}\mathbf{g}{\mathbf{a}}^{\mathbf{n}}\mathbf{,}\mathbf{}\mathbf{l}\mathbf{o}\mathbf{g}\mathbf{a}\mathbf{+}\mathbf{l}\mathbf{o}\mathbf{g}\mathbf{b}\mathbf{=}\mathbf{l}\mathbf{o}\mathbf{g}\mathbf{a}\mathbf{b}\mathbf{]} \\ \Rightarrow c^2–2ac+a^2=a^2+ac–2ab–2bc+ac+c^2 \\ \Rightarrow -2ac=-2ab+2ac–2bc \\ \Rightarrow 2ac=ab+bc \\ \Rightarrow 2ac=b(a+c) \\ \Rightarrow \frac{2}{b}=\frac{(a+c)}{ac} \end{gathered}$$

So $a,b,c$ are in HP.

Hence, the correct option is D.