Question

If $a,b,c$are in $A.P$ as well as in $G.P.$ then

• A. $a=b\ne c$
• B. $a\ne b=c$
• C. $a\ne b\ne c$
• D. $a=b=c$

Answer 1

The correct option is D

$a=b=c$

Explanation for the correct option:

Finding the value:

Since, $a,b,c$ are in Arithmetic Progression.

Therefore, $b=\frac{a+c}{2}$

Squaring, $b^2=\frac{{(a+c)}^2}{4}...\left(i\right)$

Also, $a,b,c$ are in Geometric Progression

Therefore, $b^2=ac....\left(ii\right)$

Equating $\left(i\right)\&\left(ii\right)$

$\begin{array}{rcl}\frac{{(a+c)}^2}{4}& =& ac\\ \Rightarrow a^2+c^2+2ac& =& 4ac\\ & \Rightarrow & (a^2+c^2-2ac)=0\\ \Rightarrow (a-c)& =& 0\\ \Rightarrow a& =& c\end{array}$

Put $a$in $\left(ii\right)$, $b=c$

Therefore, $a=b=c$

Hence, correct option is $\mathbf{\left(}\mathit{D}\mathbf{\right)}$.