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 $a=b=c$

Given $a,b,c$ are in A.P.
$\Rightarrow 2b=a+c\to \left(1\right)$
$\Rightarrow {\left(2b\right)}^2={(a+c)}^2$
$\Rightarrow {4b}^2=a^2+2ac+c^2\to \left(2\right)$
Also, given that $a,b,c$ are in G.P.
$\Rightarrow b^2=ac\to \left(3\right)$
Now, substitute $\left(3\right)$ in $\left(2\right)$ we get
$4ac=a^2+2ac+c^2$
$\Rightarrow a^2+c^2-2ac=0$
$\Rightarrow (a-c{)}^2=0$
$\Rightarrow (a-c)=0$
$\Rightarrow a=c$ $...\left(4\right)$
Now, substitute $\left(4\right)$ in $\left(1\right)$; we get
$2b=a+a=c+c$
$\Rightarrow 2b=2a=2c$
$\Rightarrow a=b=c$