Question

$a,b,c$ are in A.P. and the points $A(a,1),B(b,2)$ and $C(c,3)$ are such that $(OA{)}^2,(OB{)}^2$ and $(OC{)}^2$ are also in A.P; $O$ being the origin, then

• A. $a^2+c^2=2b^2-2$
• B. $ac=b^2+1$
• C. $(a+c{)}^2=4b^2$
• D. $a+b+c=3b$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A $(a+c{)}^2=4b^2$
B $ac=b^2+1$
C $a^2+c^2=2b^2-2$
D $a+b+c=3b$

a,b,c are in A.P
then
$2b=a+c$
now
${OA}^2,{OB}^2,{OC}^2$ are also in A.P
$\therefore 2{OB}^2={OA}^2+{OC}^2$
Now using distance formula,
${OA}^2=a^2+1$
${OB}^2=b^2+4$
${OA}^2=c^2+9$
$2(b^2+4)=a^2+1+c^2+9$
On solving, we get
$2b^2=a^2+c^2+2$
$a^2+c^2=2b^2-2$---------(A)
Again
$2b^2=a^2+c^2+2$
$2b^2={(a+c)}^2-2ac+2$ (since $a+c=2b$)
$2b^2={\left(2b\right)}^2-2ac+2$
on solving we get
$ac=b^2+1$ ---- (B)
Now
$2b^2=a^2+c^2+2$
$2b^2={(a+c)}^2-2ac+2$
$2b^2={(a+c)}^2-2(b^2+1)+2$ by putting the value of ac obtained in B
on solving, we get
${(a+c)}^2=4b^2$ --- (C)
$2b=a+c$
adding b to both sides then we get
$3b=a+b+c$