Question

If $a,b,c\in R$ and $a,b,c$ are in AP the match the entries from column-I and column-II \[\]
Column-I\[\]
A.${{a}^{2}},{{b}^{2}},{{c}^{2}}$ are in AP \[\]
B. ${{a}^{2}},{{b}^{2}},{{c}^{2}}$ are in GP\[\]
C. ${{a}^{2}},{{b}^{2}},{{c}^{2}}$ are in HP. \[\]
D. $a+b+c=\dfrac{3}{2}$\[\]
Column-II\[\]
1.$a=b=c$\[\]
2. $\dfrac{-1}{2}a,b,c$ are in GP\[\]
3. $a,b,\dfrac{-1}{2}c$ are in GP.\[\]
4. $b=\dfrac{1}{2}$\[\]

Answer 1

Hint: We know that in a sequence say $\left( {{x}_{n}} \right)$ any three terms ${{x}_{1}},{{x}_{2}},{{x}_{3}}$ of the sequence are related by the equation ${{x}_{2}}-{{x}_{1}}={{x}_{3}}-{{x}_{2}}$ if the sequence is in arithmetic progression or AP, $\dfrac{{{x}_{1}}}{{{x}_{2}}}=\dfrac{{{x}_{2}}}{{{x}_{1}}}\Rightarrow {{x}_{2}}^{2}={{x}_{1}}{{x}_{3}}$ if the sequence is geometric progression or GP and $\dfrac{1}{{{x}_{2}}}-\dfrac{1}{{{x}_{1}}}=\dfrac{1}{{{x}_{3}}}-\dfrac{1}{{{x}_{2}}}$ if the sequence is harmonic progression or HP. We use these relations take the statements in column-I as premise and try to reach the conclusions in column-II.\[\]

Complete step by step answer:
We are given in the question that numbers $a,b,c$ are in AP. So we have
\[\begin{align}
  & b-a=c-b...(1) \\
 & \Rightarrow 2b=a+c...(2) \\
 & \Rightarrow b=\dfrac{a+c}{2}...(3) \\
\end{align}\]

Let us take the statements in the column as premises and with help of the information that $a,b,c$ are in AP we try to reach the conclusions in column-II. So the first statement is \[\]
A.${{a}^{2}},{{b}^{2}},{{c}^{2}}$ are in AP. So we have
\[\begin{align}
  & {{b}^{2}}-{{a}^{2}}={{c}^{2}}-{{b}^{2}} \\
 & \Rightarrow \left( b-a \right)\left( b+a \right)=\left( c-b \right)\left( c+b \right) \\
\end{align}\]
We have $b-a=c-b$ as $a,b,c$ are assumed to be distinct numbers in AP. We use it and get get ,
\[\begin{align}
  & b+a=c+b \\
 & \Rightarrow a=c \\
\end{align}\]
So we have if $a=0$ we have $b=\dfrac{a+a}{2}=a$ and $b=\dfrac{c+c}{2}=c$ . So we have $a=b=c$ which is the option 1 of the column-II. So $A\to 1$.\[\]
B. It is given that ${{a}^{2}},{{b}^{2}},{{c}^{2}}$ are in GP, so we have
\[\begin{align}
  & {{\left( {{b}^{2}} \right)}^{2}}={{a}^{2}}{{c}^{2}} \\
 & \Rightarrow {{b}^{4}}={{\left( ac \right)}^{2}} \\
\end{align}\]
We take the positive square root both side to get and put the value $a+c=2b $ obtained in equation (2).We have
\[\begin{align}
  & {{b}^{2}}=ac \\
 & \Rightarrow {{\left( \dfrac{a+c}{2} \right)}^{2}}=ac \\
 & \Rightarrow {{\left( a+c \right)}^{2}}=4ac \\
 & \Rightarrow {{\left( a-c \right)}^{2}}=0 \\
\end{align}\]
So we have $a=c=b$ just like previously obtained in option A. If we take the negative square root , we have ${{b}^{2}}=-ac=\left( -a \right)c$ which means $-a,b,c$ are in GP which is in the option B. SO we have $B\to 1,B\to 2$\[\]


C. It is given that ${{a}^{2}},{{b}^{2}},{{c}^{2}}$ are in HP. We here have
\[\begin{align}
  & \dfrac{1}{{{b}^{2}}}-\dfrac{1}{{{a}^{2}}}=\dfrac{1}{{{c}^{2}}}-\dfrac{1}{{{b}^{2}}} \\
 & \Rightarrow \dfrac{\left( a-b \right)\left( b+a \right)}{{{a}^{2}}{{b}^{2}}}=\dfrac{\left( b-c \right)\left( b+c \right)}{{{b}^{2}}{{c}^{2}}} \\
 & \Rightarrow \dfrac{\left( b-a \right)\left( b+a \right)}{{{a}^{2}}}=\dfrac{\left( c-b \right)\left( b+c \right)}{{{c}^{2}}} \\
\end{align}\]
We use equation (1) to cancel the terms $b-a,c-b$ in the numerator of both side of the equation. We then cross-multiply to get,
\[\begin{align}
  & \dfrac{a+b}{{{a}^{2}}}=\dfrac{b+c}{{{c}^{2}}} \\
 & \Rightarrow a{{c}^{2}}+b{{c}^{2}}={{a}^{2}}b+{{a}^{2}}c \\
 & \Rightarrow ac\left( c-a \right)+b\left( c-a \right)\left( c+a \right)=0 \\
 & \Rightarrow \left( c-a \right)\left( ab+bc+ca \right)=0 \\
\end{align}\]
If we take $c-a=0$ then $c=a=b$.If we take $ab+bc+ca=0$ and put the value $a+c=2b$ obtained in equation (2), we get
\[\begin{align}
  & ab+bc+ca= \\
 & \Rightarrow \left( a+c \right)b=-ca \\
 & \Rightarrow 2b\cdot b=-ca \\
 & \Rightarrow {{b}^{2}}=\dfrac{-ca}{2}=a\left( \dfrac{-c}{2} \right) \\
\end{align}\]
The above obtained relation implies that $a,b,\dfrac{-c}{2}$ are in GP. The statement we concluded now is in option 3 of column-II. So $C\to 1,C\to 3.$\[\]
D. It is given that $a+b+c=\dfrac{3}{2}$. We put $a+c=2b$ obtained in equation(2) in the given equation and have,
\[\begin{align}
  & a+b+c=\dfrac{3}{2} \\
 & \Rightarrow 2b+b=\dfrac{3}{2} \\
 & \Rightarrow 3b=\dfrac{3}{2} \\
 & \Rightarrow b=\dfrac{1}{2} \\
\end{align}\]
The obtained value matches with statement-4 of the column-II. So we have $D\to 4$.\[\]

Note: We need to be careful in the exam as some options from column-I have more than one option to match in column-B. The mean of the sequence is the quotient of some of the terms when a divided number of terms. The arithmetic mean AM, geometric mean GM, harmonic mean HM are related as $\text{AM}\times \text{GM}=\text{H}{{\text{M}}^{2}}$.