Question

If a, b and c are in A.P, then which of the following is not true?
A. $ \dfrac{k}{a},\dfrac{k}{b},\dfrac{k}{c} $ are in H.P.
B. $ a+k,b+k,c+k $ are in A.P.
C. $ ka,kb,kc $ are in A.P.
D. $ {{a}^{2}},{{b}^{2}},{{c}^{2}} $ are in A.P.

Answer 1

Hint: We first use the relation between A.P. terms where any binary operation of the same number with the A.P. numbers won’t change the conditions of the A.P. We get the relation of $ a+c=2b $ and using that we find the condition which is not perfect for being in A.P.

Complete step-by-step answer:
We can also use the simple conditions of A.P. to find the right relation.
We know that any binary operation of the same number with the A.P. numbers won’t change the conditions of the A.P.
For our given terms a, b and c are in A.P. Therefore, $ a+c=2b $ .
We add $ k $ to all of them and they still remain in A.P.
So, $ a+k,b+k,c+k $ are in A.P.
Now we multiply $ k $ to all of them and they still remain in A.P.
So, $ ka,kb,kc $ are in A.P.
Now we divide by $ k $ to all of them and they still remain in A.P.
So, $ \dfrac{a}{k},\dfrac{b}{k},\dfrac{c}{k} $ are in A.P.
We know that if certain numbers are in A.P. then their reciprocals are in H.P.
So, $ \dfrac{k}{a},\dfrac{k}{b},\dfrac{k}{c} $ are in H.P.
But $ {{a}^{2}},{{b}^{2}},{{c}^{2}} $ are not in A.P as if they are then $ {{a}^{2}}+{{c}^{2}}=2{{b}^{2}} $ which cannot be proved from $ a+c=2b $ . Squaring we get
 \[\begin{align}
  & {{\left( a+c \right)}^{2}}=4{{b}^{2}} \\
 & \Rightarrow {{a}^{2}}+{{c}^{2}}+2ac=4{{b}^{2}} \\
\end{align}\] .
Therefore, option D is not true.
So, the correct answer is “Option D”.

Note: For the condition $ {{a}^{2}}+{{c}^{2}}=2{{b}^{2}} $ to satisfy, we need to have the condition of
 \[\begin{align}
  & {{a}^{2}}+{{c}^{2}}+2ac=4{{b}^{2}} \\
 & \Rightarrow ac={{b}^{2}} \\
\end{align}\]
Therefore, the terms have to be in G.P.