Question

If $a,b,c$ are in A.P., then $(a+2b-c)(2b+c-a)(c+a-b)$ equals
A. $\dfrac{1}{2}abc$
B. $abc$
C. $2abc$
D. $4abc$

Answer 1

Hint: In this question, we are to find the value of the given expression. Here the given expression has three variables that are in arithmetic progression. So, we can evaluate their arithmetic mean and apply it to the given expression, we get the required value.

Formula used: If a series or a sequence is in arithmetic progression, then the $n^{th}$ term of the series is
${{t}_{n}}=a+(n-1)d$
Where the common ratio $d={{a}_{n}}-{{a}_{n-1}}$
Here $a$ - First term; $n$ - Number of terms in the series; ${{t}_{n}}$ - Required $n^{th}$ term.
Then, the sum of the series which is the arithmetic progression is
${{S}_{n}}=\dfrac{n}{2}\left[ 2a+(n-1)d \right]$
The arithmetic mean of the series is
$\begin{align}
  & AM=b=\dfrac{a+c}{2} \\
 & \Rightarrow 2b=a+c \\
\end{align}$

Complete step by step solution: It is given that, $a,b,c$ are in arithmetic progression.
So, we can write their arithmetic mean as the average of the first and last terms which is equal to the middle term. I.e.,
$2b=a+c\text{ }...(1)$
The given expression which is to be evaluated is
$(a+2b-c)(2b+c-a)(c+a-b)$
Here in this expression, in place of $2b$ we can substitute $a+c$, since we know that from (1).
Thus, on substituting, we get
$\begin{align}
  & (a+2b-c)(2b+c-a)(c+a-b) \\
 & =(a+(a+c)-c)((a+c)+c-a)((a+c)-b) \\
 & =(a+a+c-c)(a+c+c-a)(2b-b) \\
 & =2a\cdot 2c\cdot b \\
 & \Rightarrow (a+2b-c)(2b+c-a)(c+a-b)=4abc \\
\end{align}$
Therefore, the obtained value for the given expression is $4abc$.

Thus, Option (D) is correct.

Note: Here we need to find the respective mean of the given series to find such a type of question. So, we can easily evaluate the required expression by substituting the mean. It is necessary to remember that, first we have to know the type of series given over there.