Question

Find the value of x for which $\left( 8x+4 \right),\left( 6x-2 \right)$ and $\left( 2x+7 \right)$ are in A.P.

Answer 1

Hint: For solving this question, we should know about the Arithmetic Progression (A.P). The A.P is a sequence of numbers in order, in which the difference of any two consecutive numbers is a constant value. It means the difference between them is always a constant and this is showing that all the numbers have a common difference.

Complete step-by-step solution:
In the given question it is shown that the first term is $\left( 8x+4 \right)$ and the second term is $\left( 6x-2 \right)$ and the third term is $\left( 2x+7 \right)$. We have to find the value of $x$ for which these are in arithmetic progression. By the definition of arithmetic progression, it is clear that the second element is equal to the half or addition of first and third values. So,
$b=\dfrac{a+c}{2}\ldots \ldots \ldots \left( 1 \right)$
Here,
$\begin{align}
  & a=\left( 8x+4 \right) \\
 & b=\left( 6x-2 \right) \\
 & c=\left( 2x+7 \right) \\
\end{align}$
According to the question, a, b, c should be in A.P. For them to be in A.P, the value of $x$ has to be found. So, by using equation number (1) and substituting the values of a, b and c in it we have,
$\begin{align}
  & b=\dfrac{a+c}{2} \\
 & \Rightarrow 6x-2=\dfrac{\left( 8x+4 \right)+\left( 2x+7 \right)}{2} \\
 & \Rightarrow 2\left( 6x-2 \right)=\left( 8x+4 \right)+\left( 2x+7 \right) \\
\end{align}$
By opening the brackets to solve this, we get,
$12x-4=10x+11$
By taking the $x$ terms to one side and the numbers to the other side, we get,
$\begin{align}
  & 12x-10x=11+4 \\
 & \Rightarrow 2x=15 \\
 & \Rightarrow x=\dfrac{15}{2}=7.5 \\
\end{align}$
So, the value of $x$ is 7.5 for which $\left( 8x+4 \right),\left( 6x-2 \right)$ and $\left( 2x+7 \right)$ are in A.P.

Note: The arithmetic progression properties are useful in these questions. We should remember that in any term of arithmetic progression besides the first term, all the other terms are exactly half of the addition of both values to the right and left side of the given value, so that they have a common difference.