Insert three arithmetic means between $8$ and $24$ .

Hint: Here we have to insert three arithmetic means between the two numbers given so we will use the arithmetic progression concept. Firstly we will write the general form of expressing an arithmetic progression. Then we will consider $8$ as our first term and $24$ as our ${{5}^{th}}$ term. Finally we will get the value of common difference from the formula and using it we will get our three terms.

Complete step-by-step answer:
We have to insert three arithmetic means between $8$ and $24$.
Let our arithmetic series is as follows:
$a,a+d,a+2d,a+3d,a+4d$
From the question we know that the first term is $8$ and last term is $24$ so,
$a=8$ ….$\left( 1 \right)$
$a+4d=24$….$\left( 2 \right)$
So our series is as follows:
$8,a+d,a+2d,a+3d,24$…$\left( 3 \right)$
On putting value from equation (1) to equation (2) we get,
$\Rightarrow 8+4d=24$
$\Rightarrow 4d=24-8$
Simplifying further we get,
$\Rightarrow 4d=16$
Dividing both sides by $4$ we get,
$\Rightarrow d=4$
Now substitute $a=8$ and $d=4$ in equation (3) as follows:
$8,8+4,8+2\times 4,8+3\times 4,24$
On simplifying each terms further we get,
$\Rightarrow 8,12,16,20,24$
So we get the series as $8,12,16,20,24$ .
Hence three arithmetic means between $8$ and $24$ is $12,16,20$ .
So, the correct answer is “ $12,16,20$”.

Note: Arithmetic progression is a sequence of numbers having common difference in consecutive terms. Its ${{n}^{th}}$ term formula is given as ${{a}_{n}}={{a}_{1}}+\left( n-1 \right)d$ . We can also solve this question by using arithmetic progression direct formula given as ${{a}_{n}}={{a}_{1}}+\left( n-1 \right)d$ where ${{a}_{1}}$ is the first term, $d$ is the common difference and $n$ is the position of the term. So in this question we will have:
${{a}_{1}}=8$
$n=5$
${{a}_{5}}=24$
Replace the above value in the formula ${{a}_{n}}={{a}_{1}}+\left( n-1 \right)d$ as follows:
 $\Rightarrow 24=8+\left( 5-1 \right)d$
$\Rightarrow 24-8=4d$
Simplifying further we get,
$\Rightarrow 16=4d$
Dividing both side by $4$ we get,
$\Rightarrow d=4$
As we are getting the same common difference, we will get the same answer.