Write all consecutive even numbers between 51 to71.

Hint: For solving this question we should have knowledge of even numbers. Any integer (never a fraction) that can be divided exactly by 2 is known as an even number. We can also solve this problem using Arithmetic Progression.
Formula used:
Nth term of A.P is given by ${{a}_{n}}=a+\left( n-1 \right)d$ .
 This formula can be used to find the number of integers between two numbers.
$a$ First term, ${{a}_{n}}$ is the last term, common difference $n$ numbers of terms.

Complete step-by-step answer:
We have to write all even numbers between 51 to71
First even number will be 52 & last even number will be 70
52,54,56,58,60,62,64,66,68,70
So there are “10” even numbers.

Additional Information:
Even number is that which is divisible by 2
We can use formula: ${{a}_{n}}=a+\left( n-1 \right)d$ for checking the number of terms.
In this case ${{a}_{n}}=70$ (i.e. last term)
$a=52$ (I.e. first term)
$d=$ Common difference between any two consecutive terms
$\begin{align}
  & d={{a}_{2}}-a,={{a}_{3}}-{{a}_{2}}..........={{a}_{n}}-{{a}_{n-1}} \\
 & d=54-52=2 \\
\end{align}$
${{a}_{n}}=a+\left( n-1 \right)d$
Putting values in above formula we get
$70=52+\left( n-1 \right)2$ (We have to find $n$)
 $\begin{align}
  &\Rightarrow 70-52=\left( n-1 \right)2 \\
 & \Rightarrow 18=\left( n-1 \right)2 \\
 &\Rightarrow 18=\left( n-1 \right)2 \\
 &\Rightarrow 9=n-1 \\
 & \therefore 10=n \\
\end{align}$

Note:Simplest method is to write all the even numbers between 51& 71 to get the answer . But if the difference between numbers is more than we can solve this problem by using the A. P formula.