How many 2-digit numbers are divisible by 4?

Hint: Find the first two-digit number that is divisible by 4, it will be the first term of the progression. Now find the last 2-digit number that is divisible by 4, it will be the last term of the progression. The difference between the two terms will be the common difference of the progression. Now, apply the formula ${a_n} = {a_1} + \left( {n - 1} \right)d$ to get the total 2-digit numbers which are divisible by 4.

Complete step-by-step answer:
The first two-digit number that is divisible by 4 is 12. So, the first term of the AP is 12. Then,
${a_1} = 12$
The last two-digit number that is divisible by 4 is 96. So, the last term of the AP is 96. Then,
${a_n} = 96$
The common difference is $d = 4$.
As we know that the general term of the Arithmetic Progression is,
${a_n} = {a_1} + \left( {n - 1} \right)d$
where, ${a_n}$ is the last term.
${a_1}$ is the first term.
n is the number of terms.
d is a common difference.
Now, put ${a_1} = 12,d = 4$ and ${a_n} = 96$ in the above formula. Then,
$96 = 12 + \left( {n - 1} \right) \times 4$
Move 12 on the other side and subtract it from 96.
$4\left( {n - 1} \right) = 84$
Divide both sides by 4,
$\dfrac{{4\left( {n - 1} \right)}}{4} = \dfrac{{84}}{4}$
Cancel out the common factors from numerator and denominator, we get
$n - 1 = 21$
Now, move 1 to other sides and add the terms to get the final answer,
$n = 22$

Thus, there are 22 2-digit numbers that are divisible by 4.

Note: This question can be done in another way also.
The largest two-digit number which is multiple of 4 is 96.
Divide 96 by 4.
$\dfrac{{96}}{4} = 24$
Now, count the number of multiples of 4 which are not two-digit. The numbers are 4 and 8.
So, subtract 2 from 24,
$24 - 2 = 22$
Hence, there are 22 2-digit numbers that are divisible by 4.