Hint: In this question, we use the concept of Arithmetic progression. We have to make A.P series of multiple of 4 lies between given range and then find the value of the number of terms in A.P by using the formula of the ${n}^{th}$ term of an A.P. The Formula of ${n}^{th}$ term of an A.P is ${a_n} = a + \left( {n - 1} \right)d$ where a is first term, d is common difference and n is number of terms.
Complete step-by-step answer: Now, multiple of 4 lie between 10 and 250 are 12, 16, 20………..248. In the above series all are multiple of 4 so the difference between two consecutive numbers is 4. So, it’s confirmed that the above series is an Arithmetic progression with common difference d=4. Now, 12, 16, 20………..248 First term (a) =12, common difference (d)=4 and last term $(a_n)$=248 We have to find a number of terms so we use the formula of the ${n}^{th}$ term of an A.P. $ \Rightarrow {a_n} = a + \left( {n - 1} \right)d \\ \Rightarrow 248 = 12 + \left( {n - 1} \right) \times 4 \\ \Rightarrow 248 - 12 = 4 \times \left( {n - 1} \right) \\ \Rightarrow 236 = 4 \times \left( {n - 1} \right) \\ \Rightarrow \left( {n - 1} \right) = \dfrac{{236}}{4} \\ \Rightarrow n - 1 = 59 \\ \Rightarrow n = 59 + 1 \\ \Rightarrow n = 60 \\ $ So, the number of multiples of 4 lies between 10 and 250 is 60.
Note: In such types of questions most students face problems finding the first and last term of an A.P lie between given intervals. So we use simple tips, first divide the starting value of range by 4 and add remainder to the starting value of range and also divide last value of range by 4 and subtract remainder from last value of range. Example- We have to find the first and last term of an A.P. of multiples of 4 lie between 10 and 250. So, divide 10 by 4 and add remainder to 10 then we get first term a=12. And also divide 250 by 4 and subtract remainder from 250 then we get last term an=248.