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Answer :
First multiple of 4 which lie between 10 and 250 is 12.
The last multiple of 4 which lie between 10 and 250 is 248.
Therefore, AP is of the form 12, 16, 20… ,248
First term = a = 12,
Common difference = d = 4
Using formula \(a_n\) = a + (n – 1) d, to find \(n^{th}\) term of arithmetic progression,
\(\Rightarrow \) 248 = 12 + (n - 1) (4)
\(\Rightarrow \) 248 = 12 + 4n - 4
\(\Rightarrow \) 240 = 4n
\(\Rightarrow \) n = 60
It means that 248 is the \(60^{th}\) term of AP.
So, we can say that there are 60 multiples of 4 which lie between 10 and 250.