The number divisible by 4 in between 10 to 250 are
12, 16, 20, 24, ………… 248
From the above sequence, we have
tn = 248, a = 12
t1 = 16, t2 = 20
Thus, d = t2 – t1 = 20 – 16 = 4
Now, by using nth term of an A.P. formula tn = a + (n – 1) d
we can find value of “n”
Thus, on substituting all the value in formula we get,
248 = 12 + (n – 1) × 4
⇒ 248 – 12 = (n – 1) × 4
⇒ 236 = (n – 1) × 4
⇒ n – 1 = 236/3 = 59
⇒ n = 59 + 1 = 60