How many multiple

Multiples of 4 lying between 10 and 250 are 12, 16, 20, …, 248.

Sum of these numbers forms an arithmetic series 12 + 16 + 20 + … + 248.

Here, first term = a = 12

Common difference = d = 4

We first find the number of terms in the series.

Here, last term = l = 248

∴ 248 = a + (n – 1)d

⇒ 248 = 12 + (n – 1)4

⇒ 248 – 12 = 4n – 4

⇒ 236 = 4n - 4

⇒ 236 + 4 = 4n

⇒ 4n = 240

⇒ n = 60

Now, Sum of n terms of this arithmetic series is given by:

S_n = [2a + (n - 1)d]

Therefore sum of 60 terms of this arithmetic series is given by:

∴ S₆₀ = [2(12) + (60 - 1)(4)]

= 30 × [24 + 236]

= 30 × 260

= 7800