Q. 604.8( 13 Votes )

How many multiple

Answer :

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:


Sn = [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

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