Question

please solve this
...

Answer 1

hey mate!!

I have two methods...


60 is the answer

Here are two methods...
hope you understand better...
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Method 1

The first number in the sequence will be 12 and last number will be 248

So 12,16,20…248

This is In the form 12, 12+4,12+ 2*4, 12+3*4…,12+n*4

By arithmetic progression concept

12+ n*4 = 248

n=( 248–12)/4

n=59

n+1 th term is 248

So there are n+1 terms i.e 60 multiples

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Method 2

A number is divisible by 4 if and only if its last two digits are divisble by 4

Starting from 1 and ending with 100 , 100/4 is 25 . So there exists 25 multiples of 4 in 100

But we have to exclude 4 and 8 as the series starts from 12. I.e 23

And in next 101 to 200 also there exists 25 as what bothers is last two digits.

And from 201 to 250 , we can consider them like from 1 to 50 and here we have floor of 50/4 i.e 12

So 23+25+12= 60

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.$\boxed{THANKS}$

Answer 2

hope it helps u ...
a =12 d=4 an=248 n =?
using
an=a+(n-1)d ..
find n