Question

find the no. of terms which are multiple of 4 between 10 to 250

Answer 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 that is 60....


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Answer 2

✌️✌️hey mate,


↪️60 is the answer↩️

•Here are two methods

➡️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

➡️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

thanks....

nice to help you✌️✌️