Question

Use an arithmetic sequence to find the number of multiples of 6 between 41 and 523. Justify your reasoning.

Answer 1

81

Explanation:

The numbers divisible by 6 between 41 and 523 are:

42,48,54,.....,522

We can clearly see that the first term that is a=42,

Common difference that is d=6

And
`a_n=522`

We will use the formula

`a_n=a+(n-1)d`

On substituting the values we get

`522=42+(n-1)6`

`\Rightarrow 522=42+6n-6`

`\Rightarrow 522=36+6n`

`\Rightarrow 486=6n`

`\Rightarrow 81=n`

Hence, there are 81 multiples of 6 between 41 and 523.

Answer 2

The number of multiples of 6 between 41 and 523 = 81

Explanation:

We need to find multiples of 6 between 41 and 523.

First multiple is 42.

That is first term is 42.

Common difference is 6.

The last term less than 523 and multiple of 6 is 522.

So we need to find number of terms in

42, 48, 54 , 60 , 66 .............................522

We have

a + (n-1) d = 522

42 + (n-1) 6 = 522

(n-1) 6 = 522 - 42 = 480

(n-1) = 80

n = 81

That is there are 81 terms in this arithmetic progression.

The number of multiples of 6 between 41 and 523 = 81