# Sequence of numbers divisible by 7.

• blunted
In summary, to find the first three and last three numbers in a sequence of 3-digit numbers divisible by 7, you can use the formula an = a0 + (n-1)d where a0 is the first term, an-1 is the last term, d is the common difference, and n is the number of terms. To find the number of terms, you can use the formula (an-1 - a0) / d + 1. Lastly, to find the sum of all the numbers in the sequence, you can use the formula n(a0 + an-1)/2, where n is the number of terms, a0 is the first term, and an-1 is the last term.
blunted

## Homework Statement

A sequence of 3-digit numbers divisible by 7 is given. Find:
a) first three and last three numbers.
b) how many numbers are there in that sequence.
c) sum of all the numbers in the sequence.

## The Attempt at a Solution

Cause they are 3 digit numbers that means: 100-999.
In order for a number to be divisible by 7 you need to get the last digit, double it and subtract it from the number without the last digit. If that number is divisible by 7, you stop there, if not, repeat the steps. I tried this way, but it got me nowhere. So it came to my mind that 7,14,21,28 are all divisible by 7 cause the distance between those numbers is 7. So with a bit of trial and error I found that 105 is divisible by 7. So the first three numbers of the sequence are 105, 112, 119. The last two are 994, 987, 980.
Now how do I find how many numbers between 100 and 999 are divisible by 7?
And for the sum I know the formula. Sn = ((a1 + an) / 2) * n

blunted said:

## Homework Statement

A sequence of 3-digit numbers divisible by 7 is given. Find:
a) first three and last three numbers.
b) how many numbers are there in that sequence.
c) sum of all the numbers in the sequence.

## The Attempt at a Solution

Cause they are 3 digit numbers that means: 100-999.
In order for a number to be divisible by 7 you need to get the last digit, double it and subtract it from the number without the last digit. If that number is divisible by 7, you stop there, if not, repeat the steps. I tried this way, but it got me nowhere. So it came to my mind that 7,14,21,28 are all divisible by 7 cause the distance between those numbers is 7. So with a bit of trial and error I found that 105 is divisible by 7. So the first three numbers of the sequence are 105, 112, 119. The last two are 994, 987, 980.
Now how do I find how many numbers between 100 and 999 are divisible by 7?
And for the sum I know the formula. Sn = ((a1 + an) / 2) * n

You can write the numbers as {15*7, 16*7, 17*7, ..., 142*7}. Writing them this way will make it easier to count them.

If the sequence consisted of multiples of, say, 5, like this: {15, 20, 25, 30, ..., 95} = {3*5, 4*5, 5*5, 6*5, ..., 19*5}, there are (19 - 3) + 1 = 17 elements in the set. It's important to add 1 so as to count the first number.

I shoud add that in my example, if you don't want to write the numbers as multiples of 5, then you need to divide by 5.

(95 - 15)/5 + 1 = 80/5 + 1 = 16 + 1 = 17, same as before.

You have found the first term is 105, the next is 105+7, the next 105+2*7, the next 105+3*7

You have an arithemtic sequence. Where a0 is first term, an the nth term, an-1 last term, n the number of terms and d the common difference

a0=105
an-1 = 994
d=7

an = a0 +(n-1)d use this to find n
sum of all terms is n(a0 +an-1)/2

## 1. How do you determine a sequence of numbers divisible by 7?

To determine a sequence of numbers divisible by 7, you can start with the number 7 and continue adding 7 to each subsequent number. For example, 7, 14, 21, 28, and so on.

## 2. What is the pattern of numbers divisible by 7?

The pattern of numbers divisible by 7 is that they are all multiples of 7. This means they can be divided evenly by 7 with no remainder.

## 3. How many numbers are in a sequence of numbers divisible by 7?

The number of numbers in a sequence of numbers divisible by 7 is infinite, as there is no limit to how high the numbers can go.

## 4. Can a number be divisible by 7 and another number at the same time?

Yes, a number can be divisible by 7 and another number at the same time. For example, 14 is divisible by both 7 and 2.

## 5. What is the smallest number in a sequence of numbers divisible by 7?

The smallest number in a sequence of numbers divisible by 7 is 7, as it is the starting number for the sequence.

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