Sequence of numbers divisible by 7.

[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.

Homework Equations

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

 

[Mark44]

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.

Homework Equations

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.

 

[Mark44]

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.

 

[jing2178]

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, a_n the n^th term, a_n-1 last term, n the number of terms and d the common difference

a₀=105
a_n-1 = 994
d=7

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