aritmetic series question

1. find the sum of all the integers between 1 and 1000 which are divisible by 7


okay so this means a=7 and d=7 ?

then we find the sum by doing 1000/2[2(7) + (999)(7)] ?

well this is not giving me the correct answer...

ALSOO... How do I show that the sequence 2+5+...+47 has exactly 16 positive terms?

thanks

1) There aren't 1000 terms in the series - you need to calculate the number of terms there are

2) Calculate the number of terms there are and show it's 16

how do you calculate the number of terms?

Using Un=a + (n-1)d ?

$1 \equiv \frac{7}{7}, 2 \equiv \frac{14}{7}, 3 \equiv \frac{21}{7} ...$ Notice a pattern?

yes, the numerator goes up in multiples of 7.

But what do we do with this? how do we find out how many terms there are from this?

Well yes but look at the quotient...

because your sum is effectively $7 + 14 + 21 + ...$

Sorry but I am really lost

How do we then figure out the number of terms in the series from this?

Right, if your sum was just up to 7, then your only term in the series would be 7. If it was something up to 15, it would be 7 + 14. If it was up to 25, then you would have 7 + 14 + 21. So, basically you want to find out the number closest to and less than 1000 divisible by 7 and divide it by 7 - that'll give you your number of terms, understand why?

soo... it would be 994? I think I get it now. Thanks so much dude

Yep

I do 994/2[2(7) + (993)(7)] but I am still not getting the correct answer

is this the correct calculation?

There aren't 994 terms either! :P

Think about it...if you were summing up the multiples of 7 from 1 to 10, you would have just 7 in your series and you'd have $\frac{7}{7} = 1$ term. If it was up to 15, you'd have 7 + 14 and you'd have $\frac{14}{7} = 2$ terms. Up to 25 $\Rightarrow \frac{21}{7} = 3$ terms. So if it's up to 1000 and your last term is 994, how many terms?