# Thread: Arithmetic sequences, please please look i really need help

1. ## Arithmetic sequences, please please look i really need help

i have the question

'find the sum of the artithmetic pregession 1, 4, 7, 10, ........ 1000'

and then 'every third term of the progression is removed i.e 7 and 16. the the sum of the remaining terms'

i know the formulas for the nth term and the the sum of the terms, but i cant find a way of finding out the number of terms in the sequence which i thnk will then allow me to simply put the number in the formula.

but then again i have no idea how to even start the 2nd part.

please help ive been doing this for hours and theres still another 2 questions i cant do, thanks

2. Originally Posted by alyson
i have the question

'find the sum of the artithmetic pregession 1, 4, 7, 10, ........ 1000'

and then 'every third term of the progression is removed i.e 7 and 16. the the sum of the remaining terms'

i know the formulas for the nth term and the the sum of the terms, but i cant find a way of finding out the number of terms in the sequence which i thnk will then allow me to simply put the number in the formula.

but then again i have no idea how to even start the 2nd part.

please help ive been doing this for hours and theres still another 2 questions i cant do, thanks

Part 1 : Use the formula .

$S_n=\frac{n}{2}(a+l)$

Find n :

$T_n=1000$ , $a+(n-1)d=1000$ . Thus , n=334 .

Part B : Find :

$\sum^{111}_{r=1}(9r-2)$ .

After that , use the total sum minus the sum of all the third terms .

3. An aritthmetic sequence has the nice property that the average of all the numbers is just the average of the first and last numbers. Here, the first and last numbers are 1 and 1000. 1001/2= 500.5. Because that is the average of all the numbers, their total is n(500.5) where n is the number of numbers in the sequence. If we start "labeling" the sequence at n=0, we have $a_n= 1+ 3n$ so if $1+ 3n= 1000$, $3n= 999$ so n= 333. Since we started numbering with n= 0, there are 333+1= 334 terms in the sequence.

The simplest way to find the sum with "every third number removed" is to find the sum of the numbers removed and subtract that off.

Here the numbers are 1, 4, 7, 10, 13, 16, 19, 22, 25 ... with each number 3 more than the previous. "Every third number" is 7, 16, 25 ..." with each number 3*3= 9 more than than the previous number. Again, the nth term is 7+ 3n where 7 is the 0th term. That is important in determining the "last" term. Working backwards, the last terms are 1000, 997, 994, 991, etc. 7+3n= 1000 leads to 3n= 997 so n= 332 1/3, not a whole number. 7+3n= 997 leads to 3n= 990 so n= 330 which is. 997 is the last of the "every third numbers" so the sequence is 7, 16, 25, ... 997. The first and last numbers are 7 and 997 so the average is (7+ 997)/2= 1004/2= 502. Every number in the sequence of "every third numbers" is of the form 7+ 9n. 7+ 9n= 997 gives n= 110 but since we are numbering from 0, there are 110+1= 111 such numbers. The total of all "every third number" is 110(502). Subtract that from the answer you got before.