Number sequences

bumblebee123

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I have a question that I haven't been able to figure out for ages. Can anyone help?

Question: The 2nd term of an arithmetic sequence is 80
The 7th term of this arithmetic sequence is 20

Find the common difference of the series, then find the sum of all the positive terms of this arithmetic sequence

by using simultaneous equations I found the common difference ( d ) = -12 and the first term ( a ) = 92

I'm just stuck on the second part of the question- what do I do? any help would be appreciated :)
 
a1 = a1
a2 = a1 + d
a3 = a1 + 2d
a4 = a1 + 3d
a5 = a1 + 4d
...
an = a1 + (n-1)d

What do you think Sn, the sum of the 1st n terms, would be?
 
a1 = a1
a2 = a1 + d
a3 = a1 + 2d
a4 = a1 + 3d
a5 = a1 + 4d
...
an = a1 + (n-1)d

What do you think Sn, the sum of the 1st n terms, would be?

I know there's an equation for the sum of the first n terms of an arithmetic series: Sn = n/2 ( 2a + ( n - 1 )d )
 
how am I supposed to figure anything out when I don't know how many terms there are in the series?
 
We do know how many terms there are in the series. There are an infinite amount. But why do you even need to know that? You have to add all the positive terms. Suppose for example that only the 1st 10 terms were positive. So you need to add those. Now how will the sum of the 1st 10 terms change if there were just 20 terms or 55 terms or an infinite number of terms?
 
when Sn = -1 ?
Sn is the sum of the 1st n terms. What is the exact value of n?? Since all the terms you want to add are positive I highly doubt that the sum will be -1.

How many times can you add d to a before the term is not positive?
92, 80, 68, 56, 44, 32, 20, 8, -4
So which value of n do you plug into your formula for Sn
 
sorry, I'm really bad at this. So as 92 is the first term and 80 is the second term, I'd keep on adding the common difference ( -12 ) until I got a negative number? and then I'd count which value of n it was? -4 looks like the 9th term, but would I have to put it into a formula to confirm it's definitely the 9th term?

If I can just say it's the 9th term with that evidence, would I then plug 8 into the formula for the sum of all positive integers? ( this does give me the correct answer )
 
a = 1st term, d = common difference, n = number of terms
Formulas:
SUM of first n terms: n[2a + d(n - 1)] / 2
Nth term: a + d(n - 1)

Your case:
a = 92, d = -12, n = ?

Sum positive terms = n[2a + d(n - 1)] / 2
where a + d(n - 1) >= 0 AND a + d(n - 1) + d < 0

Don't think it can be done in a simpler/shorter way...
 
sorry, I'm really bad at this. So as 92 is the first term and 80 is the second term, I'd keep on adding the common difference ( -12 ) until I got a negative number? and then I'd count which value of n it was? -4 looks like the 9th term, but would I have to put it into a formula to confirm it's definitely the 9th term? If I can just say it's the 9th term with that evidence, would I then plug 8 into the formula for the sum of all positive integers? ( this does give me the correct answer )
Do you agree that each term is \(\displaystyle a_k=92+(k-1)(-12)~?\)
So when does \(\displaystyle a_k\) first become negative?
\(\displaystyle \begin{align*}92+(k-1)(-12)&<0 \\(k-1)(-12)&<-92\\k-1&>\frac{92}{12}\\k-1&>7\\k&>8 \end{align*}\)
So now \(\displaystyle a_8>0\) while \(\displaystyle a_9<0\)
\(\displaystyle \begin{align*}\sum\limits_{k = 1}^8 {{a_k}}&=\sum\limits_{k = 1}^8 {92+(k-1)(-12)} \\&=8(92) - 12\sum\limits_{k = 1}^8 {(k - 1)}\\&=8(92)-12\left(\frac{(7)(8)}{2}\right) \end{align*}\)



 
sorry, I'm really bad at this. So as 92 is the first term and 80 is the second term, I'd keep on adding the common difference ( -12 ) until I got a negative number? and then I'd count which value of n it was? -4 looks like the 9th term, but would I have to put it into a formula to confirm it's definitely the 9th term?

If I can just say it's the 9th term with that evidence, would I then plug 8 into the formula for the sum of all positive integers? ( this does give me the correct answer )
Again Sn IS the sum of the 1st n numbers. You want to add just the positive numbers which ends with 8 which is coincidentally is the 8th term. That is you want to add a1 + a2 + a3 + a4 + a5 + a6 + a7 + a8 = S8. So yes, n=8.

Now I wrote down the terms for you in my earlier post only because you said that the sum (I assume of the positive terms) was -1. This told me that you did not understand at all what was going on. Now the real way to do this, once you truly understand that you want to sum up the positive terms, is by using the method that pka showed you.
 
I think I understand a bit more now ( thanks everyone ).

so a = 92 d = -12

Sn = 92 + ( n - 1 )( -12 )

Sn < 0

92 + ( n - 1 )( -12 ) < 0

( n - 1 )( -12 ) < - 92

( n - 1)> -92/-12
n -1 > 92/12
n > 8

now: Sn= n/2 [ 2a + ( n - 1) d ]
S8 = 8/2 [ 184 + ( 7 x -12 )]
S8 = 4 x 100
S8= 400 ( correct )

have I done the method correctly?
 
I think I understand a bit more now ( thanks everyone ).

so a = 92 d = -12

Sn = 92 + ( n - 1 )( -12 )

Sn < 0

92 + ( n - 1 )( -12 ) < 0

( n - 1 )( -12 ) < - 92

( n - 1)> -92/-12
n -1 > 92/12
n > 8

now: Sn= n/2 [ 2a + ( n - 1) d ]
S8 = 8/2 [ 184 + ( 7 x -12 )]
S8 = 4 x 100
S8= 400 ( correct )

have I done the method correctly?
Perfect!
 
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