How to find sum of arithmetic series? The 2019 Stack Overflow Developer Survey Results Are In Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)arithmetic series - solving problemHow to choose a convergence test for a given infinite series?How to find the Sum of a complex Alternating SeriesHow many terms in arithmetic series where sum < 1000?Determine whether the series $sumlimits_n=1^inftyfrac1n^2+5n+6$ is convergent or divergent. If it is convergent, find its sum.Find the sum of a series that is not a geometric or alternating seriesArithmetic Series worded questionFind all values of $k$ such that the series with terms $k^n / n^k$ converges.Find the sum of the infinite series $1+ frac12!+ frac14!+dotsb$Finding sum of none arithmetic series
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How to find sum of arithmetic series?
The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)arithmetic series - solving problemHow to choose a convergence test for a given infinite series?How to find the Sum of a complex Alternating SeriesHow many terms in arithmetic series where sum < 1000?Determine whether the series $sumlimits_n=1^inftyfrac1n^2+5n+6$ is convergent or divergent. If it is convergent, find its sum.Find the sum of a series that is not a geometric or alternating seriesArithmetic Series worded questionFind all values of $k$ such that the series with terms $k^n / n^k$ converges.Find the sum of the infinite series $1+ frac12!+ frac14!+dotsb$Finding sum of none arithmetic series
$begingroup$
Here's my question:
For the series shown, find n such that $S_n = -4290$:
42 + 34 + 26 + 18 + …
I don't want a direct answer to the question, just a way to determine $A_n$ so I can do the work on my own.
Edit: $n$ and $A_n$ are unknown at the moment.
sequences-and-series arithmetic
asked Apr 8 at 13:56
SanguineLSanguineL
63
New contributor
SanguineL is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
|
show 3 more comments
$begingroup$
Here's my question:
For the series shown, find n such that $S_n = -4290$:
42 + 34 + 26 + 18 + …
I don't want a direct answer to the question, just a way to determine $A_n$ so I can do the work on my own.
Edit: $n$ and $A_n$ are unknown at the moment.
sequences-and-series arithmetic
asked Apr 8 at 13:56
SanguineLSanguineL
63
New contributor
SanguineL is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
Use the formula: $S_n=frac12n(a_1+a_ n)$ where $S_n$ is the sum of the series, $a_1$ is the first element of the series, i.e. $42$, and $a_n$ is the $n$th element of the series, i.e. $42-(n-1)8$.
$endgroup$
– Floris Claassens
Apr 8 at 14:00
$begingroup$
Well you do know that $n$ is such that $S_n=frac12n(50-8n)=-4290$.
$endgroup$
– Floris Claassens
Apr 8 at 14:02
$begingroup$
@Amad Yes, but I can't figure out $A_n$ without knowing $n$.
$endgroup$
– SanguineL
Apr 8 at 14:08
1
$begingroup$
It isn't hard to figure out that $A_n=42-8cdot (n-1)$, then $A_1=42$, $A_2=34$, $A_3=26$, $A_4=18$ and so on
$endgroup$
– Amad
Apr 8 at 14:08
$begingroup$
@FlorisClaassens I cannot use that formula since $n$ is unknown. There are two unknown variables in this.
$endgroup$
– SanguineL
Apr 8 at 14:10
|
show 3 more comments
$begingroup$
Here's my question:
For the series shown, find n such that $S_n = -4290$:
42 + 34 + 26 + 18 + …
I don't want a direct answer to the question, just a way to determine $A_n$ so I can do the work on my own.
Edit: $n$ and $A_n$ are unknown at the moment.
sequences-and-series arithmetic
asked Apr 8 at 13:56
SanguineLSanguineL
63
New contributor
SanguineL is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Here's my question:
For the series shown, find n such that $S_n = -4290$:
42 + 34 + 26 + 18 + …
I don't want a direct answer to the question, just a way to determine $A_n$ so I can do the work on my own.
Edit: $n$ and $A_n$ are unknown at the moment.
sequences-and-series arithmetic
sequences-and-series arithmetic
asked Apr 8 at 13:56
SanguineLSanguineL
63
New contributor
SanguineL is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Apr 8 at 13:56
SanguineLSanguineL
63
New contributor
SanguineL is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Apr 8 at 14:08
SanguineL
asked Apr 8 at 13:56
SanguineLSanguineL
63
New contributor
SanguineL is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Apr 8 at 13:56
SanguineLSanguineL
63
asked Apr 8 at 13:56
SanguineLSanguineL
63
63
New contributor
SanguineL is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
SanguineL is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
SanguineL is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
Use the formula: $S_n=frac12n(a_1+a_ n)$ where $S_n$ is the sum of the series, $a_1$ is the first element of the series, i.e. $42$, and $a_n$ is the $n$th element of the series, i.e. $42-(n-1)8$.
$endgroup$
– Floris Claassens
Apr 8 at 14:00
$begingroup$
Well you do know that $n$ is such that $S_n=frac12n(50-8n)=-4290$.
$endgroup$
– Floris Claassens
Apr 8 at 14:02
$begingroup$
@Amad Yes, but I can't figure out $A_n$ without knowing $n$.
$endgroup$
– SanguineL
Apr 8 at 14:08
1
$begingroup$
It isn't hard to figure out that $A_n=42-8cdot (n-1)$, then $A_1=42$, $A_2=34$, $A_3=26$, $A_4=18$ and so on
$endgroup$
– Amad
Apr 8 at 14:08
$begingroup$
@FlorisClaassens I cannot use that formula since $n$ is unknown. There are two unknown variables in this.
$endgroup$
– SanguineL
Apr 8 at 14:10
|
show 3 more comments
$begingroup$
Use the formula: $S_n=frac12n(a_1+a_ n)$ where $S_n$ is the sum of the series, $a_1$ is the first element of the series, i.e. $42$, and $a_n$ is the $n$th element of the series, i.e. $42-(n-1)8$.
$endgroup$
– Floris Claassens
Apr 8 at 14:00
$begingroup$
Well you do know that $n$ is such that $S_n=frac12n(50-8n)=-4290$.
$endgroup$
– Floris Claassens
Apr 8 at 14:02
$begingroup$
@Amad Yes, but I can't figure out $A_n$ without knowing $n$.
$endgroup$
– SanguineL
Apr 8 at 14:08
1
$begingroup$
It isn't hard to figure out that $A_n=42-8cdot (n-1)$, then $A_1=42$, $A_2=34$, $A_3=26$, $A_4=18$ and so on
$endgroup$
– Amad
Apr 8 at 14:08
$begingroup$
@FlorisClaassens I cannot use that formula since $n$ is unknown. There are two unknown variables in this.
$endgroup$
– SanguineL
Apr 8 at 14:10
$begingroup$
Use the formula: $S_n=frac12n(a_1+a_ n)$ where $S_n$ is the sum of the series, $a_1$ is the first element of the series, i.e. $42$, and $a_n$ is the $n$th element of the series, i.e. $42-(n-1)8$.
$endgroup$
– Floris Claassens
Apr 8 at 14:00
$begingroup$
Use the formula: $S_n=frac12n(a_1+a_ n)$ where $S_n$ is the sum of the series, $a_1$ is the first element of the series, i.e. $42$, and $a_n$ is the $n$th element of the series, i.e. $42-(n-1)8$.
$endgroup$
– Floris Claassens
Apr 8 at 14:00
$begingroup$
Well you do know that $n$ is such that $S_n=frac12n(50-8n)=-4290$.
$endgroup$
– Floris Claassens
Apr 8 at 14:02
$begingroup$
Well you do know that $n$ is such that $S_n=frac12n(50-8n)=-4290$.
$endgroup$
– Floris Claassens
Apr 8 at 14:02
$begingroup$
@Amad Yes, but I can't figure out $A_n$ without knowing $n$.
$endgroup$
– SanguineL
Apr 8 at 14:08
$begingroup$
@Amad Yes, but I can't figure out $A_n$ without knowing $n$.
$endgroup$
– SanguineL
Apr 8 at 14:08
1
1
$begingroup$
It isn't hard to figure out that $A_n=42-8cdot (n-1)$, then $A_1=42$, $A_2=34$, $A_3=26$, $A_4=18$ and so on
$endgroup$
– Amad
Apr 8 at 14:08
$begingroup$
It isn't hard to figure out that $A_n=42-8cdot (n-1)$, then $A_1=42$, $A_2=34$, $A_3=26$, $A_4=18$ and so on
$endgroup$
– Amad
Apr 8 at 14:08
$begingroup$
@FlorisClaassens I cannot use that formula since $n$ is unknown. There are two unknown variables in this.
$endgroup$
– SanguineL
Apr 8 at 14:10
$begingroup$
@FlorisClaassens I cannot use that formula since $n$ is unknown. There are two unknown variables in this.
$endgroup$
– SanguineL
Apr 8 at 14:10
|
show 3 more comments
4 Answers
4
active
oldest
votes
$begingroup$
Hints:
If you can find a formula $f(n)$ for the sum of the first $n$ terms, then you have only to solve $f(n)=-4290$ for $n$.
If the first term is $a$, and the difference between successive terms is $d$, then the terms are $a$, $a+d$, $a+2d$, $a+3d$, and so on.
The sum with one term is $a =boxed1a$.
The sum with two terms is $a+ underbrace(a+d)_2textrmnd term=boxed2a+(1)d$.
The sum with three terms is $underbrace2a+(1)d_textrmsum of 2textrm terms + underbrace(a+2d)_3textrmrd term=boxed3a + (1+2)d$.
The sum with four terms is $underbrace3a + (1+2)d_textrmsum of 3textrm terms + underbrace(a+3d)_4textrmth term=boxed4a + (1+2+3)d$.
The sum with five terms is $underbrace4a + (1+2+3)d_textrmsum of 4textrm terms + underbrace(a+4d)_5textrmth term=boxed5a + (1+2+3+4)d$.
So you need to ask yourself:
- What are $a$ and $d$ for your sequence?
- Do you know the formula for $1+2+3+cdots+n$?
- What would be the pattern above for the sum of $n$ terms?
answered Apr 8 at 14:13
MPWMPW
31.2k12157
$endgroup$
add a comment |
$begingroup$
You have $a_n=42-8(n-1)$. Now use the sum formula $$begin align-4290&=frac 12n(a_1+a_n)\&=frac 12n(42+(42-8(n-1))\
&=frac 12n(84-8n+8)\&=46n-4n^2\
&4n^2-46n-4290=0end align$$
and solve the quadratic for $n$
answered Apr 8 at 14:15
Ross MillikanRoss Millikan
301k24200375
$endgroup$
add a comment |
$begingroup$
$$u_p+u_p+1+...+u_n=fracu_p+u_n2(n-p+1)$$
answered Apr 8 at 14:16
HAMIDINE SOUMAREHAMIDINE SOUMARE
2,225212
$endgroup$
add a comment |
$begingroup$
It isn't hard to figure out that $A_n=42-8cdot (n-1)$, then
$$A_1=42$$
$$A_2=34$$
$$A_3=26$$
$$A_4=18$$
and so on...
$$S_n=sum_k=1^nA_k=sum_k=1^n(42-8cdot (k-1))=sum_k=1^n42-8sum_k=1^n(k-1)=42n-8(sum_k=1^nk-sum_k=1^n1)=42n-8(sum_k=1^nk-n)$$
Since we know that $sum_k=1^nk=text [sum of all natural numbers not greater than n] = fracncdot (n+1)2$, we can simplify:
$$S_n=42n-8cdot left(fracncdot (n+1)2-nright)=42n-4ncdot(n+1)+8n=50n-4ncdot(n+1)=50n-4n^2-4n=colorred46n-4n^2$$
answered Apr 8 at 14:27
AmadAmad
5410
$endgroup$
add a comment |
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4 Answers
4
active
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4 Answers
4
active
oldest
votes
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active
oldest
votes
$begingroup$
Hints:
If you can find a formula $f(n)$ for the sum of the first $n$ terms, then you have only to solve $f(n)=-4290$ for $n$.
If the first term is $a$, and the difference between successive terms is $d$, then the terms are $a$, $a+d$, $a+2d$, $a+3d$, and so on.
The sum with one term is $a =boxed1a$.
The sum with two terms is $a+ underbrace(a+d)_2textrmnd term=boxed2a+(1)d$.
The sum with three terms is $underbrace2a+(1)d_textrmsum of 2textrm terms + underbrace(a+2d)_3textrmrd term=boxed3a + (1+2)d$.
The sum with four terms is $underbrace3a + (1+2)d_textrmsum of 3textrm terms + underbrace(a+3d)_4textrmth term=boxed4a + (1+2+3)d$.
The sum with five terms is $underbrace4a + (1+2+3)d_textrmsum of 4textrm terms + underbrace(a+4d)_5textrmth term=boxed5a + (1+2+3+4)d$.
So you need to ask yourself:
- What are $a$ and $d$ for your sequence?
- Do you know the formula for $1+2+3+cdots+n$?
- What would be the pattern above for the sum of $n$ terms?
answered Apr 8 at 14:13
MPWMPW
31.2k12157
$endgroup$
add a comment |
$begingroup$
Hints:
If you can find a formula $f(n)$ for the sum of the first $n$ terms, then you have only to solve $f(n)=-4290$ for $n$.
If the first term is $a$, and the difference between successive terms is $d$, then the terms are $a$, $a+d$, $a+2d$, $a+3d$, and so on.
The sum with one term is $a =boxed1a$.
The sum with two terms is $a+ underbrace(a+d)_2textrmnd term=boxed2a+(1)d$.
The sum with three terms is $underbrace2a+(1)d_textrmsum of 2textrm terms + underbrace(a+2d)_3textrmrd term=boxed3a + (1+2)d$.
The sum with four terms is $underbrace3a + (1+2)d_textrmsum of 3textrm terms + underbrace(a+3d)_4textrmth term=boxed4a + (1+2+3)d$.
The sum with five terms is $underbrace4a + (1+2+3)d_textrmsum of 4textrm terms + underbrace(a+4d)_5textrmth term=boxed5a + (1+2+3+4)d$.
So you need to ask yourself:
- What are $a$ and $d$ for your sequence?
- Do you know the formula for $1+2+3+cdots+n$?
- What would be the pattern above for the sum of $n$ terms?
answered Apr 8 at 14:13
MPWMPW
31.2k12157
$endgroup$
add a comment |
$begingroup$
Hints:
If you can find a formula $f(n)$ for the sum of the first $n$ terms, then you have only to solve $f(n)=-4290$ for $n$.
If the first term is $a$, and the difference between successive terms is $d$, then the terms are $a$, $a+d$, $a+2d$, $a+3d$, and so on.
The sum with one term is $a =boxed1a$.
The sum with two terms is $a+ underbrace(a+d)_2textrmnd term=boxed2a+(1)d$.
The sum with three terms is $underbrace2a+(1)d_textrmsum of 2textrm terms + underbrace(a+2d)_3textrmrd term=boxed3a + (1+2)d$.
The sum with four terms is $underbrace3a + (1+2)d_textrmsum of 3textrm terms + underbrace(a+3d)_4textrmth term=boxed4a + (1+2+3)d$.
The sum with five terms is $underbrace4a + (1+2+3)d_textrmsum of 4textrm terms + underbrace(a+4d)_5textrmth term=boxed5a + (1+2+3+4)d$.
So you need to ask yourself:
- What are $a$ and $d$ for your sequence?
- Do you know the formula for $1+2+3+cdots+n$?
- What would be the pattern above for the sum of $n$ terms?
answered Apr 8 at 14:13
MPWMPW
31.2k12157
$endgroup$
Hints:
If you can find a formula $f(n)$ for the sum of the first $n$ terms, then you have only to solve $f(n)=-4290$ for $n$.
If the first term is $a$, and the difference between successive terms is $d$, then the terms are $a$, $a+d$, $a+2d$, $a+3d$, and so on.
The sum with one term is $a =boxed1a$.
The sum with two terms is $a+ underbrace(a+d)_2textrmnd term=boxed2a+(1)d$.
The sum with three terms is $underbrace2a+(1)d_textrmsum of 2textrm terms + underbrace(a+2d)_3textrmrd term=boxed3a + (1+2)d$.
The sum with four terms is $underbrace3a + (1+2)d_textrmsum of 3textrm terms + underbrace(a+3d)_4textrmth term=boxed4a + (1+2+3)d$.
The sum with five terms is $underbrace4a + (1+2+3)d_textrmsum of 4textrm terms + underbrace(a+4d)_5textrmth term=boxed5a + (1+2+3+4)d$.
So you need to ask yourself:
- What are $a$ and $d$ for your sequence?
- Do you know the formula for $1+2+3+cdots+n$?
- What would be the pattern above for the sum of $n$ terms?
answered Apr 8 at 14:13
MPWMPW
31.2k12157
answered Apr 8 at 14:13
MPWMPW
31.2k12157
answered Apr 8 at 14:13
MPWMPW
31.2k12157
answered Apr 8 at 14:13
MPWMPW
31.2k12157
31.2k12157
add a comment |
add a comment |
$begingroup$
You have $a_n=42-8(n-1)$. Now use the sum formula $$begin align-4290&=frac 12n(a_1+a_n)\&=frac 12n(42+(42-8(n-1))\
&=frac 12n(84-8n+8)\&=46n-4n^2\
&4n^2-46n-4290=0end align$$
and solve the quadratic for $n$
answered Apr 8 at 14:15
Ross MillikanRoss Millikan
301k24200375
$endgroup$
add a comment |
$begingroup$
You have $a_n=42-8(n-1)$. Now use the sum formula $$begin align-4290&=frac 12n(a_1+a_n)\&=frac 12n(42+(42-8(n-1))\
&=frac 12n(84-8n+8)\&=46n-4n^2\
&4n^2-46n-4290=0end align$$
and solve the quadratic for $n$
answered Apr 8 at 14:15
Ross MillikanRoss Millikan
301k24200375
$endgroup$
add a comment |
$begingroup$
You have $a_n=42-8(n-1)$. Now use the sum formula $$begin align-4290&=frac 12n(a_1+a_n)\&=frac 12n(42+(42-8(n-1))\
&=frac 12n(84-8n+8)\&=46n-4n^2\
&4n^2-46n-4290=0end align$$
and solve the quadratic for $n$
answered Apr 8 at 14:15
Ross MillikanRoss Millikan
301k24200375
$endgroup$
You have $a_n=42-8(n-1)$. Now use the sum formula $$begin align-4290&=frac 12n(a_1+a_n)\&=frac 12n(42+(42-8(n-1))\
&=frac 12n(84-8n+8)\&=46n-4n^2\
&4n^2-46n-4290=0end align$$
and solve the quadratic for $n$
answered Apr 8 at 14:15
Ross MillikanRoss Millikan
301k24200375
answered Apr 8 at 14:15
Ross MillikanRoss Millikan
301k24200375
answered Apr 8 at 14:15
Ross MillikanRoss Millikan
301k24200375
answered Apr 8 at 14:15
Ross MillikanRoss Millikan
301k24200375
301k24200375
add a comment |
add a comment |
$begingroup$
$$u_p+u_p+1+...+u_n=fracu_p+u_n2(n-p+1)$$
answered Apr 8 at 14:16
HAMIDINE SOUMAREHAMIDINE SOUMARE
2,225212
$endgroup$
add a comment |
$begingroup$
$$u_p+u_p+1+...+u_n=fracu_p+u_n2(n-p+1)$$
answered Apr 8 at 14:16
HAMIDINE SOUMAREHAMIDINE SOUMARE
2,225212
$endgroup$
add a comment |
$begingroup$
$$u_p+u_p+1+...+u_n=fracu_p+u_n2(n-p+1)$$
answered Apr 8 at 14:16
HAMIDINE SOUMAREHAMIDINE SOUMARE
2,225212
$endgroup$
$$u_p+u_p+1+...+u_n=fracu_p+u_n2(n-p+1)$$
answered Apr 8 at 14:16
HAMIDINE SOUMAREHAMIDINE SOUMARE
2,225212
answered Apr 8 at 14:16
HAMIDINE SOUMAREHAMIDINE SOUMARE
2,225212
answered Apr 8 at 14:16
HAMIDINE SOUMAREHAMIDINE SOUMARE
2,225212
answered Apr 8 at 14:16
HAMIDINE SOUMAREHAMIDINE SOUMARE
2,225212
2,225212
add a comment |
add a comment |
$begingroup$
It isn't hard to figure out that $A_n=42-8cdot (n-1)$, then
$$A_1=42$$
$$A_2=34$$
$$A_3=26$$
$$A_4=18$$
and so on...
$$S_n=sum_k=1^nA_k=sum_k=1^n(42-8cdot (k-1))=sum_k=1^n42-8sum_k=1^n(k-1)=42n-8(sum_k=1^nk-sum_k=1^n1)=42n-8(sum_k=1^nk-n)$$
Since we know that $sum_k=1^nk=text [sum of all natural numbers not greater than n] = fracncdot (n+1)2$, we can simplify:
$$S_n=42n-8cdot left(fracncdot (n+1)2-nright)=42n-4ncdot(n+1)+8n=50n-4ncdot(n+1)=50n-4n^2-4n=colorred46n-4n^2$$
answered Apr 8 at 14:27
AmadAmad
5410
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add a comment |
$begingroup$
It isn't hard to figure out that $A_n=42-8cdot (n-1)$, then
$$A_1=42$$
$$A_2=34$$
$$A_3=26$$
$$A_4=18$$
and so on...
$$S_n=sum_k=1^nA_k=sum_k=1^n(42-8cdot (k-1))=sum_k=1^n42-8sum_k=1^n(k-1)=42n-8(sum_k=1^nk-sum_k=1^n1)=42n-8(sum_k=1^nk-n)$$
Since we know that $sum_k=1^nk=text [sum of all natural numbers not greater than n] = fracncdot (n+1)2$, we can simplify:
$$S_n=42n-8cdot left(fracncdot (n+1)2-nright)=42n-4ncdot(n+1)+8n=50n-4ncdot(n+1)=50n-4n^2-4n=colorred46n-4n^2$$
answered Apr 8 at 14:27
AmadAmad
5410
$endgroup$
add a comment |
$begingroup$
It isn't hard to figure out that $A_n=42-8cdot (n-1)$, then
$$A_1=42$$
$$A_2=34$$
$$A_3=26$$
$$A_4=18$$
and so on...
$$S_n=sum_k=1^nA_k=sum_k=1^n(42-8cdot (k-1))=sum_k=1^n42-8sum_k=1^n(k-1)=42n-8(sum_k=1^nk-sum_k=1^n1)=42n-8(sum_k=1^nk-n)$$
Since we know that $sum_k=1^nk=text [sum of all natural numbers not greater than n] = fracncdot (n+1)2$, we can simplify:
$$S_n=42n-8cdot left(fracncdot (n+1)2-nright)=42n-4ncdot(n+1)+8n=50n-4ncdot(n+1)=50n-4n^2-4n=colorred46n-4n^2$$
answered Apr 8 at 14:27
AmadAmad
5410
$endgroup$
It isn't hard to figure out that $A_n=42-8cdot (n-1)$, then
$$A_1=42$$
$$A_2=34$$
$$A_3=26$$
$$A_4=18$$
and so on...
$$S_n=sum_k=1^nA_k=sum_k=1^n(42-8cdot (k-1))=sum_k=1^n42-8sum_k=1^n(k-1)=42n-8(sum_k=1^nk-sum_k=1^n1)=42n-8(sum_k=1^nk-n)$$
Since we know that $sum_k=1^nk=text [sum of all natural numbers not greater than n] = fracncdot (n+1)2$, we can simplify:
$$S_n=42n-8cdot left(fracncdot (n+1)2-nright)=42n-4ncdot(n+1)+8n=50n-4ncdot(n+1)=50n-4n^2-4n=colorred46n-4n^2$$
answered Apr 8 at 14:27
AmadAmad
5410
answered Apr 8 at 14:27
AmadAmad
5410
answered Apr 8 at 14:27
AmadAmad
5410
answered Apr 8 at 14:27
AmadAmad
5410
5410
add a comment |
add a comment |
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$begingroup$
Use the formula: $S_n=frac12n(a_1+a_ n)$ where $S_n$ is the sum of the series, $a_1$ is the first element of the series, i.e. $42$, and $a_n$ is the $n$th element of the series, i.e. $42-(n-1)8$.
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– Floris Claassens
Apr 8 at 14:00
$begingroup$
Well you do know that $n$ is such that $S_n=frac12n(50-8n)=-4290$.
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– Floris Claassens
Apr 8 at 14:02
$begingroup$
@Amad Yes, but I can't figure out $A_n$ without knowing $n$.
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– SanguineL
Apr 8 at 14:08
1
$begingroup$
It isn't hard to figure out that $A_n=42-8cdot (n-1)$, then $A_1=42$, $A_2=34$, $A_3=26$, $A_4=18$ and so on
$endgroup$
– Amad
Apr 8 at 14:08
$begingroup$
@FlorisClaassens I cannot use that formula since $n$ is unknown. There are two unknown variables in this.
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– SanguineL
Apr 8 at 14:10