Find the Largest possible value of the sum of the arithmetic sequence 85, 78, 71,… [closed]
Obviously $a=85$ and $d=-7$ but I cannot figure out how to calculate the maximum value given this?
arithmetic
edited Nov 28 '18 at 21:15
dantopa
6,42932042
asked Nov 28 '18 at 20:31
Max Moscrop
6
closed as off-topic by Shaun, Henrik, RRL, Jyrki Lahtonen, Alexander Gruber♦ Nov 30 '18 at 3:25
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Shaun, Henrik, RRL, Jyrki Lahtonen, Alexander Gruber
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
Obviously $a=85$ and $d=-7$ but I cannot figure out how to calculate the maximum value given this?
arithmetic
edited Nov 28 '18 at 21:15
dantopa
6,42932042
asked Nov 28 '18 at 20:31
Max Moscrop
6
closed as off-topic by Shaun, Henrik, RRL, Jyrki Lahtonen, Alexander Gruber♦ Nov 30 '18 at 3:25
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Shaun, Henrik, RRL, Jyrki Lahtonen, Alexander Gruber
If this question can be reworded to fit the rules in the help center, please edit the question.
Hint. Once the numbers are negative the sum so far decreases. When does that happen?
– Ethan Bolker
Nov 28 '18 at 20:32
So the last positive term is the 13th no do I need to do 13/2(2*85+(13-1)*-7)
– Max Moscrop
Nov 28 '18 at 20:36
Yes you did....
– amWhy
Nov 28 '18 at 20:42
Yes, it's correct. (Didn't see the $n-1$ there...)
– KM101
Nov 28 '18 at 20:52
add a comment |
Obviously $a=85$ and $d=-7$ but I cannot figure out how to calculate the maximum value given this?
arithmetic
edited Nov 28 '18 at 21:15
dantopa
6,42932042
asked Nov 28 '18 at 20:31
Max Moscrop
6
Obviously $a=85$ and $d=-7$ but I cannot figure out how to calculate the maximum value given this?
arithmetic
arithmetic
edited Nov 28 '18 at 21:15
dantopa
6,42932042
asked Nov 28 '18 at 20:31
Max Moscrop
6
edited Nov 28 '18 at 21:15
dantopa
6,42932042
asked Nov 28 '18 at 20:31
Max Moscrop
6
edited Nov 28 '18 at 21:15
dantopa
6,42932042
edited Nov 28 '18 at 21:15
dantopa
6,42932042
edited Nov 28 '18 at 21:15
dantopa
6,42932042
6,42932042
asked Nov 28 '18 at 20:31
Max Moscrop
6
asked Nov 28 '18 at 20:31
Max Moscrop
6
asked Nov 28 '18 at 20:31
Max Moscrop
6
6
closed as off-topic by Shaun, Henrik, RRL, Jyrki Lahtonen, Alexander Gruber♦ Nov 30 '18 at 3:25
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Shaun, Henrik, RRL, Jyrki Lahtonen, Alexander Gruber
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Shaun, Henrik, RRL, Jyrki Lahtonen, Alexander Gruber♦ Nov 30 '18 at 3:25
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Shaun, Henrik, RRL, Jyrki Lahtonen, Alexander Gruber
If this question can be reworded to fit the rules in the help center, please edit the question.
Hint. Once the numbers are negative the sum so far decreases. When does that happen?
– Ethan Bolker
Nov 28 '18 at 20:32
So the last positive term is the 13th no do I need to do 13/2(2*85+(13-1)*-7)
– Max Moscrop
Nov 28 '18 at 20:36
Yes you did....
– amWhy
Nov 28 '18 at 20:42
Yes, it's correct. (Didn't see the $n-1$ there...)
– KM101
Nov 28 '18 at 20:52
add a comment |
Hint. Once the numbers are negative the sum so far decreases. When does that happen?
– Ethan Bolker
Nov 28 '18 at 20:32
So the last positive term is the 13th no do I need to do 13/2(2*85+(13-1)*-7)
– Max Moscrop
Nov 28 '18 at 20:36
Yes you did....
– amWhy
Nov 28 '18 at 20:42
Yes, it's correct. (Didn't see the $n-1$ there...)
– KM101
Nov 28 '18 at 20:52
Hint. Once the numbers are negative the sum so far decreases. When does that happen?
– Ethan Bolker
Nov 28 '18 at 20:32
Hint. Once the numbers are negative the sum so far decreases. When does that happen?
– Ethan Bolker
Nov 28 '18 at 20:32
So the last positive term is the 13th no do I need to do 13/2(2*85+(13-1)*-7)
– Max Moscrop
Nov 28 '18 at 20:36
So the last positive term is the 13th no do I need to do 13/2(2*85+(13-1)*-7)
– Max Moscrop
Nov 28 '18 at 20:36
Yes you did....
– amWhy
Nov 28 '18 at 20:42
Yes you did....
– amWhy
Nov 28 '18 at 20:42
Yes, it's correct. (Didn't see the $n-1$ there...)
– KM101
Nov 28 '18 at 20:52
Yes, it's correct. (Didn't see the $n-1$ there...)
– KM101
Nov 28 '18 at 20:52
add a comment |
3 Answers
3
active
oldest
votes
sum increases while the terms are positive, so compute the number of terms to include the last positive term. Then compute the actual sum.
answered Nov 28 '18 at 20:33
gt6989b
33.1k22452
add a comment |
Hint: Your common difference is negative. Therefore, the largest sum will be the sum of all positive terms. Recall that in an arithmetic sequence with the initial term $u_1$ and common difference $d$, the $n^{th}$ term is found by
$$u_n = u_1+(n-1)d$$
Try to find the greatest value of $n$ such that
$$u_n > 0 implies u_1+(n-1)d > 0$$
so that you can cover all the positive terms, followed by plugging the value of $n$ to find that $u_n$. Finally sum up all the terms from the initial term to the $n^{th}$ term. You can use
$$S_n = frac{n(u_1+u_n)}{2}$$
answered Nov 28 '18 at 20:38
KM101
5,3811423
85 + (13-1)*-7 = 1 does this not mean that n = 13?
– Max Moscrop
Nov 28 '18 at 20:47
Yes yes, my bad. (Didn't pay attention to the $n-1$ you wrote.)
– KM101
Nov 28 '18 at 20:48
when I use 12 I get a value one off the answer given in my book but 13 is way off leading me to believe that 12 is the correct n...
– Max Moscrop
Nov 28 '18 at 20:49
No, $13$ is the correct $n$ as you've written. Your formula looks fine as well (for $S_n$), so maybe you miscalculated for the final answer?... Otherwise, the book is incorrect.
– KM101
Nov 28 '18 at 20:51
You're right I miscalculated xD! Thank you for all of your help!
– Max Moscrop
Nov 28 '18 at 20:52
|
show 1 more comment
We have that $85=12cdot 7 +1$, then we need to evaluate
$$sum_{k=0}^{12}(1+7k)=sum_{k=0}^{12} 1+7sum_{k=0}^{12}k=13+7frac{12cdot 13}{2}=13+7cdot6cdot 13=559$$
where we have used that
- $sum_{k=0}^{m} 1=m+1$
- $sum_{k=0}^{n}k=frac{n(n+1)}{2}$
answered Nov 28 '18 at 20:35
gimusi
1
1
A large part of the problem here is to find that $12$, I think.
– Arthur
Nov 28 '18 at 20:37
@Arthur Sorry but I don't understand your observation.
– gimusi
Nov 28 '18 at 20:39
Yeah, that' crucial to the answer, and not explaining it leaves a deficit in this answer.
– amWhy
Nov 28 '18 at 20:39
@Arthur Ah ok, you are claiming that it is difficult see what the sequence is. I'm not sure the main doubt was that. Anyway I've suggested a way to easily see that.
– gimusi
Nov 28 '18 at 20:42
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
sum increases while the terms are positive, so compute the number of terms to include the last positive term. Then compute the actual sum.
answered Nov 28 '18 at 20:33
gt6989b
33.1k22452
add a comment |
sum increases while the terms are positive, so compute the number of terms to include the last positive term. Then compute the actual sum.
answered Nov 28 '18 at 20:33
gt6989b
33.1k22452
add a comment |
sum increases while the terms are positive, so compute the number of terms to include the last positive term. Then compute the actual sum.
answered Nov 28 '18 at 20:33
gt6989b
33.1k22452
sum increases while the terms are positive, so compute the number of terms to include the last positive term. Then compute the actual sum.
answered Nov 28 '18 at 20:33
gt6989b
33.1k22452
answered Nov 28 '18 at 20:33
gt6989b
33.1k22452
answered Nov 28 '18 at 20:33
gt6989b
33.1k22452
answered Nov 28 '18 at 20:33
gt6989b
33.1k22452
33.1k22452
add a comment |
add a comment |
Hint: Your common difference is negative. Therefore, the largest sum will be the sum of all positive terms. Recall that in an arithmetic sequence with the initial term $u_1$ and common difference $d$, the $n^{th}$ term is found by
$$u_n = u_1+(n-1)d$$
Try to find the greatest value of $n$ such that
$$u_n > 0 implies u_1+(n-1)d > 0$$
so that you can cover all the positive terms, followed by plugging the value of $n$ to find that $u_n$. Finally sum up all the terms from the initial term to the $n^{th}$ term. You can use
$$S_n = frac{n(u_1+u_n)}{2}$$
answered Nov 28 '18 at 20:38
KM101
5,3811423
85 + (13-1)*-7 = 1 does this not mean that n = 13?
– Max Moscrop
Nov 28 '18 at 20:47
Yes yes, my bad. (Didn't pay attention to the $n-1$ you wrote.)
– KM101
Nov 28 '18 at 20:48
when I use 12 I get a value one off the answer given in my book but 13 is way off leading me to believe that 12 is the correct n...
– Max Moscrop
Nov 28 '18 at 20:49
No, $13$ is the correct $n$ as you've written. Your formula looks fine as well (for $S_n$), so maybe you miscalculated for the final answer?... Otherwise, the book is incorrect.
– KM101
Nov 28 '18 at 20:51
You're right I miscalculated xD! Thank you for all of your help!
– Max Moscrop
Nov 28 '18 at 20:52
|
show 1 more comment
Hint: Your common difference is negative. Therefore, the largest sum will be the sum of all positive terms. Recall that in an arithmetic sequence with the initial term $u_1$ and common difference $d$, the $n^{th}$ term is found by
$$u_n = u_1+(n-1)d$$
Try to find the greatest value of $n$ such that
$$u_n > 0 implies u_1+(n-1)d > 0$$
so that you can cover all the positive terms, followed by plugging the value of $n$ to find that $u_n$. Finally sum up all the terms from the initial term to the $n^{th}$ term. You can use
$$S_n = frac{n(u_1+u_n)}{2}$$
answered Nov 28 '18 at 20:38
KM101
5,3811423
85 + (13-1)*-7 = 1 does this not mean that n = 13?
– Max Moscrop
Nov 28 '18 at 20:47
Yes yes, my bad. (Didn't pay attention to the $n-1$ you wrote.)
– KM101
Nov 28 '18 at 20:48
when I use 12 I get a value one off the answer given in my book but 13 is way off leading me to believe that 12 is the correct n...
– Max Moscrop
Nov 28 '18 at 20:49
No, $13$ is the correct $n$ as you've written. Your formula looks fine as well (for $S_n$), so maybe you miscalculated for the final answer?... Otherwise, the book is incorrect.
– KM101
Nov 28 '18 at 20:51
You're right I miscalculated xD! Thank you for all of your help!
– Max Moscrop
Nov 28 '18 at 20:52
|
show 1 more comment
Hint: Your common difference is negative. Therefore, the largest sum will be the sum of all positive terms. Recall that in an arithmetic sequence with the initial term $u_1$ and common difference $d$, the $n^{th}$ term is found by
$$u_n = u_1+(n-1)d$$
Try to find the greatest value of $n$ such that
$$u_n > 0 implies u_1+(n-1)d > 0$$
so that you can cover all the positive terms, followed by plugging the value of $n$ to find that $u_n$. Finally sum up all the terms from the initial term to the $n^{th}$ term. You can use
$$S_n = frac{n(u_1+u_n)}{2}$$
answered Nov 28 '18 at 20:38
KM101
5,3811423
Hint: Your common difference is negative. Therefore, the largest sum will be the sum of all positive terms. Recall that in an arithmetic sequence with the initial term $u_1$ and common difference $d$, the $n^{th}$ term is found by
$$u_n = u_1+(n-1)d$$
Try to find the greatest value of $n$ such that
$$u_n > 0 implies u_1+(n-1)d > 0$$
so that you can cover all the positive terms, followed by plugging the value of $n$ to find that $u_n$. Finally sum up all the terms from the initial term to the $n^{th}$ term. You can use
$$S_n = frac{n(u_1+u_n)}{2}$$
answered Nov 28 '18 at 20:38
KM101
5,3811423
edited Nov 28 '18 at 20:54
answered Nov 28 '18 at 20:38
KM101
5,3811423
answered Nov 28 '18 at 20:38
KM101
5,3811423
answered Nov 28 '18 at 20:38
KM101
5,3811423
5,3811423
85 + (13-1)*-7 = 1 does this not mean that n = 13?
– Max Moscrop
Nov 28 '18 at 20:47
Yes yes, my bad. (Didn't pay attention to the $n-1$ you wrote.)
– KM101
Nov 28 '18 at 20:48
when I use 12 I get a value one off the answer given in my book but 13 is way off leading me to believe that 12 is the correct n...
– Max Moscrop
Nov 28 '18 at 20:49
No, $13$ is the correct $n$ as you've written. Your formula looks fine as well (for $S_n$), so maybe you miscalculated for the final answer?... Otherwise, the book is incorrect.
– KM101
Nov 28 '18 at 20:51
You're right I miscalculated xD! Thank you for all of your help!
– Max Moscrop
Nov 28 '18 at 20:52
|
show 1 more comment
85 + (13-1)*-7 = 1 does this not mean that n = 13?
– Max Moscrop
Nov 28 '18 at 20:47
Yes yes, my bad. (Didn't pay attention to the $n-1$ you wrote.)
– KM101
Nov 28 '18 at 20:48
when I use 12 I get a value one off the answer given in my book but 13 is way off leading me to believe that 12 is the correct n...
– Max Moscrop
Nov 28 '18 at 20:49
No, $13$ is the correct $n$ as you've written. Your formula looks fine as well (for $S_n$), so maybe you miscalculated for the final answer?... Otherwise, the book is incorrect.
– KM101
Nov 28 '18 at 20:51
You're right I miscalculated xD! Thank you for all of your help!
– Max Moscrop
Nov 28 '18 at 20:52
85 + (13-1)*-7 = 1 does this not mean that n = 13?
– Max Moscrop
Nov 28 '18 at 20:47
85 + (13-1)*-7 = 1 does this not mean that n = 13?
– Max Moscrop
Nov 28 '18 at 20:47
Yes yes, my bad. (Didn't pay attention to the $n-1$ you wrote.)
– KM101
Nov 28 '18 at 20:48
Yes yes, my bad. (Didn't pay attention to the $n-1$ you wrote.)
– KM101
Nov 28 '18 at 20:48
when I use 12 I get a value one off the answer given in my book but 13 is way off leading me to believe that 12 is the correct n...
– Max Moscrop
Nov 28 '18 at 20:49
when I use 12 I get a value one off the answer given in my book but 13 is way off leading me to believe that 12 is the correct n...
– Max Moscrop
Nov 28 '18 at 20:49
No, $13$ is the correct $n$ as you've written. Your formula looks fine as well (for $S_n$), so maybe you miscalculated for the final answer?... Otherwise, the book is incorrect.
– KM101
Nov 28 '18 at 20:51
No, $13$ is the correct $n$ as you've written. Your formula looks fine as well (for $S_n$), so maybe you miscalculated for the final answer?... Otherwise, the book is incorrect.
– KM101
Nov 28 '18 at 20:51
You're right I miscalculated xD! Thank you for all of your help!
– Max Moscrop
Nov 28 '18 at 20:52
You're right I miscalculated xD! Thank you for all of your help!
– Max Moscrop
Nov 28 '18 at 20:52
|
show 1 more comment
We have that $85=12cdot 7 +1$, then we need to evaluate
$$sum_{k=0}^{12}(1+7k)=sum_{k=0}^{12} 1+7sum_{k=0}^{12}k=13+7frac{12cdot 13}{2}=13+7cdot6cdot 13=559$$
where we have used that
- $sum_{k=0}^{m} 1=m+1$
- $sum_{k=0}^{n}k=frac{n(n+1)}{2}$
answered Nov 28 '18 at 20:35
gimusi
1
1
A large part of the problem here is to find that $12$, I think.
– Arthur
Nov 28 '18 at 20:37
@Arthur Sorry but I don't understand your observation.
– gimusi
Nov 28 '18 at 20:39
Yeah, that' crucial to the answer, and not explaining it leaves a deficit in this answer.
– amWhy
Nov 28 '18 at 20:39
@Arthur Ah ok, you are claiming that it is difficult see what the sequence is. I'm not sure the main doubt was that. Anyway I've suggested a way to easily see that.
– gimusi
Nov 28 '18 at 20:42
add a comment |
We have that $85=12cdot 7 +1$, then we need to evaluate
$$sum_{k=0}^{12}(1+7k)=sum_{k=0}^{12} 1+7sum_{k=0}^{12}k=13+7frac{12cdot 13}{2}=13+7cdot6cdot 13=559$$
where we have used that
- $sum_{k=0}^{m} 1=m+1$
- $sum_{k=0}^{n}k=frac{n(n+1)}{2}$
answered Nov 28 '18 at 20:35
gimusi
1
1
A large part of the problem here is to find that $12$, I think.
– Arthur
Nov 28 '18 at 20:37
@Arthur Sorry but I don't understand your observation.
– gimusi
Nov 28 '18 at 20:39
Yeah, that' crucial to the answer, and not explaining it leaves a deficit in this answer.
– amWhy
Nov 28 '18 at 20:39
@Arthur Ah ok, you are claiming that it is difficult see what the sequence is. I'm not sure the main doubt was that. Anyway I've suggested a way to easily see that.
– gimusi
Nov 28 '18 at 20:42
add a comment |
We have that $85=12cdot 7 +1$, then we need to evaluate
$$sum_{k=0}^{12}(1+7k)=sum_{k=0}^{12} 1+7sum_{k=0}^{12}k=13+7frac{12cdot 13}{2}=13+7cdot6cdot 13=559$$
where we have used that
- $sum_{k=0}^{m} 1=m+1$
- $sum_{k=0}^{n}k=frac{n(n+1)}{2}$
answered Nov 28 '18 at 20:35
gimusi
1
We have that $85=12cdot 7 +1$, then we need to evaluate
$$sum_{k=0}^{12}(1+7k)=sum_{k=0}^{12} 1+7sum_{k=0}^{12}k=13+7frac{12cdot 13}{2}=13+7cdot6cdot 13=559$$
where we have used that
- $sum_{k=0}^{m} 1=m+1$
- $sum_{k=0}^{n}k=frac{n(n+1)}{2}$
answered Nov 28 '18 at 20:35
gimusi
1
edited Nov 28 '18 at 20:46
answered Nov 28 '18 at 20:35
gimusi
1
answered Nov 28 '18 at 20:35
gimusi
1
answered Nov 28 '18 at 20:35
gimusi
1
1
1
A large part of the problem here is to find that $12$, I think.
– Arthur
Nov 28 '18 at 20:37
@Arthur Sorry but I don't understand your observation.
– gimusi
Nov 28 '18 at 20:39
Yeah, that' crucial to the answer, and not explaining it leaves a deficit in this answer.
– amWhy
Nov 28 '18 at 20:39
@Arthur Ah ok, you are claiming that it is difficult see what the sequence is. I'm not sure the main doubt was that. Anyway I've suggested a way to easily see that.
– gimusi
Nov 28 '18 at 20:42
add a comment |
1
A large part of the problem here is to find that $12$, I think.
– Arthur
Nov 28 '18 at 20:37
@Arthur Sorry but I don't understand your observation.
– gimusi
Nov 28 '18 at 20:39
Yeah, that' crucial to the answer, and not explaining it leaves a deficit in this answer.
– amWhy
Nov 28 '18 at 20:39
@Arthur Ah ok, you are claiming that it is difficult see what the sequence is. I'm not sure the main doubt was that. Anyway I've suggested a way to easily see that.
– gimusi
Nov 28 '18 at 20:42
1
1
A large part of the problem here is to find that $12$, I think.
– Arthur
Nov 28 '18 at 20:37
A large part of the problem here is to find that $12$, I think.
– Arthur
Nov 28 '18 at 20:37
@Arthur Sorry but I don't understand your observation.
– gimusi
Nov 28 '18 at 20:39
@Arthur Sorry but I don't understand your observation.
– gimusi
Nov 28 '18 at 20:39
Yeah, that' crucial to the answer, and not explaining it leaves a deficit in this answer.
– amWhy
Nov 28 '18 at 20:39
Yeah, that' crucial to the answer, and not explaining it leaves a deficit in this answer.
– amWhy
Nov 28 '18 at 20:39
@Arthur Ah ok, you are claiming that it is difficult see what the sequence is. I'm not sure the main doubt was that. Anyway I've suggested a way to easily see that.
– gimusi
Nov 28 '18 at 20:42
@Arthur Ah ok, you are claiming that it is difficult see what the sequence is. I'm not sure the main doubt was that. Anyway I've suggested a way to easily see that.
– gimusi
Nov 28 '18 at 20:42
add a comment |
Hint. Once the numbers are negative the sum so far decreases. When does that happen?
– Ethan Bolker
Nov 28 '18 at 20:32
So the last positive term is the 13th no do I need to do 13/2(2*85+(13-1)*-7)
– Max Moscrop
Nov 28 '18 at 20:36
Yes you did....
– amWhy
Nov 28 '18 at 20:42
Yes, it's correct. (Didn't see the $n-1$ there...)
– KM101
Nov 28 '18 at 20:52