Arithmetic Sequence problem

[Shu]

Hello, this is my first time asking a question here so sorry if I make any mistakes. So essentially, this is a question from my math exam in June. I horribly failed the exam but apparently they will be letting us retake it. Since that is the case I wanted to make sure I have a clear understanding of every question I previously failed.

The Question in exact wording is
"An arithmetic sequence has the first term a and common difference d. The sixth term of the sequence is 10. The sum of the first eight terms of the sequence is 62.
Find a and d"

The teacher gave me no real indication of what I did wrong after the exam for this question so I'd be appreciative is somebody can just show me the first step or how I should even tackle this problem.

The only part he marked right is
N/2 (2(u1) + (n-1)d) = 62 = 8/2( 2(u1) + (8-1)d)

I'll also attach my work from that question in totality. Again any help is appreciated.

 

[Deleted member 4993]

Hello, this is my first time asking a question here so sorry if I make any mistakes. So essentially, this is a question from my math exam in June. I horribly failed the exam but apparently they will be letting us retake it. Since that is the case I wanted to make sure I have a clear understanding of every question I previously failed.

The Question in exact wording is
"An arithmetic sequence has the first term a and common difference d. The sixth term of the sequence is 10. The sum of the first eight terms of the sequence is 62.
Find a and d"

The teacher gave me no real indication of what I did wrong after the exam for this question so I'd be appreciative is somebody can just show me the first step or how I should even tackle this problem.

The only part he marked right is
N/2 (2(u1) + (n-1)d) = 62 = 8/2( 2(u1) + (8-1)d)

I'll also attach my work from that question in totality. Again any help is appreciated.

If an arithmetic sequence ha s the first term 'a' and common difference 'd' :

What would be the expression for the 'n' th term of the sequence above?

What would be the expression for the sum of the first 'n' tems?

 

[canvas]

[math]\text{Solve this system of equations}: \begin{cases}a_6=a+5d=10\\S_8=\frac{a+a+7d}{2}\cdot8=62\end{cases}[/math]

 

[JeffM]

Thank you very much for showing your work and giving us the exact problem Those are the two most important things to do when asking for help here.

I must admit, however, that I have difficulty following your logic. You seem to start by guessing that a = 1 and d = 2, and 1 + 5(2) = 11. But 11 is not 10.

Guessing is not a great strategy.

What you could do is use algebra. You have two unknowns, a and d. To find two unknowns, you need two equations.

You are told that the sixth term = 10. You knew the proper formula for the nth term so that gives you one equation:

[math]a + (6 - 1)d = 10 \implies a + 5d = 10 \implies a = 10 - 5d.[/math]
Now toward the middle of your thought process you did try to use that formula, but wrote it as a + 6d = 10. That is always a danger with formulas: you forget them. Then you made life hard for yourself by using fractions and solving for d. From that point on, I completely lose your train of thought.

I truly believe that if you had recognized two unknowns means two equations, you would have arrived at the rather simple equation of a = 10 - 5d.

You also saw that the sum of the first eight terms being 62 gives an equation. Wow that is two equations, which is all you need. Again you made life hard for yourself by introducing some new and unnecessary notation. You could have written

[math]62= \dfrac{8}{2} * \{2a + (8 - 1)d\} = 4(2a + 7d) = 8a + 28d \implies 8a = 62 - 28d.[/math]
Can you solve a = 10 - 5d and 8a = 62 - 28d?

Does the answer check?

Things to learn.

Count the number of unknowns. That tells you how many equations you need.

Be sure you get your formulas right. In an exam you feel time’s winged chariot pressing on your heels, but do not let the pressure of time drive you to careless errors.

Make things simple. a = 10 - 5d and 8a = 62 - 28d are quite simple and avoid fractions.

Check your work against the original problem if you can spare the time.

 

[Harry_the_cat]

Hi Shu,

After reading the suggestions above, here is a similar problem to try, so that you can test your understanding.

"An arithmetic sequence has the first term a and common difference d. The fourth term is 11. The sum of the first ten terms is 155. Find a and d."

Please post your fully worked solution.

 

[Shu]

If an arithmetic sequence ha s the first term 'a' and common difference 'd' :

What would be the expression for the 'n' th term of the sequence above?

What would be the expression for the sum of the first 'n' tems?

Hey, Thanks for taking the time to give me some pointers. I think what you were trying to do was confirm I knew the formulas? Sorry if I am misunderstanding your intentions. Either way I took a few minutes to write it down on a piece of paper. Hopefully this is what you were looking for.

 

[Shu]

[math]\text{Solve this system of equations}: \begin{cases}a_6=a+5d=10\\S_8=\frac{a+a+7d}{2}\cdot8=62\end{cases}[/math]

Hello Canvas, Thank you too for taking some time out of your day to help me. If i'm being honest my answer to your problem is the one I am most unsure about. I'll post my work along with this reply so you can see what I did. I'll be going to sleep after posting all this but I'll check later to see if my answer was sufficient. If not I'll retry it in the morning.

 

[Shu]

Thank you very much for showing your work and giving us the exact problem Those are the two most important things to do when asking for help here.

I must admit, however, that I have difficulty following your logic. You seem to start by guessing that a = 1 and d = 2, and 1 + 5(2) = 11. But 11 is not 10.

Guessing is not a great strategy.

What you could do is use algebra. You have two unknowns, a and d. To find two unknowns, you need two equations.

You are told that the sixth term = 10. You knew the proper formula for the nth term so that gives you one equation:

[math]a + (6 - 1)d = 10 \implies a + 5d = 10 \implies a = 10 - 5d.[/math]
Now toward the middle of your thought process you did try to use that formula, but wrote it as a + 6d = 10. That is always a danger with formulas: you forget them. Then you made life hard for yourself by using fractions and solving for d. From that point on, I completely lose your train of thought.

I truly believe that if you had recognized two unknowns means two equations, you would have arrived at the rather simple equation of a = 10 - 5d.

You also saw that the sum of the first eight terms being 62 gives an equation. Wow that is two equations, which is all you need. Again you made life hard for yourself by introducing some new and unnecessary notation. You could have written

[math]62= \dfrac{8}{2} * \{2a + (8 - 1)d\} = 4(2a + 7d) = 8a + 28d \implies 8a = 62 - 28d.[/math]
Can you solve a = 10 - 5d and 8a = 62 - 28d?

Does the answer check?

Things to learn.

Count the number of unknowns. That tells you how many equations you need.

Be sure you get your formulas right. In an exam you feel time’s winged chariot pressing on your heels, but do not let the pressure of time drive you to careless errors.

Make things simple. a = 10 - 5d and 8a = 62 - 28d are quite simple and avoid fractions.

Check your work against the original problem if you can spare the time.

Good Afternoon JeffM, Thank you for taking time to write this response. Itt was really helpful and I realize that if I had taken some time to think about substitution and looked at the question properly I might have done a bit better. Thanks for the helpful advice. I took a few minutes to solve what you gave me. I hope that it is correct. If not please let me know, I'll try to do it again in the morning.

 

[Shu]

Hi Shu,

After reading the suggestions above, here is a similar problem to try, so that you can test your understanding.

"An arithmetic sequence has the first term a and common difference d. The fourth term is 11. The sum of the first ten terms is 155. Find a and d."

Please post your fully worked solution.

Good Afternoon Harry_the_cat, Thank you for going out of your way to write a response and even give me another problem to try out. Before this I had been looking for examples of problems like this but with no luck. I tried it after reading the other comments and I managed to finish it. Like you asked I'll post my work. Please let me know if I made any mistakes. Like I commented to others I will be going to sleep but I will check back in the morning or whenever you have time to see my answer. Thanks again, I hope you all have a good night!

 

[canvas]

Hello Canvas, Thank you too for taking some time out of your day to help me. If i'm being honest my answer to your problem is the one I am most unsure about. I'll post my work along with this reply so you can see what I did. I'll be going to sleep after posting all this but I'll check later to see if my answer was sufficient. If not I'll retry it in the morning.

[math]\text{You should get:} \begin{cases}a=\frac{5}{2}\\d=\frac{3}{2}\end{cases}[/math]

 

[canvas]

Good Afternoon Harry_the_cat, Thank you for going out of your way to write a response and even give me another problem to try out. Before this I had been looking for examples of problems like this but with no luck. I tried it after reading the other comments and I managed to finish it. Like you asked I'll post my work. Please let me know if I made any mistakes. Like I commented to others I will be going to sleep but I will check back in the morning or whenever you have time to see my answer. Thanks again, I hope you all have a good night!

That's correct! Great job!

 

[canvas]

[math]\frac{a+a+7d}{2}\cdot8=62 \Rightarrow 4(2a+7d)=62 \Rightarrow 8a+28d=62\\\text{That's your fault, because you got:}\,\,\,32a+112d=62[/math]

 

[JeffM]

Good Afternoon JeffM, Thank you for taking time to write this response. Itt was really helpful and I realize that if I had taken some time to think about substitution and looked at the question properly I might have done a bit better. Thanks for the helpful advice. I took a few minutes to solve what you gave me. I hope that it is correct. If not please let me know, I'll try to do it again in the morning.

That is perfect.

I know this was a test, and you may not have time to check your own work during a test. But checking your own work is helpful when you practice and essential for applying math to important practical problems.. If a is 2.5 and d = 1.5

The sequence is 2.5, 4, 5.5, 7, 8.5, 10,11.5, 13, …

The partial sums are 2.5, 6.5, 12, 19, 27.5, 37.5, 49, 62.

It checks.

There are two reasons to make a habit of checking your work.

First, it is embarrassing to have someone else point out your errors.

Second, if you can notice your own errors and correct them, it gives you confidence that you really can do math. Everyone who tries math makes blunders; doing math means correcting them.

 

[Harry_the_cat]

Good Afternoon Harry_the_cat, Thank you for going out of your way to write a response and even give me another problem to try out. Before this I had been looking for examples of problems like this but with no luck. I tried it after reading the other comments and I managed to finish it. Like you asked I'll post my work. Please let me know if I made any mistakes. Like I commented to others I will be going to sleep but I will check back in the morning or whenever you have time to see my answer. Thanks again, I hope you all have a good night!

Well done Shu. Perfect solution! Make sure you understand each step.

Just one thing I would have done differently, and this comes with experience:

When you had 155 = 5(2a +9d) and then 155 = 10a + 45d,

I would have divided both sides by 5 to get 31 = 2a +9d instead.

What you have done is completely correct but you are dealing with numbers bigger than they need to be. You can't always do this, but I find if you work with smaller numbers there is less chance of making an arithmetical error later.

Well done. Keep working hard. It is a pleasure helping students like you!

 

[Shu]

[math]\frac{a+a+7d}{2}\cdot8=62 \Rightarrow 4(2a+7d)=62 \Rightarrow 8a+28d=62\\\text{That's your fault, because you got:}\,\,\,32a+112d=62[/math]

If I'm correct what I did wrong was try to cancel the division of a+a+7d by 2 by multiplying the whole equation by two? So am I correct to assume I should have put the 8 against the 2 to cancel it that way? I did it again just to be sure, here is my work. Sorry if this question was a bit unclear. I'm not quite sure how to word it.

 

[Shu]

I think with that last comment that was all I needed to do for this thread. Thank you all for the helpful comments and questions. I feel like I understand this type of problem a lot more than before.

 

[lev888]

If I'm correct what I did wrong was try to cancel the division of a+a+7d by 2 by multiplying the whole equation by two? So am I correct to assume I should have put the 8 against the 2 to cancel it that way? I did it again just to be sure, here is my work. Sorry if this question was a bit unclear. I'm not quite sure how to word it.

The problem is that you did not multiply "the whole equation" by 2 correctly. This is very important to understand 100%, otherwise you'll have problems with fractions and equations.
1. Fractions: You can multiply the numerator and denominator by the same number. This produces an equivalent fraction which is the same value as the original fraction.
2. Equations: You can multiply both sides by the same number.

In your case you broke both rules.
1. You multiplied only the numerator by 2 and omitted the 2 in the denominator. As a result the left side of the equation got multiplied by 4.
2. You multiplied only the left side of the equation - the right side remained 62.

Regarding your question about getting rid of the denominator using the 8 in the numerator - yes, that's what I would do. It's not wrong to multiply the equation by 2, but any time you can just reduce fractions, you should do it - it leaves you with smaller numbers (similar to the idea in post 14).