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SolomonZelman

  • one year ago

sequence notations. ~ arithmetic sequence.

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  1. SolomonZelman
    • one year ago
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    PLEASE DON'T TYPE. IF TYPED, DELETE. TNX

  2. SolomonZelman
    • one year ago
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    I am posting this thread to get some idea of the notations that people use when it comes to sequences. Some of you go nuts when they see these notations, and for this reason I want to make a basic recap of notations for you, if you do not understand something. Lets start from this sequence: `1, 4, 7, 10, 13, 16, 19 ....` (I am sure that you noticed that I am adding a 3 every time.) The first term is 1, the second term is 4, the third term is 7, the fourth term is 10, and on ... The notation for a first term of a sequence is \(\large a_1\). So, lets say someone asks you to label the \(a_1\) in this sequence, then you know that they are asking you to give the first term of the sequence. In this case, the first term (the \(a_1\)) is equivalent to 1. So you can write \(a_1=1\) THEN, here are some general notations. notation \(\Large |\) meaning \(\text{__________________________________________________}\) ` ` \(a_2\) \(\Large |\) 2nd term of a sequence \(\tiny~~~~~~~\) ` ` \(\tiny~~~~~\) \(\Large |\) \(\tiny~~~~~~\) ` ` \(a_3\) \(\Large |\) 3rd term of a sequence \(\tiny~~~~~~\) ` ` \(\tiny~~~~~\) \(\Large |\) \(\tiny~~~~~~\) ` ` \(a_4\) \(\Large |\) 4th term of a sequence \(\tiny~~~~\) ` ` \(\tiny~~~~~\) \(\Large |\) \(\tiny~~~~~~\) ` ` \(a_5\) \(\Large |\) 5th term of a sequence \(\tiny~~~~~~~~\) ` ` \(\tiny~~~~~\) \(\Large |\) \(\tiny~~~~~~\) ` ` \(\bf \huge ...\) \(\Large |\) \(\bf \huge ...\) \(\tiny~~~~~~\) ` ` \(\tiny~~~~~\) \(\Large |\) \(\tiny~~~~~~\) ` ` \(a_n\) \(\tiny~~~~~~~~~~~\) \(\Large |\) \(\large \rm some ~~{\LARGE \bf n}th~~term~~of~~a~~sequence \) \(\tiny~~~~~~~\) ` ` \(\tiny~~~~~\) \(\Large |\) \(\tiny~~~~~~\) ` ` \(\text{__________________________________________________}\) ` ` NOW, going back to our sequence, `1, 4, 7, 10, 13, 16, 19 ....` can you label the \(a_6\) ? (In your head)

  3. SolomonZelman
    • one year ago
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    Now, that we got an understanding of what \(a_1\) , \(a_2\) and any \(a_n\) means, we can go on. We will stay with the same example for your convinience `1, 4, 7, 10, 13, 16, 19 .... ` (this is our example of a sequence) ------------------------------------------------------- Here, you add 3 every time, to get to the next term (right?) ((Not in every sequence or pattern you add a number to get to the next term, but in this case it is so. )) This number that you add to get to the next term is called the "common difference" and is denoted with a letter d. In our case, it is true that whenever you want to find some \(a_n\) in our sequence, you would add a common difference (d), to the term before \(a_n\). The term before \(a_n\) (some nth term)\(,\) can be defined as \(a_{n-1}\) ('n-1' th term) \(.\) (( Do you agree? For example, 5th term is right before the 6th term\(.\) 8th term is before 9th term. And same way, 'n-1'th term is right before the 'n'th term.)) So, this is why the "common difference" (d), has the following definition: \(\large {\rm d}=a_{n}~-~a_{n-1}\) ((That is, that to find the common difference d, you need to subtract some 'n-1'th term from 'n'th term. Hope this is clear.))

  4. SolomonZelman
    • one year ago
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    You can use the d to find any term in a sequence. Lets introduce a different example this is the example: `-3, 4, 11, 18 .... ` In this case you can tell we are not multiplying to find the next, term because this sequence would be growing much faster if it was indeed multiplication. This is probably an addition, but lets go ahead an verify that it is so. If this is an addition pattern (i\(\small.\)e\(\small.\) you add some number d to obtain the next term), then all term would satisfy the definition: \({\rm d}=a_{n}~-~a_{n-1}\) lets plug in 2 3 and 4 for n \(\color{red}{\rm pay~~close~~ATTENTION~~to~~every~~step~~please!!}\) \(\large a_{n-1}~-~a_{n}~~~~~~~{\small( \rm now,~I~am~plugging~n=4)}\\[0.5em] \large =a_{4}~-~a_{4-1}=a_4-a_3=18-11=7\) So the difference between the 4th and 3rd term is 7. Now, I am going to check the difference when n=3, and that is going to be between 3rd and 2nd term) \(\large a_{3}~-~a_{3-1}=a_3-a_2=11-4=7\) and lastly I will plug in n=2 \(\large a_{2}~-~a_{2-1}=a_2-a_1=4-(-3)=4+3=7\) \((\)you have probably recalled the rule \(\color{blue}{\rm a--b=a+b}\) \()\) So, you can see that the difference between any \(a_{n}\) and \(a_{n-1}\) is (in our example) equal to 7.

  5. SolomonZelman
    • one year ago
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    if you wanted to get from \(a_1\) to \(a_2\) you would add the common difference once. THAT IS: \(\color{green}{a_1+{\rm d}=a_2}\) if you wanted to get from \(a_1\) to \(a_3\) you would add the common difference twice (1 less than a 3). THAT IS: \(\color{green}{a_1+2{\rm d}=a_3}\) if you wanted to get from \(a_1\) to \(a_4\) you would add the common difference 3 times (1 less than a 4). THAT IS: \(\color{green}{a_1+{\rm 3d}=a_4}\) if you wanted to get from \(a_1\) to \(a_5\) you would add the common difference 4 times (1 less than a 5). THAT IS: \(\color{green}{a_1+{\rm 4d}=a_5}\) and same way, if you wanted to get from \(a_1\) to \(a_n\) you would then add the common difference 'n-1' times (1 less than a 'n') THAT IS: \(\color{green}{a_1+(n-1){\rm d}=a_n}\)

  6. SolomonZelman
    • one year ago
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    So from here we conclude that \(\color{green}{a_n=a_1+(n-1){\rm d}}\) (this formula is now yours)

  7. SolomonZelman
    • one year ago
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    And same way if you wanted to find \(a_n\) from some \(a_k\) term, then you would apply \(\color{green}{a_n=a_k+(n-k){\rm d}}\) (Why this is so is an intellectual exercise for you. You can always ask a question on this site or message me if you want to know.)

  8. SolomonZelman
    • one year ago
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    you should be able to confirm at this point that. 1. \(a_n\) is some nth term in a sequence 2. \(\rm d\) is the common difference in a sequence

  9. SolomonZelman
    • one year ago
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    \({{\Huge\bbox[5pt, lightyellow ,border:2px solid black ]{ \color{lightyellow}{\frac{\frac{\color{black}{\rm NOTE:~~~the~~formulas~~I~~applied~~here~~only } }{ \color{black}{\rm apply~~to~~an~~arithmetic~~sequence.~~(This }~~~~ } }{ \frac{\color{black}{\rm whole~~time~~we~~were~~exploring~~arithmetic } }{ \color{black}{\rm sequence.~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ ~~} } }} }} }\)

  10. SolomonZelman
    • one year ago
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    The only knowledge that applies to every sequence is the fact that ` (1) ` \(a_1\) is the 1st term of a sequence \(a_2\) is the 2nd term of a sequence \(a_3\) is the 3rd term of a sequence --------------------------------- ` (2) ` \(a_n\) is some nth term of a sequence \(a_{n-1}\) is some 'n-1'th term of a sequence ---------------------------------- \(\bf \huge \color{red}{ BUT}\) ` (1) ` \(\rm d\) (the "common difference" between all terms) doesn't exist/apply in other sequences, only in geometric sequences. --------------------------------------- ` (2) ` And these formulas apply only to ARITHMETIC sequence, NOT OTHERS. ~ \(a_{n}=a_{1}+{\rm d}(n-1) \) ~ \(a_{n}=a_{k}+{\rm d}(n-k) \) ~ \({\rm d}=a_{n}-a_{n-1} \) ---------------------------------------

  11. SolomonZelman
    • one year ago
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    \(\large{\bbox[5pt, yellow ,border:2px solid black ]{ \rm Question~~\text{#}1 }}\) The first term of the arithmetic sequence is 5 and the 3rd term of this sequence is 11. What is the 56th term? \(\large{\bbox[5pt, lightcyan ,border:2px solid black ]{ \rm Answer~~\text{#}1 }}\) \(\large \rm \color{blue}{given:} \\[0.5em]\) 1) \(\large\color{black}{ \displaystyle a_1=5 }\) and \(\large\color{black}{ \displaystyle a_3=11 }\) 2) formula, \(\color{black}{ \displaystyle a_n=a_1+{\rm d}(n-1) }\) \(\large \rm \color{blue}{solution :} \\[0.5em]\) We are going to use the formula \(\color{black}{ \displaystyle a_n=a_1+{\rm d}(n-1) }\) (we know the 3rd term and the 1st, so lets work the d with which we would find the 3rd term. Working the formula backwards) \(\color{black}{ \displaystyle a_n=a_1+{\rm d}(n-1) }\) plugging n=3 \(\color{black}{ \displaystyle a_3=a_1+{\rm d}(3-1) }\) (this is also true, when we are looking for the 3rd term) we know our \(a_1=5\) and \(a_3=11\) so, \(\color{black}{ \displaystyle11=5+{\rm d}(3-1) }\) now we can solve for d algebraically. \(\color{black}{ \displaystyle11\color{red}{-5}=5+{\rm d}(3-1)\color{red}{-5} }\) \(\color{black}{ \displaystyle11\color{red}{-5}=\cancel{~5~}+{\rm d}(3-1)\cancel{\color{red}{-5}} }\) \(\color{black}{ \displaystyle 6={\rm d}(3-1) }\) \(\color{black}{ \displaystyle 6=2{\rm d} }\) \(\color{black}{ \displaystyle 6\color{red}{\div 2}=2{\rm d} \color{red}{\div 2} }\) \(\color{black}{ \displaystyle 6\color{red}{\div 2}=\cancel{2}{\rm d} \color{red}{\div }\cancel{\color{red}{ 2} }}\) \(\color{black}{ \displaystyle 3={\rm d} }\) We know all of the information, we got the common difference, but the initial question was to find the 56th term of this sequence. Let's apply our formula again. \(\color{black}{ \displaystyle a_n=a_1+{\rm d}(n-1) }\) we know \({\rm d}=3\) and \(a_1\)=5. That is all we need. \(\color{black}{ \displaystyle a_{56}=a_1+{\rm d}(56-1) }\) (plugged 56 for n, since we are looking for 56th term) \(\color{black}{ \displaystyle a_{56}=5+{\rm 3}(56-1) }\) (plugged in 5 for \(a_1\) and 3 for d, because we know that is true) \(\color{black}{ \displaystyle a_{56}=5+{\rm 3}(56-1) }\) \(\color{black}{ \displaystyle a_{56}=5+{\rm 3}(55) }\) \(\color{black}{ \displaystyle a_{56}=5+165 }\) \(\color{black}{ \displaystyle a_{56}=170 }\) `----------ALTERNATIVELY----------` Also, we could have found the 56th term using the 3rd term, and here we would apply the formula: \(\color{black}{ \displaystyle a_n=a_k+{\rm d}(n-k) }\) (where k is smaller than n) We are starting from 3rd term, so instead of \(a_k\) we have \(a_3\). (because k=3 in this case) \(\color{black}{ \displaystyle a_n=a_3+{\rm d}(n-3) }\) we know \({\rm d}=3\) and \(a_3=11\), so \(\color{black}{ \displaystyle a_n=11+{\rm 3}(n-3) }\) Now, we are looking for the 56th term, so we plug in n=56: \(\color{black}{ \displaystyle a_{56}=11+{\rm 3}(56-3) }\) \(\color{black}{ \displaystyle a_{56}=11+{\rm 3}(53) }\) \(\color{black}{ \displaystyle a_{56}=11+159 }\) \(\color{black}{ \displaystyle a_{56}=170 }\) `--------------------------------------` So whether we apply \(\color{black}{ \displaystyle a_n=a_k+{\rm d}(n-k) }\) or \(\color{black}{ \displaystyle a_n=a_1+{\rm d}(n-1) }\) [[which is another version of the first formula with \(a_k\), except that we start from the first term, from \(a_1\)]], we then get the same result - the same answer. \(\large \rm \color{blue}{Answer:} \\[0.5em]\) \(\color{black}{ \displaystyle a_{56}=170 }\)

  12. SolomonZelman
    • one year ago
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    (I might post more examples, but that will happen later)

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