008A Sample Final A, Question 8 - Revision history

MathAdmin at 06:54, 26 May 2015

2015-05-26T06:54:31Z

MathAdmin at 19:36, 23 May 2015

2015-05-23T19:36:12Z

MathAdmin at 19:35, 23 May 2015

2015-05-23T19:35:56Z

MathAdmin at 19:34, 23 May 2015

2015-05-23T19:34:59Z

MathAdmin at 19:32, 23 May 2015

2015-05-23T19:32:59Z

MathAdmin: Created page with "'''Question: ''' Given a sequence $27, 23, 19, 15, \ldots$ use formulae to compute $S_{10}$ and $A_{15}$ . {| class="mw-collapsible mw-colla..."

2015-04-28T21:10:04Z

Created page with "'''Question: ''' Given a sequence <math> 27, 23, 19, 15, \ldots </math> use formulae to compute <math>S_{10}</math> and <math>A_{15}</math>. {| class="mw-collapsible mw-colla..."

New page

'''Question: ''' Given a sequence <math> 27, 23, 19, 15, \ldots </math> use formulae to compute <math>S_{10}</math> and <math>A_{15}</math>.

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations
|-
|1) Which of the <math>S_n</math> formulas should you use?
|-
|2) What is the common ratio or difference?
|-
|3) How do you find the values you need to use the formula?
|-
|Answer:
|-
|1) The variables in the formulae give a bit of a hint. The r stands for ratio, and ratios are associated to geometric series. This sequence is arithmetic, so we want the formula that does not involve r.
|-
|2) Take two adjacent terms in the sequence, say <math>A_1</math> and <math>A_2</math>, and d = <math>A_2 - A_1</math>
|-
|3) Since we have a value for d, we want to use the formula for <math>A_n</math> that involves d.
|}

Solution:

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
! Step 1:
|-
|The formula for <math>S_n</math> that involves a common difference, d, is the one we want. The other formula involves a common ratio, r. So we have to determine the value of n, <math>A_1</math>, and <math>A_n</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
! Step 2:
|-
|Now we determine <math>A_n</math> by finding d. To do this we use the formula <math>A_n = A_1 + d(n - 1)</math> with n = 2, <math>A_1 = 27</math>, and<math>A_2 = 23</math>. This yields d = -4.
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
! Step 3:
|-
|Now we have d, and we can use the same formula for <math>A_n</math> to get <math>A_{10}</math> and <math>A_{15}</math>. Using these formulas with the appropriate values will yield
|-style = "text-align:center"
|
<math>\begin{array}{rcl}
A_{15} &=& 27 + (-4)(15 - 1) \\
&=& 27 - 56\\
& =& -39
\end{array}</math>
|-
|and
|-style = "text-align:center"
|
<math>\begin{array}{rcl}
A_{10} &=& 27 + (-4)(10 - 1)\\
& =& 27 -36\\
& = &-9
\end{array}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
! Step 4:
|-
|Since we found <math>A_{15}</math> in the last step, and we found the necessary pieces, we find <math>S_{10}</math> by using the formula <math>S_{10} = \frac{10}{2}(27 + -9) = 5 (-18) = -90</math>
|- style = "text-align:center"
|
<math >\begin{array}{rcl}
A_{15} &= &27 + (-4)(15 - 1)\\
& =& 27 - 56\\
& =& -39
\end{array}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
! Final Answer:
|-
|<math>S_{10} = -90, A_{15} = -39</math>
|}
[[008A Sample Final A|<u>'''Return to Sample Exam</u>''']]

@@ Line 2: / Line 2: @@
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Foundations
+!Foundations: &nbsp;
 |-
 |1) Which of the <math>S_n</math> formulas should you use?
@@ Line 22: / Line 22: @@
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-! Step 1:
+!Step 1: &nbsp;
 |-
 |The formula for <math>S_n</math> that involves a common difference, d, is the one we want. The other formula involves a common ratio, r. So we have to determine the value of n, <math>A_1</math>, and <math>A_n</math>
@@ Line 28: / Line 28: @@
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-! Step 2:
+!Step 2: &nbsp;
 |-
 |Now we determine <math>A_n</math> by finding d. To do this we use the formula <math>A_n = A_1 + d(n - 1)</math> with n = 2, <math>A_1 = 27</math>, and<math>A_2 = 23</math>. This yields d = -4.
@@ Line 34: / Line 34: @@
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-! Step 3:
+!Step 3: &nbsp;
 |-
 |Now we have d, and we can use the same formula for <math>A_n</math> to get <math>A_{10}</math> and <math>A_{15}</math>. Using these formulas with the appropriate values will yield
@@ Line 56: / Line 56: @@
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-! Step 4:
+!Step 4: &nbsp;
 |-
 |Since we found <math>A_{15}</math> in the last step, and we found the necessary pieces, we find <math>S_{10}</math> by using the formula <math>S_{10} = \frac{10}{2}(27 + -9) = 5 (-18) = -90</math>
@@ Line 69: / Line 69: @@
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-! Final Answer:
+!Final Answer: &nbsp;
 |-
 |<math>S_{10} = -90, A_{15} = -39</math>
 |}
 [[008A Sample Final A|<u>'''Return to Sample Exam</u>''']]

@@ Line 61: / Line 61: @@
 |-
+|
-<math >\begin{array}{rcl}
+::<math >\begin{array}{rcl}
 A_{15} &= &27 + (-4)(15 - 1)\\
 & =& 27 - 56\\

@@ Line 37: / Line 37: @@
 |-
 |Now we have d, and we can use the same formula for <math>A_n</math> to get <math>A_{10}</math> and <math>A_{15}</math>. Using these formulas with the appropriate values will yield
+|
 <math>\begin{array}{rcl}
 A_{15} &=& 27 + (-4)(15 - 1) \\
 &=& 27 - 56\\
@@ Line 46: / Line 46: @@
 |-
 |and
-|-style = "text-align:center"
+|-
+|
- <math>\begin{array}{rcl}
+::<math>\begin{array}{rcl}
 A_{10} &=& 27 + (-4)(10 - 1)\\
 & =& 27 -36\\
@@ Line 59: / Line 59: @@
 |-
 |Since we found <math>A_{15}</math> in the last step, and we found the necessary pieces, we find <math>S_{10}</math> by using the formula <math>S_{10} = \frac{10}{2}(27 + -9) = 5 (-18) = -90</math>
-|- style = "text-align:center"
+|
 <math >\begin{array}{rcl}

@@ Line 14: / Line 14: @@
 |1) The variables in the formulae give a bit of a hint. The r stands for ratio, and ratios are associated to geometric series. This sequence is arithmetic, so we want the formula that does not involve r.
 |-
-|2) Take two adjacent terms in the sequence, say <math>A_1</math> and <math>A_2</math>, and d = <math>A_2 - A_1</math>
+|2) We determine the common difference by taking two adjacent terms in the sequence, say <math>A_1</math> and <math>A_2</math>, and finding their difference <math> d = A_2 - A_1 = -4</math>
 |-
 |3) Since we have a value for d, we want to use the formula for <math>A_n</math> that involves d.