Difference between revisions of "8A F11 Q8"
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\end{array}</math> | \end{array}</math> | ||
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− | and | + | |and |
|-style = "text-align:center" | |-style = "text-align:center" | ||
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− | <math >\begin{array}{rcl} | + | <math>\begin{array}{rcl} |
− | A_{15} &= &27 + (-4)(15 - 1)\\ | + | A_{15} &=& 27 + (-4)(15 - 1)\\ |
− | & =& 27 - 56\\ | + | & = &27 - 56\\ |
& =& -39 | & =& -39 | ||
\end{array}</math> | \end{array}</math> | ||
|} | |} | ||
+ | |||
+ | {| class="mw-collapsible mw-collapsed" style = "text-align:left;" | ||
+ | ! Final Answer: | ||
+ | |- | ||
+ | |<math>S_{10} = -90, A_{15} = -39</math> | ||
+ | |} | ||
+ | [[8AF11Final|<u>'''Return to Sample Exam</u>''']] |
Latest revision as of 13:54, 13 May 2015
Question: Given a sequence use formulae to compute and .
Foundations |
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1) Which of the 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 and , and d = |
3) Since we have a value for d, we want to use the formula for that involves d. |
Solution:
Step 1: |
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The formula for 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, , and |
Step 2: |
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Now we determine by finding d. To do this we use the formula with n = 2, , and. This yields d = -4. |
Step 3: |
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Now we have d, and we can use the same formula for to get and . Using these formulas with the appropriate values will yield |
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and |
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Step 4: |
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Since we found in the last step, and we found the necessary pieces, we find by using the formula |
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Final Answer: |
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