<h3>Given</h3>
A geometric sequence such that ...
![a_2=-2\quad\text{and}\quad a_5=16](https://tex.z-dn.net/?f=a_2%3D-2%5Cquad%5Ctext%7Band%7D%5Cquad%20a_5%3D16)
<h3>Find</h3>
![a_{14}](https://tex.z-dn.net/?f=a_%7B14%7D)
<h3>Solution</h3>
We can use the ratio of the given terms to find the common ratio of the sequence, then use that to find the desired term from one of the given terms. We don't actually need the common ratio (-2). All we need is its cube (-8).
![a_2=a_1r^{(2-1)}=a_1r^1\\a_5=a_1r^{(5-1)}=a_1r^4\\a_{14}=a_1r^{(14-1)}=a_1r^13=a_5r^9\\\\\dfrac{a_5}{a_2}=\dfrac{a_1r^4}{a_1r^1}=r^3=\dfrac{16}{-2}=-8\\\\r^9=\left(r^3\right)^3=(-8)^3=-512\\a_{14}=a_5(-512)\\\\a_{14}=-8192](https://tex.z-dn.net/?f=a_2%3Da_1r%5E%7B%282-1%29%7D%3Da_1r%5E1%5C%5Ca_5%3Da_1r%5E%7B%285-1%29%7D%3Da_1r%5E4%5C%5Ca_%7B14%7D%3Da_1r%5E%7B%2814-1%29%7D%3Da_1r%5E13%3Da_5r%5E9%5C%5C%5C%5C%5Cdfrac%7Ba_5%7D%7Ba_2%7D%3D%5Cdfrac%7Ba_1r%5E4%7D%7Ba_1r%5E1%7D%3Dr%5E3%3D%5Cdfrac%7B16%7D%7B-2%7D%3D-8%5C%5C%5C%5Cr%5E9%3D%5Cleft%28r%5E3%5Cright%29%5E3%3D%28-8%29%5E3%3D-512%5C%5Ca_%7B14%7D%3Da_5%28-512%29%5C%5C%5C%5Ca_%7B14%7D%3D-8192)
Answer:
See below. <u><em>I assume that (x) = 8x2 - 7x + 3 is really (x) = 8x^2 - 7x + 3</em></u>
Step-by-step explanation:
Substitute the value of x given in f(x) into the equation f(x) = 8x^2 - 7x + 3
For example, f(0) would be f(0) = 8(0)^2 - 7(0) + 3. f(0) = 3
f(-2) would be f(-2) = 8(-2)^2 - 7(-2) + 3.
= 8*4 + 14 +3
= 32 + 17 therefore f(-2) = 49
<u>x</u> <u>f(x)</u>
-2 49
-1 18
0 3
1 4
2 21
x^3 - 3x^2 - 18x
Using the FOIL method, I arrived at my solution!
Answer:
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