Identify the sequence graphed below and the average rate of change from n = 1 to n = 3. coordinate plane showing the point 1, 8, point 2, 4, point 4, 1, and point 5, .5

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Identify the sequence graphed below and the average rate of change from n = 1 to n = 3. coordinate plane showing the point 1, 8, point 2, 4, point 4, 1, and point 5, .5

Algebra
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\[a_1=8=2^3\] \[a_2=4=2^2\] \[a_4=1=2^0\] \[a_5=0.5=2^{-1}\] this one's not in there but you should see the pattern well enough to write the sequence \[a_3=2=2^1\]
@peachpi those look like none of my question choices. im so confued

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the pattern between points is multiplying by ½, so that's the common ratio, r. The sequence is geometric. The formula for a geometric sequence is \[a_n=a_1 r^{n-1}\]
r is ½. a_1 is the first term of the sequence
Identify the sequence graphed below and the average rate of change from n = 1 to n = 3. coordinate plane showing the point 1, 8, point 2, 4, point 4, 1, and point 5, .5 A) an = 8(one half)n − 1; average rate of change is −3 B) an = 10(one half)n − 1; average rate of change is 3 C)an = 8(one half)n − 1; average rate of change is 3 D) an = 10(one half)n − 1; average rate of change is −3 Those are my answer choices @peachpi
ok. use the formula I just put up to get a formula for a_n
|dw:1435250004835:dw|
okay then then what
@peachpi what would the correct answer be
your terms represent a geometric sequence, whose first term is 8 and the constant is 1/2. Now the general term of a geometric sequence is given by the subsequent formula: \[\Large {a_n} = {a_1}{q^{n - 1}}\] please substitute q=1/2 and a_1=8, what do you get?
4?
hint: \[\Large {a_n} = {a_1}{q^{n - 1}} = 8 \times {\left( {\frac{1}{2}} \right)^{n - 1}}\] am I right?
yes
ok! that is your answer!
A) an = 8(one half)n − 1; average rate of change is −3 B) an = 10(one half)n − 1; average rate of change is 3 C)an = 8(one half)n − 1; average rate of change is 3 D) an = 10(one half)n − 1; average rate of change is −3 out of these which woud it be?
for example if we set n=2, we get: \[\Large {a_2} = 8 \times {\left( {\frac{1}{2}} \right)^{2 - 1}} = 8 \times \frac{1}{2} = 4\] similarly for other terms
if we set n=3, we get: \[\Large {a_3} = 8 \times {\left( {\frac{1}{2}} \right)^{3 - 1}} = 8 \times \frac{1}{{{2^2}}} = 8 \times \frac{1}{4} = 2\]
still confused for what the avg. rate of change would be
now the requested change rate, can be this:
- 3 or 3??
\[\large r = \frac{{{a_1} + {a_2} + {a_3} + {a_4}}}{4} = \frac{{8 + 4 + 2 + 0.5}}{4} = ...\]
3.625
that's right!
so the correct answer is C?
yes I think so!

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