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Erwin Kalvelagen · Answer 1 · 2023-10-30T12:56:44+0000

Given:

$\begin{gathered} a3=36 \\ \\ r=-(3)/(4) \end{gathered}$

Required:

To find the rule for the nth term. Then graph the first six terms of the sequence.

Step-by-step explanation:

The nth term of the geometric sequence is given by

Here

$\begin{gathered} a_n=36 \\ n=3 \\ r=-(3)/(4) \end{gathered}$

Now we have to find the value of a,

$\begin{gathered} 36=a(-(3)/(4))^(3-1) \\ \\ 36=a(-(3)/(4))^2 \\ \\ 36=a((9)/(16)) \\ \\ a=(576)/(9) \\ \\ a=64 \end{gathered}$

The first term is 64, therefore the formula is

Now the first six terms of the sequence are

$\begin{gathered} a1=64(-(3)/(4))^0 \\ \\ a1=64 \\ \\ a2=64(-(3)/(4))^1 \\ \\ =64(-(3)/(4)) \\ \\ =-48 \\ \\ a3=64(-(3)/(4))^2 \\ \\ =64((9)/(16)) \\ \\ =36 \\ \\ a4=64(-(3)/(4))^3 \\ \\ =64(-(27)/(64)) \\ \\ =-27 \\ \\ a5=64(-(3)/(4))^4 \\ \\ =64((81)/(256)) \\ \\ =(81)/(4) \\ \\ a6=64(-(3)/(4))^5 \\ \\ =-(243)/(16) \end{gathered}$

Therefore the first 6th terns are

Final Answer:

The rule for the nth term is

The first 6th terns are :

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