Answered

What is the 6th term of the Geometric Sequence where a1 = 128 and a3=8?


Explicit formula:

a_n(r)^n-1


Please can you explain step-by-step to me how to find this?

Verified Answer :

Answer:

The 6th term of the sequence is 0.125.

Step-by-step explanation:

Given : Geometric Sequence where\(a_1=128\) and \(a_3=8\)

To find : The 6th term.

Solution :

The formula of nth term of geometric sequence is

\(a_n=a_1\times r^{n-1}\)  ........[1]

We have given,

\(a_1=128\)

\(a_3=8\)

To find ratio we put n=3 in [1]

\(a_3=a_1\times r^{3-1}\)

\(8=128\times r^{2}\)

\(\frac{8}{128}= r^{2}\)

\(0.0625= r^{2}\)

\(r=\sqrt{0.0625}\)

\(r=0.25\)

Now, we know first term \(a_1=128\) and common ratio \(r=0.25\)

Put n=6 in [1] to get 6th term of the sequence is

\(a_6=a_1\times r^{6-1}\)

\(a_6=128\times (0.25)^{5}\)

\(a_6=128\times 0.000976\)

\(a_6=0.125\)

Therefore, The 6th term of the sequence is 0.125.

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