Which formula can be used to find the nth term of a geometric sequence where the fifth term is 1/16 and the common ratio is 1/4?

Any geometric sequence can be expressed as:

a(n)=ar^(n-1), a=initial term, r=common ratio, n=term number

We are given that a(5)=1/16 and r=1/4 so we can say:

1/16=a(1/4)^(5-1)

1/16=a(1/4)^4

1/16=a/256

256/16=a

16=a

So the initial term is 16 so our formula is:

a(n)=16(1/4)^(n-1)

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dubeyshailja64 dubeyshailja64 · Answer 2 · 2022-01-20T14:24:15+01:00

A geometric sequence can be defined as a sequence in which the ratio of a term to its preceding term is equal.

The formula to express a geometric sequence is [tex]\rm T_{n} = ar^{n-1}[/tex], where a is the initial term and r is the common ratio between terms of the geometric sequence.

The formula to find [tex]\rm n^{th}[/tex] term of geometric sequence where fifth term is [tex]\dfrac{1}{16}[/tex] and common ratio is [tex]\dfrac{1}{4}[/tex] is:

[tex]\rm T_{n} = 16(\dfrac{1}{4})^ {(n-1)[/tex]

Explanation:

The formula to obtain terms of geometric sequence is [tex]\rm T_{n} = ar^{n-1}[/tex].

Given:

[tex]\begin{aligned} \rm T_{5} &= \dfrac{1}{16}\\\\r&=\dfrac{1}{4} \end[/tex]

Therefore value of a can be calculated as:

[tex]\begin{aligned} \rm T_{5}&= a(\dfrac{1}{4})^{(5-1)}\\\\T_n &= a(\dfrac{1}{4})^4\\\\T_5&= a(\dfrac{1}{256})\\\\\dfrac{1}{16}&= \dfrac{a}{256}\\\\a&= \dfrac{256}{16}\\\\a&=16\end[/tex]

Hence formula for the geometric sequence will be:

[tex]\rm T_{n} = 16(\dfrac{1}{4})^ {(n-1)[/tex]

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