Find the fifth term of a geometric sequence whose fourth term is { \frac{32}{27} } and whose...

Question:

Find the fifth term of a geometric sequence whose fourth term is {eq}\frac{32}{27} {/eq} and whose seventh term is {eq}\frac{256}{729} {/eq}.

Geometric Series: Identifying Terms

A geometric series is a sequence of numbers in which the ratio of two successive numbers is always constant and known as the Common Ratio.

Let's consider a G.P.:

{eq}\displaystyle a, ar, ar^2, ar^3,...................ar^{n-1},................ {/eq}

Here 'a' = first term.

'r' = common ratio

'n' = number of terms.

nth term of a G.P.:

{eq}\displaystyle a_n = ar^{n-1} {/eq}

Sum of first 'n' numbers of the given G.P.:

{eq}\displaystyle S_n = a(\frac{r^n - 1}{r - 1}) {/eq}

Here {eq}\displaystyle r > 1 {/eq}

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Given:

The fourth term of the G.P.:

{eq}\displaystyle a_4 = ar^3 = \frac{32}{27} {/eq}

And

The seventh term of the G.P.:

{eq}\displaystyle a_7 =...