# Write an expression for the nth term of the sequence. \frac{5}{2}, \frac{5}{4}, \frac{5}{8},...

## Question:

Write an expression for the nth term of the sequence.

{eq}\frac{5}{2} {/eq}, {eq}\frac{5}{4} {/eq}, {eq}\frac{5}{8} {/eq}, _____, _____, _____, ...

## The nth term of a Geometric Sequence

The general terms of a standard geometric sequence can be represented as:

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

Here {eq}a {/eq} is the first term and {eq}r {/eq} is the common ratio.

In the above geometric terms, it is very clear that the ratio of two successive terms is constant which known as the common ratio (r). It means each term of a geometric sequence is the same multiple of the preceding terms.

#### nth term of a geometric sequence:

$$\displaystyle a_{n} = a.r^{n-1}$$

Given:

The geometric sequence:

$$\displaystyle \frac{5}{2},\frac{5}{4},\frac{5}{8},..............$$

Here

{eq}\text{First term} (a) = \displaystyle \frac{5}{2} {/eq}

and

{eq}\displaystyle \text{Common ratio} (r) = \frac{\displaystyle \frac{5}{4}}{\displaystyle \frac{5}{2}} = \frac{\displaystyle \frac{5}{8}}{\displaystyle \frac{5}{4}} = \frac{1}{2} {/eq}

Now the {eq}n {/eq}th term of the given geometric sequence:

\begin{align} a_{n} & = a.r^{n-1} \\[0.2 cm] & = \frac{5}{2}.\left(\frac{1}{2}\right)^{n-1} \\[0.2 cm] & = \frac{5}{2} \times \frac{1}{2^{n-1} } \\[0.2 cm] a_{n} & = \frac{5}{2^{n} } \\[0.2 cm] \end{align}