nth Term of an Geometric Sequence – Examples and Practice

In this case, we know the values of the first term and the common ratio directly. Then, we can see the following information:

• First term $latex a=5$
• Common ratio: $latex r=2$
• Position of term: $latex n=4$

Now, we apply the formula for the nth term of a geometric sequence with the given values:

$latex a_{n}=ar^{n-1}$

$latex a_{4}=(5)(2)^{4-1}$

$latex a_{4}=(5)(2)^3$

$latex a_{4}=(5)(8)$

$latex a_{4}=40$

As in the previous example, here we also know the values of the first term and the common ratio. Then, we have:

• $latex a=3$
• $latex r=3$
• $latex n=6$

When we use the formula for the nth term of a geometric sequence, we have:

$latex a_{n}=ar^{n-1}$

$latex a_{6}=(3)(3)^{6-1}$

$latex a_{6}=(3)(3)^5$

$latex a_{6}=(3)(243)$

$latex a_{6}=729$

We can observe the following values:

• $latex a=6$
• $latex r=-2$
• $latex n=6$

We see that we have a negative common ratio. However, the formula for the nth term applies in any case:

$latex a_{n}=ar^{n-1}$

$latex a_{6}=(6)(-2)^{6-1}$

$latex a_{6}=(6)(-2)^5$

$latex a_{6}=(6)(-32)$

$latex a_{6}=-192$

In this example, we know the value of the first term, but we don’t know the value of the common ratio of the sequence.

We can find the value of the common ratio by dividing any term by its previous term. For example, $latex \frac{6}{3}=2$. Therefore, we have:

• $latex a=3$
• $latex r=2$
• $latex n=6$

Now, we use these values in the formula for the nth term:

$latex a_{n}=ar^{n-1}$

$latex a_{6}=(3)(2)^{6-1}$

$latex a_{6}=(3)(2)^5$

$latex a_{6}=(3)(32)$

$latex a_{6}=96$

We have to start by finding the value of the common ratio. Then, we have $latex \frac{15}{-5}=-3$:

• $latex a=-5$
• $latex r=-3$
• $latex n=6$

Now, we can apply the formula for the nth term:

$latex a_{n}=ar^{n-1}$

$latex a_{6}=(-5)(2)^{6-1}$

$latex a_{6}=(-5)(-3)^5$

$latex a_{6}=(-5)(-243)$

$latex a_{6}=1215$

The common ratio of the progression is $latex \frac{4}{2}=2$. Then, we have the following values:

• $latex a=2$
• $latex r=2$
• $latex n=10$

Now, we can apply the formula for the nth term with these values:

$latex a_{n}=ar^{n-1}$

$latex a_{10}=(2)(2)^{10-1}$

$latex a_{10}=(2)(2)^9$

$latex a_{10}=(2)(512)$

$latex a_{10}=1024$

The value of the common ratio of the sequence is $latex \frac{-3}{1}=-3$. Then, we have the following:

• $latex a=1$
• $latex r=-3$
• $latex n=7$

Now, we apply the formula for the nth term with the given values:

$latex a_{n}=ar^{n-1}$

$latex a_{7}=(1)(-3)^{7-1}$

$latex a_{7}=(1)(-3)^6$

$latex a_{7}=(1)(729)$

$latex a_{7}=729$

We find the common ratio by dividing a term by its previous term: $latex \frac{3}{2}$. Therefore, we have:

• $latex a=2$
• $latex r=\frac{3}{2}$
• $latex n=9$

Now, we use the formula for the nth term:

$latex a_{n}=ar^{n-1}$

$$a_{9}=(2)\left(\frac{3}{2}\right)^{9-1}$$

$$a_{9}=(2)\left(\frac{3}{2}\right)^8$$

$$a_{9}=(2)\left(\frac{6561}{256}\right)$$

$$a_{9}=\frac{6561}{128}$$

$$a_{9}=51\frac{33}{128}$$

The common ratio of the sequence is $latex \frac{-54}{81}=-\frac{2}{3}$. Then, we have the following values:

• $latex a=81$
• $latex r=-\frac{2}{3}$
• $latex n=8$

Applying the formula for the nth term with the given values, we have:

$latex a_{n}=ar^{n-1}$

$$a_{8}=(81)\left(-\frac{2}{3}\right)^{8-1}$$

$$a_{8}=(81)\left(-\frac{2}{3}\right)^7$$

$$a_{8}=(81)\left(-\frac{128}{2187}\right)$$

$$a_{8}=-\frac{128}{27}$$

$$a_{8}=-4\frac{20}{27}$$

The common ratio of the sequence is $latex \frac{2}{5}\div 2= \frac{1}{5}$. Therefore, we have the following:

• $latex a=2$
• $latex r=\frac{1}{5}$
• $latex n=5$

Now, we use these values in the formula for the nth term of a geometric sequence:

$latex a_{n}=ar^{n-1}$

$$a_{5}=(2)\left(\frac{1}{5}\right)^{5-1}$$

$$a_{5}=(2)\left(\frac{1}{5}\right)^4$$

$$a_{5}=(2)\left(\frac{1}{625}\right)$$

$$a_{5}=\frac{2}{625}$$