Geometric Sequences – Examples and Practice Problems

First, we have to find the common ratio of the geometric progression. To do this, we divide a term by the previous term:

• $latex \frac{32}{16}=2$
• $latex \frac{16}{8}=2$
• $latex \frac{8}{4}=2$

Therefore, the common ratio is 2. To find the next term, we multiply the last term by the common ratio: $latex 32\times 2=64$.

We start by finding the common ratio for the geometric progression. Then, we divide each term by its previous term:

• $latex \frac{375}{75}=5$
• $latex \frac{75}{15}=5$
• $latex \frac{15}{3}=5$

We see that the common ratio is 5. We find the next term by multiplying the last term by the common ratio : $latex 375 \times 5=1875$.

Again, we start by finding the common ratio in the progression:

• $latex \frac{6}{12}=0.5$
• $latex \frac{12}{24}=0.5$
• $latex \frac{24}{48}=0.5$

In this case, we see that the common ratio is between 0 and 1, so the progression is slowing down. The next term in the geometric progression is $latex 6\times 0.5=3$.

We have the following values:

• First term: $latex a_{1}=3$
• Common ratio: $latex r=2$
• Position of term: $latex n=6$

Then, we can use the formula for geometric sequences with the given values:

$latex a_{n}=a_{1}(r^{n-1})$

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

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

$latex a_{6}=5(32)$

$latex a_{6}=160$

In this case, we have to use the formula of geometric progressions $latex a_{n}=a_{1}({{r}^{n-1}})$. Therefore, we have to identify the first term, the common reason and the position of the term:

• First term: $latex a_{1}=5$
• Common ratio: $latex r=3$
• Position of term: $latex n=12$

Now, we substitute this data into the formula:

$latex a_{n}=a_{1}({{r}^{n-1}})$

$latex a_{12}=5({{3}^{12-1}})$

$latex a_{12}=5({{3}^{11}})$

$latex a_{12}=5(177147)$

$latex a_{12}=885 735$

We see that we have a very large number. Geometric progressions tend to grow rapidly depending on the common proportion.

Again, we start by identifying the first term, the common ratio, and the position of the term to be used with the formula:

• First term: $latex a_{1}=8$
• Common ratio: $latex r=4$
• Position of term: $latex n=8$

Now, we use the formula with these values:

$latex a_{n}=a_{1}({{r}^{n-1}})$

$latex a_{8}=8({{4}^{8-1}})$

$latex a_{8}=8({{4}^{7}})$

$latex a_{8}=8(16384)$

$latex a_{8}=131072$

In this case, we have a decreasing geometric progression, so we expect the common ratio to be between 0 and 1:

• First term: $latex a_{1}=168$
• Common ratio: $latex r=0.5$
• Position of term: $latex n=10$

We use the formula to find the term 10:

$latex a_{n}=a_{1}({{r}^{n-1}})$

$latex a_{10}=168({{0.5}^{10-1}})$

$latex a_{10}=168({{0.5}^{9}})$

$latex a_{10}=168(0.001953)$

$latex a_{10}=0.328$

Similar to the previous example, here we have a decreasing geometric progression, so the common ratio must be between 0 and 1:

• First term: $latex a_{1}=540$
• Common ratio: $latex r=\frac{1}{3}$
• Possiion of term: $latex n=7$

We use these values to substitute in the formula:

$latex a_{n}=a_{1}({{r}^{n-1}})$

$latex a_{7}=540({{\left( \frac{1}{3}\right)}^{7-1}})$

$latex a_{7}=540({{\left( \frac{1}{3}\right)}^{6}})$

$latex a_{7}=540(0.0013717)$

$latex a_{7}=0.7407$

In this case, we know neither the value of the first term nor the common ratio. However, we can start by forming the following equations:

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

$latex 16=a_{1}(r^{3})~~~[1]$

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

$latex 128=a_{1}(r^{6})~~~[2]$

If we divide equation 2 by equation 1, we have:

$$\frac{128}{16}=\frac{a_{1}(r^{6})}{a_{1}(r^{3})$$

$latex 8=r^{3}$

$latex r=2$

If we consider the 7th term as the 1st term, the 11th term is now the 5th term:

• First term: $latex a_{1}=128$
• Common ratio: $latex r=2$
• Position of term: $latex n=5$

Using these values in the formula, we have:

$latex a_{n}=a_{1}(r^{n-1})$

$latex a_{5}=128(2^{5-1})$

$latex a_{5}=128(2^4)$

$latex a_{5}=128(16)$

$latex a_{5}=2048$

Then, the 11th term of the given sequence is 2048.

Similar to the previous example, we can find the common ratio by forming the following equations:

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

$latex 256=a_{1}(r^{2})~~~[1]$

$latex a_{8}=a_{1}(r^{8-1})$

$latex -8=a_{1}(r^{7})~~~[2]$

Now we divide them to obtain:

$$\frac{-8}{256}=\frac{a_{1}(r^{7})}{a_{1}(r^{2})$$

$$-\frac{1}{32}=r^{5}$$

$$r=-\frac{1}{2}$$

Considering the 8th term as the first term, the 14th term corresponds to the 7th term. Then:

• First term: $latex a_{1}=-8$
• Common ratio: $latex r=-\frac{1}{2}$
• Position of term: $latex n=7$

Using the formula, we have:

$latex a_{n}=a_{1}(r^{n-1})$

$$a_{7}=-8(-\frac{1}{2}^{7-1})$$

$$a_{7}=-8(-\frac{1}{2}^6)$$

$$a_{7}=-8(\frac{1}{64})$$

$$a_{7}=-\frac{1}{8}$$

Then, the 14th term of the given sequence is $latex -\frac{1}{8}$.