Chapter 12: Q 15. (page 824)

Show that each sequence is geometric. Then find the common ratio and write out the first four terms $(t_{n}) = \{\frac{3^{n - 1}}{2^{n}}\}$

Short Answer

Expert verified

The common ratio is $\frac{3}{2}$ and the first four terms are $\frac{1}{2}, \frac{3}{4}, \frac{9}{8}, \frac{27}{16}$ .

Step by step solution

Step 1. Given information.

We have given sequence $t_{n} = \{\frac{3^{n - 1}}{2^{n}}\}$

The $n t h$ and $(n - 1) t h$ terms are

role="math" localid="1646929838851" $t_{n} = \{\frac{3^{n - 1}}{2^{n}}\} t_{n - 1} = \{\frac{3^{(n - 1) - 1}}{2^{n - 1}}\}$

Step 2. To find the common ratio.

We have,

$r = \frac{t_{n}}{t_{n - 1}} r = \frac{\frac{3^{n - 1}}{2^{n}}}{\frac{3^{(n - 1) - 1}}{2^{n - 1}}} r = \frac{3^{n - 1 - n + 2}}{2^{n - n + 1}} r = \frac{3}{2}$

Since the ratio of the successive term is the non zero constant $\frac{3}{2}$ therefore $(t_{n})$ is a geometric sequence with common ratio is $\frac{3}{2}$ .

Step 3. To find the terms.

We have,

$t_{1} = \frac{3^{1 - 1}}{2^{1}} t_{1} = \frac{1}{2} t_{2} = \frac{3^{2 - 1}}{2^{2}} t_{2} = \frac{3}{4} t_{3} = \frac{3^{3 - 1}}{2^{3}} t_{3} = \frac{9}{8} t_{4} = \frac{3^{4 - 1}}{2^{4}} t_{4} = \frac{27}{16}$

Thus, the first four terms are $\frac{1}{2}, \frac{3}{4}, \frac{9}{8}, \frac{27}{16}$

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Show that each sequence is geometric. Then find the common ratio and write out the first four terms $(t_{n}) = \{\frac{3^{n - 1}}{2^{n}}\}$

Short Answer

Step by step solution

Step 1. Given information.

Step 2. To find the common ratio.

Step 3. To find the terms.

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Show that each sequence is geometric. Then find the common ratio and write out the first four termstn=3n-12n

Short Answer

Step by step solution

Step 1. Given information.

Step 2. To find the common ratio.

Step 3. To find the terms.

One App. One Place for Learning.

Most popular questions from this chapter

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Show that each sequence is geometric. Then find the common ratio and write out the first four terms $(t_{n}) = \{\frac{3^{n - 1}}{2^{n}}\}$