Question

The common ratio in a geometric sequence is $\dfrac{3}{2}$ , and the fifth term is $1$ . How do you find the first three terms?

Answer 1

Hint: For solving this particular question you must know that for a geometric sequence with the first term ${a_1}$and the common ratio $r$ , the ${n^{th}}$ (or general) term is given by ${a_n} = {a_1} \times {r^{n - 1}}$ . With the help of this formula , we can find the first three terms.

Formula Used:Given a geometric sequence with the first term ${a_1}$ and the common ratio $r$ , the ${n^{th}}$ (or general) term is given by
 ${a_n} = {a_1} \times {r^{n - 1}}$ .

Complete step-by-step solution:
Given a geometric sequence with the first term ${a_1}$and the common ratio $r$ , the ${n^{th}}$ (or general) term is given by
 $ \Rightarrow {a_n} = {a_1} \times {r^{n - 1}}$ .
We know that ,
$ \Rightarrow {a_n} = {a_1} \times {r^{n - 1}}$
Where ,
$
  \Rightarrow {a_1} = ? \\
  \Rightarrow {a_5} = 1 \\
  \Rightarrow r = \dfrac{3}{2} \\
 $
Now , substitute theses values, we will get ,
$
   \Rightarrow {a_5} = {a_1} \times {\left( {\dfrac{3}{2}} \right)^{5 - 1}} \\
   \Rightarrow 1 = {a_1} \times {\left( {\dfrac{3}{2}} \right)^4} \\
 $
Hence ,
$
   \Rightarrow {a_1} = \dfrac{1}{{{{\left( {\dfrac{3}{2}} \right)}^4}}} \\
   \Rightarrow {a_1} = \dfrac{1}{{\dfrac{{81}}{{16}}}} = \dfrac{{16}}{{81}} \\
   \Rightarrow {a_2} = \dfrac{{16}}{{81}} \times \dfrac{3}{2} = \dfrac{8}{{27}} \\
   \Rightarrow {a_3} = \dfrac{8}{{27}} \times \dfrac{3}{2} = \dfrac{4}{9} \\
 $

Therefore, we have $\dfrac{{16}}{{81}},\dfrac{8}{{27}},\dfrac{4}{9}$ as our first three terms.
Additional Information:
A geometric sequence may be a sequence within which any element after the very first is obtained by multiplying the preceding element by a fixed number called the common ratio which is denoted by $r$ .
The common ratio is obtained by dividing any term by the preceding term:
$r = \dfrac{{{a_n}}}{{{a_{n - 1}}}}$ .

Note:
If the common ratio is:
-Negative: the result will alternate between positive and negative.
-Greater than one: there'll be an exponential growth towards infinity (positive).
-Less than minus one: there'll be an exponential growth towards infinity (positive and negative).
- Between one and minus one: there'll be a decay towards zero.
- Zero: the result will remain at zero .