Question

How do you find the missing terms in the geometric sequence $32,\\_,72,\\_,162$ ?

Answer 1

When we start the problem, we first write the general expression for the ${{n}^{th}}$ term of the geometric sequence which is ${{a}_{n}}=a{{r}^{n-1}}$ . We are provided with the data that the first, third and fifth terms are known. So, we put, $a=32$ , ${{a}_{3}}=72$ and then $n=3$ in the general expression to get the constant ratio, which can be used to find the missing terms.

Complete step by step solution:
A sequence is an enumerated collection of objects, especially numbers, in which repetitions are allowed and in which the order of objects matters. A sequence may be finite or infinite depending on the number of objects in the sequence. Sequences can be of various types such as arithmetic sequence, geometric sequence and so on. Sequences can be completely random as well. The ${{n}^{th}}$ term of a sequence is sometimes written as a function of n.
A geometric sequence is the one in which the ratio between the consecutive objects is a constant. It is called the common ratio of the geometric sequence. The general expression of the ${{n}^{th}}$ term will be,
${{a}_{n}}=a{{r}^{n-1}}....\left( 1 \right)$ where, a is the first term of the sequence and r is the common ratio.
In this problem, we are provided with the first, third and the fifth terms of the sequence. The first term is $32$ , the third term is $72$ and the fifth term is $162$ . This information gives us,
$a=32,{{a}_{3}}=72,{{a}_{5}}=162$
Putting $n=3$ in the equation $\left( 1 \right)$ ,
$\begin{aligned} & 72=\left( 32 \right){{r}^{3-1}} \\\ & \Rightarrow {{r}^{2}}=\dfrac{72}{32}=\dfrac{36}{16}={{\left( \dfrac{6}{4} \right)}^{2}} \\\ & \Rightarrow r=\dfrac{6}{4}=\dfrac{3}{2} \\\ \end{aligned}$
The second term will be,
$\begin{aligned} & {{a}_{2}}=\left( 32 \right){{\left( \dfrac{3}{2} \right)}^{2-1}} \\\ & \Rightarrow {{a}_{2}}=\left( 32 \right)\times \left( \dfrac{3}{2} \right) \\\ & \Rightarrow {{a}_{2}}=48 \\\ \end{aligned}$
The fourth term will be,
$\begin{aligned} & {{a}_{4}}=\left( 32 \right){{\left( \dfrac{3}{2} \right)}^{4-1}} \\\ & \Rightarrow {{a}_{4}}=\left( 32 \right)\times {{\left( \dfrac{3}{2} \right)}^{3}} \\\ & \Rightarrow {{a}_{4}}=108 \\\ \end{aligned}$

Therefore, we can conclude that the missing terms are $48,108$

Note: In the problems of sequence, we must be extra cautious while writing the general expression for the ${{n}^{th}}$ term as a lot of students make a mistake here. We can also solve this problem in another way. We know that the geometric mean of two numbers, say a and b is given by $\sqrt{ab}$ . As in these problems, we need to find the middle terms and the terms are in geometric progression, we can use $\sqrt{32\times 72}=48$ and $\sqrt{72\times 162}=108$ to get the missing terms.