Question

Insert two numbers between 3 and 81 so that the resulting sequence forms a GP.

Answer 1

Hint: Assume the four terms of the GP as $a$, $ar$, $a{{r}^{2}}$ and $a{{r}^{3}}$ where ‘a’ is the first term and ‘r’ is the common ratio. Now, consider 3 as the first term and 81 as the last term using which to find the value of r by substituting the value a = 3 in the relation $a{{r}^{3}}=81$. Once the value of r is found, substitute it in the second and third term to get the answer.

Complete step by step answer:
Here we have been provided with numbers 3 and 81 and it is asked to determine two numbers between 3 and 81 such that the sequence of combined four terms forms a GP.
Now, four terms in a GP can be assumed as $a$, $ar$, $a{{r}^{2}}$ and $a{{r}^{3}}$ where ‘a’ is the first term and ‘r’ is the common ratio. So considering 3 as the first term and 81 as the fourth term we have a = 3 and $a{{r}^{3}}=81$. Therefore we have,
$\begin{align}
  & \Rightarrow 3\times {{r}^{3}}=81 \\
 & \Rightarrow {{r}^{3}}=27 \\
 & \Rightarrow {{r}^{3}}={{3}^{3}} \\
\end{align}$
Taking cube root both the sides we get,
$\Rightarrow r=3$
Substituting the values of a and r in the assumed expression for the second and third term we get,
(1) For the second term we have,
$\begin{align}
  & \Rightarrow ar=3\times 3 \\
 & \therefore ar=9 \\
\end{align}$
(2) For the third term we have,
$\begin{align}
  & \Rightarrow a{{r}^{2}}=3\times {{3}^{2}} \\
 & \Rightarrow a{{r}^{2}}=3\times 9 \\
 & \therefore a{{r}^{2}}=27 \\
\end{align}$
Hence, the two terms that must be included between 3 and 81 are 9 and 27 in the same order.

Note: Never assume the terms of a GP as different variables otherwise you may get confused in several variables, so it is better to work with two variables a and r only. In case of an AP we have to assume the terms as a, (a + d), (a + 2d),… and so on where a is the first term and d is the common difference. Also remember the formulas for the ${{n}^{th}}$ term and the sum of n terms of these important sequences.