Question

What is a recursive formula for a geometric sequence?

Answer 1

Hint: A recursive formula is a relation between the preceding term of a sequence and the next term of the sequence. So it essentially defines the next term of a sequence using the previous term. A geometric sequence is the sequence of terms in which the ratio of the two consecutive terms is always a constant. The ${{n}^{th}}$ term is given by ${{a}_{n}}=a{{r}^{n-1}}$. From this formula, we can derive the required recursive relation for a geometric sequence.

Complete step-by-step solution:
We know that a geometric sequence is the sequence of terms in which the ratio of the two consecutive terms is always a constant. Therefore, if the first term of a geometric sequence is a, and the constant ratio of the geometric sequence is equal to r, then the next terms can be written as $ar,a{{r}^{2}},a{{r}^{3}}........$. Observing this sequence of the terms of the geometric sequence, we can generalise the ${{n}^{th}}$term of the sequence as
$\Rightarrow {{a}_{n}}=a{{r}^{n-1}}.......\left( i \right)$
But we also know that a recursive formula for a sequence is a relation between the next term and the previous term. Therefore, for writing the recursive relation for the geometric sequence, we need to define the next term with respect to the previous term. This means that we have to define the ${{n}^{th}}$ term with respect to the ${{\left( n-1 \right)}^{th}}$ term. For the ${{\left( n-1 \right)}^{th}}$ term, we substitute $n=n-1$ in the equation (i) to get
$\begin{align}
  & \Rightarrow {{a}_{n-1}}=a{{r}^{n-1-1}} \\
 & \Rightarrow {{a}_{n-1}}=a{{r}^{n-2}}........\left( ii \right) \\
\end{align}$
Dividing the equation (i) by (ii) we get
\[\begin{align}
  & \Rightarrow \dfrac{{{a}_{n}}}{{{a}_{n-1}}}=\dfrac{a{{r}^{n-1}}}{a{{r}^{n-2}}} \\
 & \Rightarrow \dfrac{{{a}_{n}}}{{{a}_{n-1}}}=\dfrac{{{r}^{n-1}}}{{{r}^{n-2}}} \\
\end{align}\]
Using the laws of exponents, we can write the above equation as
\[\begin{align}
  & \Rightarrow \dfrac{{{a}_{n}}}{{{a}_{n-1}}}={{r}^{\left( n-1 \right)-\left( n-2 \right)}} \\
 & \Rightarrow \dfrac{{{a}_{n}}}{{{a}_{n-1}}}={{r}^{n-1-n+2}} \\
 & \Rightarrow \dfrac{{{a}_{n}}}{{{a}_{n-1}}}={{r}^{1}} \\
 & \Rightarrow \dfrac{{{a}_{n}}}{{{a}_{n-1}}}=r \\
\end{align}\]
Multiplying both sides by \[{{a}_{n-1}}\], we get
\[\Rightarrow {{a}_{n}}={{a}_{n-1}}\left( r \right)\]
Hence, the recursive formula for the geometric sequence is \[{{a}_{n}}={{a}_{n-1}}\left( r \right)\].

Note: A recursive relation is a relation between two consecutive terms. It is not necessary to define it as the expression for the present term with respect to the previous term. We can define it as the relation between the next term and the present term also. Therefore, the recursive relation for a geometric series can also be written as ${{a}_{n+1}}={{a}_{n}}\left( r \right)$.