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Find the $n^{th}$ term of the $G.P. 7, 49, 343, .......$

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Answer:

$7^{ n }$

Step by Step Explanation:

1. A geometric progression $(G.P.)$ is of the form, $a, ar, ar^2, ar^3, ......,$ where $a$ is called the first term and $r$ is called the common ratio of the $G.P.$
   The $n^{ th }$ term of a $G.P.$ is given by, $a_n= ar^{n-1}$
2. Here, the first term, $a = 7$
   The common ratio, $r = \dfrac{ a_{k+1} }{ a_k }$ where $k \ge 1$
   $\implies r = \dfrac{a_{1+1} }{ a_1 } = \dfrac{ a_2 }{ a_1 } = \dfrac{ 49 }{ 7 } = 7$
3. Now, we need to find the $n^{th}$ term of the $G.P., i.e. a_n.$
   $$\begin{align} & a_{ n } = ar^{ n-1 } \\ \implies & a_n = 7(7)^{ n-1 } \\ \implies & a_n = 7 ^{ 1 + n-1 } \\ \implies & a_n = 7^n \end{align}$$
4. Hence, the $n^{th}$ term of the given $G.P.$ is $7^{ n }$.

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