Question

Find the ninth term of the G.P. \[1, 4, 16, 64,...\]

Answer 1

Hint: In this problem we need to find $9^{th}$ term of the G.P. for this first term should be analysed using the given information d is found by dividing second term from first term using formula of GP, each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio
A geometric progression (G.P) is of the form \[a,ar,a{r^2},a{r^3}....\]
The $n^{th}$ term of G.P is ${a_n}$
\[ \Rightarrow \] ${a_n}$ \[ = a{r^{n - 1}}\]
Where ${a_n}$ = last term of the G.P
              $a$ = first term of the G.P
              $r$ = common ratio
Hence, a geometric series with common ratio r and first term a
      $n^{th}$ term $ = a \times {r^{n - 1}}$
                ${a_n} = a{r^{n - 1}}$

Complete step-by-step answer:
Given G.P,
$ 1, 4, 16, 64 $
We conclude $a = 1$, because $a$ is the first term of the series
$r = \dfrac{4}{1} = 4$ i.e. common ratio.
To find $n^{th}$ term in a geometric series with common ratio ‘r’ and first term ‘a’ we have
${a_n} = a{r^{n - 1}}\,\,......\,eq.(i)$
${a_n} = {n^{th}}$ term of the series
$a$ = first term of the series
$r$ = common ratio
To find ${9^{th}}$ term with $a = 1$ and $r = 4$
We have, $n = 9$
Putting all the values in eq. (i)
             ${a_9} = (1){(4)^{9 - 1}}$
             \[{a_9} = {4^8}\]
              \[{a_9} = 65,536\] [Multiplying \[4\] eight times]
Hence, the ninth term of the G.P is \[65536\]

Note: Here students make mistakes by considering the given Geometric Progression as Arithmetic Progression. They will find the common difference instead of common ratios.