Find the position of given term in GP

In this section we will discuss to find the position of given term in GP
The position of a given term in geometric progression of a given term in geometric progression is denoted by 'n'.
If the first term of G.P is 'a' and the common ratio is 'r' then the G.P can be written as
a,ar,a$r^{2}$, a$r^{3}$,...,a$r^{n -1}$
First term = a
second term = ar
third term = a$r^{2}$
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nth term = a$r^{n -1}$
Example : The given progression is $\frac{1}{2},\frac{1}{3},\frac{2}{9},\frac{4}{27},\frac{8}{81}$
So, First term = $\frac{1}{2}$
second term = $\frac{1}{3}$
third term = $\frac{2}{9}$
forth term = $\frac{4}{27}$
fifth term = $\frac{8}{81}$
In this way there are five terms and here the 5th term is nth term.

Examples to find the position of given term in GP

1) Which term of the G.P. 2,1,$\frac{1}{2},\frac{1}{4}$ , .... is $\frac{1}{128}$?

Solution : The given progression is in G.P with first term = a = 2 and
common ratio = r = $\frac{1}{2}$
Let the nth term be $\frac{1}{128}$
nth term is given by
$a_{n} = ar^{n - 1}$
$\frac{1}{128} = 2\left ( \frac{1}{2} \right )^{n - 1}$

$\frac{1}{128} =\left ( \frac{1}{2} \right )^{n - 2}$

$\left ( \frac{1}{2} \right )^{7} = \left ( \frac{1}{2} \right )^{n - 2}$
as the bases are same so the exponents are equal
7 = n - 2
∴ n = 9
Thus 9th term of the given G.P is $\frac{1}{128}$

2) Which term of the given G.P 5,10,20,40 is 5120 ?
Solution : The given progression is in G.P with first term = a = 5 and
common ratio = r = 2
Let the nth term be 5120
nth term is given by
$a_{n} = ar^{n - 1}$
5120 = 5$(2)^{n - 1}$
1024 =5$(2)^{n - 1}$
$2^{10} = (2)^{n - 1}$
as the bases are same so the exponents are equal
10 = n - 1
∴ n = 11
Thus 11th term of the given G.P is 5120

3) Which term of the G.P 2,8,32,... is 131072 ?
Solution : The given progression is in G.P with first term = a = 2 and
common ratio = r = 4
Let the nth term be 131072
nth term is given by
$a_{n} = ar^{n - 1}$
131072 = 2$(4)^{n - 1}$
65536 =$(4)^{n - 1}$
$4^{8} = (4)^{n - 1}$
as the bases are same so the exponents are equal
8 = n - 1
∴ n = 9
Thus 9th term of the given G.P is 131072.


11th grade math

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