Question

If the nth term of GP \[5,\dfrac{{ - 5}}{2},\dfrac{5}{4},\dfrac{{ - 5}}{8},...\] is \[\dfrac{5}{{1024}}\] then the value of n is
a) 11
b) 10
c) 9
d) 4

Answer 1

Hint: Here we have to the nth term of the sequence. The sequence is an infinite sequence. First we have to determine the kind of a sequence and then we have a formula for the nth term and then by substituting the values we determine the solution for the question.

Complete step by step answer:
In mathematics we have three types of series namely, arithmetic series, geometric series and harmonic series.
First we have to determine the kind of the sequence.
Suppose if the sequence is an arithmetic sequence, then we have to check the common difference. The \[{a_1} = 5,\,\,{a_2} = \dfrac{{ - 5}}{2},\,\,{a_3} = \dfrac{5}{4},\,{a_4} = \dfrac{{ - 5}}{8}\].Let we determine the difference between first two terms \[{d_1} = {a_2} - {a_1} = - \dfrac{5}{2} - 5 = \dfrac{{ - 15}}{2}\]and we determine the difference between the next two terms \[{d_2} = {a_3} - {a_2} = \dfrac{5}{4} + \dfrac{5}{2} = \dfrac{{30}}{8}\]. Hence \[{d_1} \ne {d_2}\]. Therefore the given sequence is not an arithmetic sequence.
Suppose if the sequence is a geometric sequence, then we have to check the common ratio. The \[{a_1} = 5,\,\,{a_2} = \dfrac{{ - 5}}{2},\,\,{a_3} = \dfrac{5}{4},\,{a_4} = \dfrac{{ - 5}}{8}\].Let we determine the ratio between first two terms \[{r_1} = \dfrac{{{a_2}}}{{{a_1}}} = \dfrac{{\dfrac{{ - 5}}{2}}}{5} = \dfrac{{ - 5}}{2} \times \dfrac{1}{5} = \dfrac{{ - 1}}{2}\]and we determine the difference between the next two terms \[{r_2} = \dfrac{{{a_3}}}{{{a_2}}} = \dfrac{{\dfrac{5}{4}}}{{\dfrac{{ - 5}}{2}}} = \dfrac{5}{4} \times \dfrac{2}{{ - 5}} = \dfrac{{ - 1}}{2}\]. Hence \[{r_1} = {r_2}\]. Therefore the given sequence is a geometric sequence.

The geometric series is defined as the series with a constant ratio between the two successive terms. The finite geometric series is generally represented as \[a,ar,a{r^2},...,a{r^n}\], where a is first term and r is a common ratio.
The nth term of the given sequence is given by \[{T_n} = a{r^{n - 1}}\], where \[{T_n}\]represents the nth term. The \[a\] is the first term and it is 5. The \[r\] is the common ratio and it is \[r = - \dfrac{1}{2}\].
Therefore the nth term is given by \[{T_n} = 5.{\left( { - \dfrac{1}{2}} \right)^{n - 1}}\]. We have the term \[\dfrac{5}{{1024}}\]
So we have
\[ \Rightarrow \dfrac{5}{{1024}} = 5{\left( { - \dfrac{1}{2}} \right)^{n - 1}}\]
On cancelling the number 5 we have
\[ \Rightarrow \dfrac{1}{{1024}} = 1{\left( { - \dfrac{1}{2}} \right)^{n - 1}}\]
\[ \Rightarrow {\left( {\dfrac{1}{2}} \right)^{10}} = {\left( { - \dfrac{1}{2}} \right)^{n - 1}}\]
\[ \Rightarrow {\left( { - \dfrac{1}{2}} \right)^{10}} = {\left( { - \dfrac{1}{2}} \right)^{n - 1}}\]
On equating the powers we have
\[ \Rightarrow n - 1 = 10\]
\[ \Rightarrow n = 11\]
Therefore, $n=11$. Hence, option (a) is the correct option.

Note:
We must know about the geometric progression arrangement and it is based on the first term and common ratio. The common ratio of the geometric progression is defined as \[\dfrac{{{a_2}}}{{{a_1}}}\] where \[{a_2}\] represents the second term and \[{a_1}\] represents the first term. The sum of n terms is defined on the basis of common ratio.