Question

If the \[(p + q)\] th term of a geometric series is $m$ and the $(p - q)$ th term is $n$ , then the $p$ th term is
A) $\sqrt {mn} $
B) $mn$
C) $m + n$
D) $m - n$

Answer 1

Hint: A geometric series is a series for which the ratio of each two consecutive terms is a constant function. Formula for finding the $n$ th term in geometric sequence is ${T_n} = a \times {r^{n - 1}}$, where $a = $ start term and $r = $ common ratio. First we find the given terms then multiply them and calculate the multiplication after that we get the required result.

Complete step by step answer:
In the given data there are not given the start term and the common ratio of a geometric series
 then we take $a = $ start term and $r = $ common ratio
First we find the $(p + q)$ th term in geometric series
${T_{p + q}} = a \times {r^{p + q - 1}}$
From the given data , we get
${T_{p + q}} = a \times {r^{p + q - 1}} = m$
Now we find the $(p - q)$ th term in geometric series
${T_{p - q}} = a \times {r^{p - q - 1}}$
From the given data , we get
${T_{p - q}} = a \times {r^{p - q - 1}} = n$
Multiply the $(p + q)$ th term and $(p - q)$ th term then we have
$({T_{p + q}}) \times ({T_{p - q}}) = mn$
$ \Rightarrow (a{r^{p + q - 1}}) \times (a{r^{p - q - 1}}) = mn$
We know if ${r^a}$ multiply with ${r^b}$ then it becomes ${r^{a + b}}$ , we use this in above equation and we get
$ \Rightarrow {a^2} \times {r^{(p + q - 1) + (p - q - 1)}} = mn$
$ \Rightarrow {a^2} \times {r^{p + q - 1 + p - q - 1}} = mn$
$ \Rightarrow {a^2} \times {r^{2p - 2}} = mn$
We take common $2$ from the power of \[r\] , we get
$ \Rightarrow {a^2} \times {r^{2(p - 1)}} = mn$
Taking square root both sides of the above equation and get
$ \Rightarrow \sqrt {{a^2} \times {r^{2(p - 1)}}} = \sqrt {mn} $
$ \Rightarrow a{r^{p - 1}} = \sqrt {mn} $ ………………………………..(i)
Now we find the $p$ th term of geometric series
${T_p} = a \times {r^{p - 1}}$
From the equation (i) , we get the value of $p$ th term and we substitute this and get
${T_p} = a{r^{p - 1}} = \sqrt {mn} $
$\therefore $ The $p^{th}$ term of the geometric series is ${T_p} = \sqrt {mn} $. So, Option (A) is correct.

Note:
If you forget the formula you can establish at the time . In a geometric progression there are $a = $ start term and $r = $ common ratio then we get the members are \[a,ar,a{r^2},a{r^3},...,a{r^n},...\] , then we need to find the $s$ th term then we take the sequence ${T_1} = a = a{r^{(1 - 1)}}$ ,${T_2} = ar = a{r^{(2 - 1)}}$ ,${T_3} = a{r^2} = a{r^{(3 - 1)}}$ ,……., ${T_n} = a{r^{n - 1}}$ ,…..
${T_s} = $ $a \times {r^{s - 1}}$ . This is the required formula and you can establish it easily when you need it.