Question

If \[{(p + q)^{th}}\] term of a G.P. is \[m\] and \[{(p - q)^{th}}\] term is \[n\], then the \[{p^{th}}\] term will be
A. \[m/n\]
B. \[\sqrt {mn} \]
C. \[mn\]
D. \[0\]

Answer 1

Hint:
A geometric series is a set of terms where the ratio between any two of them is a fixed function \[{T_n} = a \times {r^{n - 1}}\], where ‘a’ is the start term and ‘r’ is the common ratio, is the formula for determining the nth term in a geometric series. Finding the given terms is followed by multiplying them and computing the multiplication to arrive at the desired outcome.
Formula use:
In G.P, ‘a’ for the start term and ‘r’ for the common ratio.
 \[{T_n} = a \times {r^{n - 1}}\]
Complete step-by-step solution
When the start term and common ratio of a geometric series are absent from the given data, we substitute ‘a’ for the start term and ‘r’ for the common ratio.
First, we determine the geometric series' \[{(p + q)^{th}}\] term.
\[{T_{p + q}} = a \times {r^{p + q - 1}}\]
Now, from the given information, we get
\[{T_{p + q}} = a \times {r^{p + q - 1}} = m\]
Now, let’s determine \[{(p - q)^{th}}\] term in geometric series, we obtain
\[{T_{p - q}} = a \times {r^{p - q - 1}}\]
From the above date, we have
\[{T_{p - q}} = a \times {r^{p - q - 1}} = n\]
Now, we have to multiply \[{(p + q)^{th}}\] and \[{(p - q)^{th}}\] term, we have
\[({T_{p - q}}) \times ({T_{p - q}}) = mn\]
\[ \Rightarrow (a{r^{p + q - 1}})(a{r^{p - q - 1}}) = mn\]
Knowing that \[{r^a}\] multiplies by \[{r^b}\] to form \[{r^{a + b}}\], we apply this knowledge to the previous equation to obtain,
\[{a^2} \times {r^{(p + q - 1) + (p - q - 1)}} = mn\]
\[ \Rightarrow {a^2} \times {r^{p + q - 1 + p - q - 1}} = mn\]
\[ \Rightarrow {a^2} \times {r^{2p - 2}} = mn\]
Now, we take \[2\] as common from the power of \[r\], we have
\[{a^2} \times {r^{2(p - 1)}} = mn\]
Let’s take square root on both sides we get
\[\sqrt {{a^2} \times {r^{2(p - 1)}}} = \sqrt {mn} \]
\[a{r^{(p - 1)}} = \sqrt {mn} \]-------- (1)
Let’s find the \[{p^{th}}\] term of Geometric series,
\[{T_p} = a \times {r^{p - 1}}\]
We get the value of \[{p^{th}}\] term from equation (1), on substituting that, we get
\[{T_p} = a{r^{p - 1}} = \sqrt {mn} \]
Therefore, the \[{p^{th}}\] term will be\[\sqrt {mn} \].
Hence, option B is correct.
Note:
If you misplace the formula, you can decide then. The members of a geometric progression are\[a,ar,a{r^2},a{r^3},......a{r^n},....\]and when we need to get the \[{s^{th}}\] term, we use the sequence \[{T_1} = a = a{r^{(1 - 1)}},{T_2} = ar = a{r^{(2 - 1)}},{T_3} = a{r^2} = a{r^{(3 - 1)}}.......{T_n} = a{r^{n - 1}}\]. You can easily establish this formula when you need it because it is necessary.