
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
163.2k+ views
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.
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.
Recently Updated Pages
Geometry of Complex Numbers – Topics, Reception, Audience and Related Readings

JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Displacement-Time Graph and Velocity-Time Graph for JEE

Degree of Dissociation and Its Formula With Solved Example for JEE

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Instantaneous Velocity - Formula based Examples for JEE

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

NCERT Solutions for Class 11 Maths Chapter 4 Complex Numbers and Quadratic Equations

NCERT Solutions for Class 11 Maths Chapter 6 Permutations and Combinations

NCERT Solutions for Class 11 Maths In Hindi Chapter 1 Sets

JEE Advanced 2025 Notes
