
In a GP of \[3{\rm{n}}\] terms, \[{S_1}\] denotes the sum of first \[{\rm{n}}\] terms, \[{S_2}\] the sum of the second block of \[{\rm{n}}\] terms and \[{S_3}\] the sum of last \[{\rm{n}}\] terms. Then \[{S_1},{S_2},{S_3}\] are in
A. A.P.
B. G.P.
C. H.P.
D. None of these
Answer
162k+ views
Hint: In our case, we are provided that in a G.P of \[3{\rm{n}}\] terms \[{S_1}\] denotes the sum of first \[{\rm{n}}\] terms, \[{S_2}\] the sum of the second block of \[{\rm{n}}\] terms and \[{S_3}\] the sum of last \[{\rm{n}}\] terms. And we have to find the series it has. For that, we have to write the given information in the form of series. Then we have to solve for the condition given to get the desired solution.
Formula Used: \[ = \dfrac{{a\left( {1 - {r^n}} \right)}}{{1 - r}}\]
Complete step by step solution: We have been provided in the question that,
In a GP of \[3{\rm{n}}\] terms, \[{S_1}\] denotes the sum of first \[{\rm{n}}\] terms, \[{S_2}\] the sum of the second block of \[{\rm{n}}\] terms and \[{S_3}\] the sum of last \[{\rm{n}}\] terms.
And we are questioned to find the value of \[{S_1},{S_2},{S_3}\]
The \[3{\rm{n}}\] terms of G.P is considered as
\[a,ar,a{r^2}, \ldots ,a{r^{n - 1}},a{r^n},a{r^{n + 1}},a{r^{n + 2}}, \ldots ,a{r^{2n - 1}},a{r^{2n}},a{r^{2n + 1}},a{r^{2n + 2}}, \ldots ,a{r^{3n - 1}}\]
Now, the values of \[{{\rm{S}}_1}{\rm{,}}{{\rm{S}}_2}{\rm{,}}{{\rm{S}}_3}\] has to be determined
It is given that \[{S_1}\] denotes the sum of first \[{\rm{n}}\] terms
Then the expression for n terms of the given series for \[{{\rm{S}}_1}\] is
\[{{\rm{S}}_1} = a + ar + a{r^2} + \ldots + a{r^{n - 1}} = \dfrac{{a\left( {1 - {r^n}} \right)}}{{1 - r}}\]------- (1)
It is given that \[{S_2}\] is the sum of the second block of \[{\rm{n}}\] terms
Then the expression for n terms of the given series for \[{{\rm{S}}_2}\] is
\[{{\rm{S}}_2} = a{r^n} + a{r^{n + 1}} + a{r^{n + 2}} + \ldots + a{r^{2n - 1}} = \dfrac{{a{r^n}\left( {1 - {r^n}} \right)}}{{1 - r}}\]------- (2)
It is given that \[{S_3}\] is the sum of last \[{\rm{n}}\] terms
Then the expression for n terms of the given series for \[{{\rm{S}}_3}\] is
\[{{\rm{S}}_3} = a{r^{2n}} + a{r^{2n + 1}} + a{r^{2n + 2}} + \ldots + a{r^{3n - 1}} = \dfrac{{a{r^{2n}}\left( {1 - {r^n}} \right)}}{{1 - r}}\]------- (3)
Now, we have to calculate the value of \[S_2^2\] we have
Now, \[S_2^2 = \dfrac{{{a^2}{r^{2n}}{{\left( {1 - {r^n}} \right)}^2}}}{{{{(1 - r)}^2}}}\]
Now, we have to split denominator for each term in the numerator by expansion, we get
\[ \Rightarrow S_2^2 = \dfrac{{a\left( {1 - {r^n}} \right)}}{{(1 - r)}} \times \dfrac{{a{r^{2n}}\left( {1 - {r^n}} \right)}}{{(1 - r)}}\]
From equation (1) and equation (3), we can write the above expression as
\[ \Rightarrow S_2^2 = {S_1} \times {S_3}\]
Therefore, then \[{S_1},{S_2},{S_3}\] are in G.P
Option ‘B’ is correct
Note: Students got confused while solving these types of problems because series is not given and they didn’t know what formula to apply and what method to follow. So, one should be completely thorough with the progression formulas and concepts to solve these types of problem in order to get the solution.
Formula Used: \[ = \dfrac{{a\left( {1 - {r^n}} \right)}}{{1 - r}}\]
Complete step by step solution: We have been provided in the question that,
In a GP of \[3{\rm{n}}\] terms, \[{S_1}\] denotes the sum of first \[{\rm{n}}\] terms, \[{S_2}\] the sum of the second block of \[{\rm{n}}\] terms and \[{S_3}\] the sum of last \[{\rm{n}}\] terms.
And we are questioned to find the value of \[{S_1},{S_2},{S_3}\]
The \[3{\rm{n}}\] terms of G.P is considered as
\[a,ar,a{r^2}, \ldots ,a{r^{n - 1}},a{r^n},a{r^{n + 1}},a{r^{n + 2}}, \ldots ,a{r^{2n - 1}},a{r^{2n}},a{r^{2n + 1}},a{r^{2n + 2}}, \ldots ,a{r^{3n - 1}}\]
Now, the values of \[{{\rm{S}}_1}{\rm{,}}{{\rm{S}}_2}{\rm{,}}{{\rm{S}}_3}\] has to be determined
It is given that \[{S_1}\] denotes the sum of first \[{\rm{n}}\] terms
Then the expression for n terms of the given series for \[{{\rm{S}}_1}\] is
\[{{\rm{S}}_1} = a + ar + a{r^2} + \ldots + a{r^{n - 1}} = \dfrac{{a\left( {1 - {r^n}} \right)}}{{1 - r}}\]------- (1)
It is given that \[{S_2}\] is the sum of the second block of \[{\rm{n}}\] terms
Then the expression for n terms of the given series for \[{{\rm{S}}_2}\] is
\[{{\rm{S}}_2} = a{r^n} + a{r^{n + 1}} + a{r^{n + 2}} + \ldots + a{r^{2n - 1}} = \dfrac{{a{r^n}\left( {1 - {r^n}} \right)}}{{1 - r}}\]------- (2)
It is given that \[{S_3}\] is the sum of last \[{\rm{n}}\] terms
Then the expression for n terms of the given series for \[{{\rm{S}}_3}\] is
\[{{\rm{S}}_3} = a{r^{2n}} + a{r^{2n + 1}} + a{r^{2n + 2}} + \ldots + a{r^{3n - 1}} = \dfrac{{a{r^{2n}}\left( {1 - {r^n}} \right)}}{{1 - r}}\]------- (3)
Now, we have to calculate the value of \[S_2^2\] we have
Now, \[S_2^2 = \dfrac{{{a^2}{r^{2n}}{{\left( {1 - {r^n}} \right)}^2}}}{{{{(1 - r)}^2}}}\]
Now, we have to split denominator for each term in the numerator by expansion, we get
\[ \Rightarrow S_2^2 = \dfrac{{a\left( {1 - {r^n}} \right)}}{{(1 - r)}} \times \dfrac{{a{r^{2n}}\left( {1 - {r^n}} \right)}}{{(1 - r)}}\]
From equation (1) and equation (3), we can write the above expression as
\[ \Rightarrow S_2^2 = {S_1} \times {S_3}\]
Therefore, then \[{S_1},{S_2},{S_3}\] are in G.P
Option ‘B’ is correct
Note: Students got confused while solving these types of problems because series is not given and they didn’t know what formula to apply and what method to follow. So, one should be completely thorough with the progression formulas and concepts to solve these types of problem in order to get the solution.
Recently Updated Pages
JEE Advanced Course 2025 - Subject List, Syllabus, Course, Details

Crack JEE Advanced 2025 with Vedantu's Live Classes

JEE Advanced Maths Revision Notes

JEE Advanced Chemistry Revision Notes

Download Free JEE Advanced Revision Notes PDF Online for 2025

The students S1 S2 S10 are to be divided into 3 groups class 11 maths JEE_Advanced

Trending doubts
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

IIT CSE Cutoff: Category‐Wise Opening and Closing Ranks

IIT Fees Structure 2025

Top IIT Colleges in India 2025

IIT Roorkee Average Package 2025: Latest Placement Trends Updates

Other Pages
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

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

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

JEE Advanced 2025: Dates, Registration, Syllabus, Eligibility Criteria and More
