Question

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 1

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.