
If \[{t_n}\] denotes the nth term of a G.P. whose common ratio is r, then the progression whose nth term \[\dfrac{1}{{t_n^2 + t_{n + 1}^2}}\] is
A. A.P.
B. G.P.
C. H.P.
D. None of these
Answer
164.4k+ views
Hint: In our case, we are provided that \[{t_n}\] denotes the nth term of a G.P. whose common ratio is r and are asked to determine the nth term \[\dfrac{1}{{t_n^2 + t_{n + 1}^2}}\]for that we have to use the G.P formula for n terms \[G.P{\rm{ }} = {\rm{ }}a{r^{n - 1}}\] and now we have to replace the value given in the place of n terms and solve further to obtain the desired solution.
Formula Used: The nth term of G.P can be determined by
\[G.P{\rm{ }} = {\rm{ }}a{r^{n - 1}}\]
Complete step by step solution: We have been provided in the question that,
If \[{t_n}\] denotes the nth term of a G.P. whose common ratio is r
The nth term is given by
\[\dfrac{1}{{t_n^2 + t_{n + 1}^2}}\]
We have been already known that the nth term of G.P is
\[G.P{\rm{ }} = {\rm{ }}a{r^{n - 1}}\]
Here, ‘r’ is said to common ratio and ‘a’ is said to the first term
\[{x_n} = \dfrac{1}{{t_n^2 + t_{n + 1}^2}}\]
Now, we have to write the above equation in terms of nth term of G.P we get
\[ = \dfrac{{{\rm{ }}1}}{{({a^2}{r^{2(n - 1)}}\; + {\rm{ }}{a^2}{r^{2n}})}}\]
Now, let’s take \[{a^2}{r^{2n}}\] as common from the above expression, we obtain
\[ = \dfrac{1}{{{a^2}{r^{2n}}({r^{ - 2}}\; + {\rm{ }}1)}}\]
On splitting the terms above expression we get
\[ = \dfrac{1}{{{a^2}{r^{2n}}}} \cdot \dfrac{{{r^2}}}{{1 + {r^2}}}\]
From the above calculation, it is obtained that
\[{x_{n - 1}}\; = {\rm{ }}\dfrac{{{r^2}}}{{{a^2}{r^{2n - 2}}(1{\rm{ }} + {\rm{ }}{r^2})}}\]
And thus we concluded that,
\[\dfrac{{{x_n}\;}}{{{x_{n - 1}}}} = \dfrac{1}{{{r^2}\;}} = \]Constant
Therefore, the progression whose nth term \[\dfrac{1}{{t_n^2 + t_{n + 1}^2}}\] is G.P
Option ‘B’ is correct
Note: Mostly students got confused while solving these types of problems because it has some difficult calculations that include fractions and powers. So, one should be cautious while solving these types of problems to avoid mistakes and not to get wrong solution. And should be thorough will all the progression formulas to get the correct solution.
Formula Used: The nth term of G.P can be determined by
\[G.P{\rm{ }} = {\rm{ }}a{r^{n - 1}}\]
Complete step by step solution: We have been provided in the question that,
If \[{t_n}\] denotes the nth term of a G.P. whose common ratio is r
The nth term is given by
\[\dfrac{1}{{t_n^2 + t_{n + 1}^2}}\]
We have been already known that the nth term of G.P is
\[G.P{\rm{ }} = {\rm{ }}a{r^{n - 1}}\]
Here, ‘r’ is said to common ratio and ‘a’ is said to the first term
\[{x_n} = \dfrac{1}{{t_n^2 + t_{n + 1}^2}}\]
Now, we have to write the above equation in terms of nth term of G.P we get
\[ = \dfrac{{{\rm{ }}1}}{{({a^2}{r^{2(n - 1)}}\; + {\rm{ }}{a^2}{r^{2n}})}}\]
Now, let’s take \[{a^2}{r^{2n}}\] as common from the above expression, we obtain
\[ = \dfrac{1}{{{a^2}{r^{2n}}({r^{ - 2}}\; + {\rm{ }}1)}}\]
On splitting the terms above expression we get
\[ = \dfrac{1}{{{a^2}{r^{2n}}}} \cdot \dfrac{{{r^2}}}{{1 + {r^2}}}\]
From the above calculation, it is obtained that
\[{x_{n - 1}}\; = {\rm{ }}\dfrac{{{r^2}}}{{{a^2}{r^{2n - 2}}(1{\rm{ }} + {\rm{ }}{r^2})}}\]
And thus we concluded that,
\[\dfrac{{{x_n}\;}}{{{x_{n - 1}}}} = \dfrac{1}{{{r^2}\;}} = \]Constant
Therefore, the progression whose nth term \[\dfrac{1}{{t_n^2 + t_{n + 1}^2}}\] is G.P
Option ‘B’ is correct
Note: Mostly students got confused while solving these types of problems because it has some difficult calculations that include fractions and powers. So, one should be cautious while solving these types of problems to avoid mistakes and not to get wrong solution. And should be thorough will all the progression formulas to get the correct solution.
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