Sequence And Series Question 444
Question: If $ t_{n} $ denotes the nth term of a G.P. whose common ratio is r, then the progression whose nth term is $ \frac{1}{t_n^{2}+t_{n+1}^{2}} $ is
Options:
A) A.P.
B) G.P.
C) H.P.
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
[b] If a be the first term of GP. then given $ x_{n}=\frac{1}{t_n^{2}+t_{n+1}^{2}}=\frac{1}{a^{2}{r^{2(n-1)}}+a^{2}r^{2n}} $ $ =\frac{1}{a^{2}r^{2n}}.\frac{1}{{r^{-2}}+1}=\frac{1}{a^{2}r^{2n}}.\frac{r^{2}}{1+r^{2}} $
$ \therefore {x_{n-1}}=\frac{1}{a^{2}{r^{2,n-2}}}.\frac{r^{2}}{1+r^{2}} $
$ \therefore \frac{x_{n}}{{x_{n-1}}}=\frac{1}{r^{2}}= $ constant
$ \therefore $ The sequence $ \langle x_{n} \rangle $ is a GP.
BETA
