A) A.P.
B) G.P.
C) H.P.
D) None of these
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 \[\left\langle {{x}_{n}} \right\rangle \] is a GP.
You need to login to perform this action.
You will be redirected in
3 sec
