Sequence-And-Series Question 648
Question: The expression $ \frac{{a^{n+1}}+{b^{n+1}}}{a^{n}+b^{n}} $ is $ [a\ne b\ne 0] $ is (where a and b are unequal non-zero numbers)
Options:
A) A.M. between a and b if $ n=-1 $
B) G.M. between a and b if $ n=-\frac{1}{2} $
C) H.M. between a and b if n = 0
D) All are correct
Show Answer
Answer:
Correct Answer: B
Solution:
[b] Let $ \frac{{a^{n+1}}+{b^{n+1}}}{a^{n}+b^{n}}=\frac{a+b}{2} $
$ \Rightarrow ,2{a^{n+1}}+2{b^{n+1}}={a^{n+1}}+{b^{n+1}}+ab^{n}+ba^{n} $
$ \Rightarrow (a-b)(a^{n}-b^{n})=0 $
$ \Rightarrow a^{n}=b^{n} $ it is possible for unequal numbers a and b if $ n=0 $ Let $ \frac{{a^{n+1}}+{b^{n+1}}}{a^{n}+b^{n}}=\sqrt{ab} $
$ \Rightarrow {a^{n+1}}+{b^{n+1}}={a^{n+\frac{1}{2}}},{b^{\frac{1}{2}}}+{a^{\frac{1}{2}}},{b^{n+\frac{1}{2}}} $
$ \Rightarrow ,( {a^{n+\frac{1}{2}}}-{b^{n+\frac{1}{2}}} ),( \sqrt{a}-\sqrt{b} )=0 $
$ \Rightarrow {a^{n+\frac{1}{2}}}-{b^{n+\frac{1}{2}}}=0, $ which holds true if $ n+\frac{1}{2}=0\Rightarrow n=-\frac{1}{2} $ Let $ \frac{{a^{n+1}}+{b^{n+1}}}{a^{n}+b^{n}}=\frac{2ab}{a+b} $
$ \Rightarrow ,{a^{n+2}}+{a^{n+1}}b+a{b^{n+1}}+b_{=2a}^{n+2},{{,}^{n+1}}b+2a{b^{n+1}} $
$ \Rightarrow ,(a-b)({a^{n+1}}-{b^{n+1}})=0 $
$ \Rightarrow ,{a^{n+1}}-{b^{n+1}}=0\Rightarrow n=-1 $
BETA
