If \[{{a}_{1}}=1\] and \[{{a}_{n+1}}=\frac{4+3{{a}_{n}}}{3+2{{a}_{n}}},\ n\ge 1\] and if \[-\frac{1}{3}\], then the value of a is

If \[{{a}_{1}}=1\] and \[{{a}_{n+1}}=\frac{4+3{{a}_{n}}}{3+2{{a}_{n}}}

1.	If \[{{a}_{1}}=1\] and \[{{a}_{n+1}}=\frac{4+3{{a}_{n}}}{3+2{{a}_{n}}},\ n\ge 1\] and if \[-\frac{1}{3}\], then the value of a is
A.	\[\sqrt{2}\]
B.	\[-\sqrt{2}\]
C.	2
D.	None of these
Answer» B. \[-\sqrt{2}\]

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If \[{{a}_{1}}=1\] and \[{{a}_{n+1}}=\frac{4+3{{a}_{n}}}{3+2{{a}_{n}}},\ n\ge 1\] and if \[-\frac{1}{3}\], then the value of a is

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