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The Fibonacci sequence is defined by
$1 = {a_1} = {a_2}{\rm{ }}$ and ${a_n} = {a_{n - 1}} + {a_{n - 2}},n\, > \,2$
Find $\frac{a_{n+1}}{a_{n}},$ for $n=1,2,3,4,5$
$1,2,\frac{3}{2},\frac{5}{3},\frac{8}{5}$
$1,2,\frac{3}{2},\frac{5}{3},\frac{8}{5}$
$1,2,\frac{3}{2},\frac{5}{3},\frac{8}{5}$
$1,2,\frac{3}{2},\frac{5}{3},\frac{8}{5}$
Solution
$1=a_{1}=a_{2}$
$a_{n}=a_{n-1}+a_{n-2}, n\,>\,2$
$\therefore a_{3}=a_{2}+a_{1}=1+1=2$
$a_{4}=a_{3}+a_{2}=2+1=3$
$a_{5}=a_{4}+a_{3}=3+2=5$
$a_{6}=a_{5}+a_{4}=5+3=8$
For $n=1, \frac{a_{n+1}}{a_{n}}=\frac{a_{2}}{a_{1}}=\frac{1}{1}=1$
For $n=2, \frac{a_{n+1}}{a_{n}}=\frac{a_{3}}{a_{2}}=\frac{2}{1}=2$
For $n=3, \frac{a_{n+1}}{a_{n}}=\frac{a_{4}}{a_{3}}=\frac{3}{2}$
For $n=4, \frac{a_{n+1}}{a_{n}}=\frac{a_{5}}{a_{4}}=\frac{5}{3}$
For $n=5, \frac{a_{n+1}}{a_{n}}=\frac{a_{6}}{a_{5}}=\frac{8}{5}$
