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 $n=1,2,3,4,5$
$1=a_{1}=a_{2}$
$a_{n}=a_{n-1}+a_{n-2}, n\,>\,2$
$\therefore a_{3}=a_{2}+a_{1}=1+1=2$
$a_{4}=a_{3}+a_{2}=2+1=3$
$a_{5}=a_{4}+a_{3}=3+2=5$
$a_{6}=a_{5}+a_{4}=5+3=8$
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 $n=5, \frac{a_{n+1}}{a_{n}}=\frac{a_{6}}{a_{5}}=\frac{8}{5}$
How many terms of the $A.P.$ $-6,-\frac{11}{2},-5, \ldots \ldots$ are needed to give the sum $-25 ?$
For any three positive real numbers $a,b,c$ ; $9\left( {25{a^2} + {b^2}} \right) + 25\left( {{c^2} - 3ac} \right) = 15b\left( {3a + c} \right)$ then
If the sum of the series $54 + 51 + 48 + .............$ is $513$, then the number of terms are
The value of $\sum\limits_{r = 1}^n {\log \left( {\frac{{{a^r}}}{{{b^{r - 1}}}}} \right)} $ is
If $a$ and $b$ are the roots of $x^{2}-3 x+p=0$ and $c, d$ are roots of $x^{2}-12 x+q=0$ where $a, b, c, d$ form a $G.P.$ Prove that $(q+p):(q-p)=17: 15$