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Arithmetic Progressions — MATHS STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceMATHSArithmetic Progressions2 Marks
Question
The fibonacci sequence is defined by $\text{a}_1=1=\text{a}_2,\ \text{a}_\text{n}=\text{a}_{\text{n}-1}+\text{a}_{\text{n}-2}$ from $\text{n}>2.$ Find $\frac{\text{a}_{\text{n}+1}}{\text{a}_\text{n}}$ for $\text{n}=1,\ 2,\ 3,\ 4,\ 5.$
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Answer

$\text{a}_\text{n}=\text{a}_{\text{n}-1}+\text{a}_{\text{n}-2}$ For $​​\text{n}>2$ $\Rightarrow​​\text{a}_3=​​\text{a}_{3-1}+​\text{a}_{3-2}=​​\text{a}_2+​​\text{a}_1=1+1=2$ $\Rightarrow​​\text{a}_4=​​\text{a}_{4-1}+​\text{a}_{4-2}=​​\text{a}_3+​​\text{a}_2=2+1=3$ $\Rightarrow​​\text{a}_5=​​\text{a}_{5-1}+​​​\text{a}_{5-2}=\text{a}_4+​​\text{a}_3=3+2=3$ $\Rightarrow​​\text{a}_6=​​\text{a}_{6-1}+​​​\text{a}_{6-2}=\text{a}_5+​​\text{a}_4=5+31=8$ $\therefore$ for $​​\text{n}=1$ $\frac{​​\text{a}_{​​\text{n}-1}}{​​\text{a}​​_\text{n}}=\frac{​​\text{a}_2}{​​\text{a}_1}=\frac{1}{1}=1$ For $​​\text{n}=2$ $\frac{​​\text{a}^3}{​​\text{a}^2}=\frac{2}{1}=2$ For $​​\text{n}=3$ $\frac{​​\text{a}^3}{​​\text{a}^2}=\frac{3}{2}=1.5$ For $​​\text{n}=4$ and $\text{n}=5$ $\frac{​​\text{a}_5}{​​\text{a}_4}=\frac{5}{3}$ and $\frac{​​\text{a}_6}{​​\text{5}_5}=\frac{8}{5}$ $\therefore$ The reqquired series is $1,2,\frac{3}{2},\frac{5}{3},\frac{8}{5},...$

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