Question

Let $a_1=p$ and $b_1=q$ where p and q are positive quantities. Define $a_n=p{b}_{n-1}$ and $b_n=q{b}_{n-1}$, for even $n>1$ and $a_n=p{a}_{n-1},bn=q{a}_{n-1}$, for odd $n>1$. If $p=\frac{1}{3}$ and $q=\frac{2}{3}$, then what is the smallest odd n such that $a_n+b_n<0.01?$

• A. 13
• B. 11
• C. 9
• D. 15

Answer 1

The correct option is C 9

If $p=\frac{1}{3}$ and $q=\frac{2}{3}$, then $p+q=1$ and $pq=\frac{2}{9}$
$a_1+b_1=p+q=1$
$a_3+b_3=p^{2}q+p{q}^2=pq(p+q)=\frac{2}{9}.$
$a_5+b_5=p^3{q}^2+p^2{q}^3=p^2{q}^2(p+q)=\frac{4}{81}.$
So, in general, for odd 'n' we can write -
$a_n+b_n={\left(\frac{2}{9}\right)}^{\left(\frac{n-1}{2}\right)}$; we need to find least value of n such that ${\left(\frac{2}{9}\right)}^{\left(\frac{n-1}{2}\right)}<0.01$.
Using the options given, if $n=7,a_7+b_7={\left(\frac{2}{9}\right)}^3>0.01;$ if $n=9,a_9+b_9={\left(\frac{2}{9}\right)}^4<0.01$.