Mathematics - Sequence & Series Question with Solution | TestHub

$$\begin{array}{rl}\frac{1}{a_{n+1}}& =\frac{n-a_n}{(n-1)a_n}n=1,2,3,4,5\dots \\ & =\left(\frac{n}{n-1}\right)\frac{1}{a_n}-\frac{1}{n-1}\\ \Rightarrow \frac{1}{n{a}_{n+1}}& =\frac{1}{(n-1)a_n}-\frac{1}{n(n-1)}\dots \dots \dots \dots (1)\\ Let{b}_n& =\frac{1}{(n-1)a_n}\\ then,b_{n+1}-b_n& =\frac{1}{n}-\frac{1}{n-1}n=2,3,4\dots \\ \therefore b_n-b_2& =\frac{1}{n-1}-1+\frac{1}{a_2}\\ & =\frac{1}{n-1}+3=\frac{3n-2}{(n-1)}\\ \therefore a_n& =\frac{1}{3n-2}\forall n\ge 2\\ 100a_9& =100\left(\frac{1}{25}\right)=4\end{array}$$