Year in a geometric series

Let the geometric series be $a,ar,ar^2.\cdots$ , where $a$ and $r$ are the first term and common ratio respectively. Therefore, by the formula of summation of infinite geometric series,we have $\frac{a}{1-r}=2017$ Also, $a^2+a^2r^2+a^2r^4+ \cdots = 2017$ $\implies a^2 \left(1+r^2+r^4+r^6+\cdots \right)=2017$ $\implies \frac{a^2}{1-r^2}=2017$ , (because first term is $1$ and common ratio is $r^2$ ). Now since $\large \frac{a \times a}{(1+r)(1-r)}=2017$ , we get $\large \frac{a}{1+r}=1$ $\implies a=1+r$ . Now by simply solving the 2 linear equations, we will get $\large r=\frac{2016}{2018}$ and $\large a=\boxed{\frac{4034}{2018}}$

We are given that the series $a + ar + ar^{2} + ar^{3} + ..... = \dfrac{a}{1 - r} = 2017$ and that $a^{2} + a^{2}r^{2} + a^{2}r^{4} + a^{2}r^{6} + .... = \dfrac{a^{2}}{1 - r^{2}} = 2017$ .

Then $\dfrac{a}{1 - r} = \dfrac{a^{2}}{1 - r^{2}} \Longrightarrow \dfrac{1 - r^{2}}{1 - r} = a \Longrightarrow \dfrac{(1 - r)(1 + r)}{1 - r} = a \Longrightarrow a = 1 + r \Longrightarrow r = a - 1$ .

Plugging this result into $\dfrac{a}{1 - r} = 2017$ gives us that $\dfrac{a}{1 - (a - 1)} = 2017 \Longrightarrow a = 2017(2 - a) \Longrightarrow 2018a = 4034 \Longrightarrow a = \boxed{\dfrac{4034}{2018}}$ .