Find the first term

Let the first term of the infinite geometric progression be $a$ , its common ratio be $r$ and $b=2014$ . Then,

$\begin{cases} \dfrac a{1-r} & = b & \implies a = b - br & ...(1) \\ \dfrac {a^2}{1-r^2} & = b & \implies a^2 = b - br^2 & ...(2) \end{cases}$

$\begin{aligned} \implies (1)^2 = (2): \quad (b-br)^2 & = b - br^2 & \small \color{#3D99F6} \text{Divide both sides by }b \\ b(1-r)^2 & = 1 - r^2 \\ br^2 - 2br + b & = 1-r^2 \\ (b+1)r^2 - 2br + b-1 & = 0 \end{aligned}$

$\implies r = \dfrac {2b \pm \sqrt{4b^2-4(b+1)(b-1)}}{2(b+1)} = \begin{cases} 1 & \small \color{#D61F06} \text{Series diverges -- rejected} \\ \dfrac {b-1}{b+1} < 1 & \small \color{#3D99F6} \text{Series converses -- accepted} \end{cases}$

Therefore, the first term $a = b - br = b - \dfrac {b(b-1)}{b+1} = \dfrac {2b}{b+1}$ . For $b = 2014$ , $a = \boxed{\dfrac {4028}{2015}}$ .

The first series is a + ar + ar^2 +...…. with S 1 = a/(1-r) = 2014. The second series is a^2 + a^2r^2 +...…. with S 2 = a^2/(1 - r^2) = 2014. Is S 1 = S 2, a/(1 - r) = a^2/(1 - r^2). Multiplying by (1 -r)/a, 1 = a/(1 + r), so a = 1 + r. Then 2014 = (1+r)/(1 - r), or 2014 - 2014r = 1 + r, 2013 = 2015r, and r = 2013/2015. a = 1 +r = 1 + 2013/2015 = 4028/2015. Ed Gray

$\sum _{i=0}^{\infty } \left(\frac{a}{x^i}\right)^1 \Longrightarrow \frac{a x}{x-1}$

$\sum _{i=0}^{\infty } \left(\frac{a}{x^i}\right)^2 \Longrightarrow \frac{a^2 x^2}{x^2-1}$

$\frac{a x}{x-1}=\frac{a^2 x^2}{x^2-1}=2014 \Longrightarrow a\to \frac{4028}{2015},x\to \frac{2015}{2013}$

The value of $a$ is the answer. If the series were both started from i=1 instead, then the answer is the same: $\frac{4024}{2015}$ . In general, the answer is $\frac{2\ \text{year}}{\text{year}+1}$ .