The first $n$ terms of a series are given by $\log_a 3 + \log_a 27 + \log_a 243 + \ldots + \log_a 3^{2^{n-1}}$, where $a$ is a positive constant. Show that the series is an arithmetic series. Given that sum of the first 30 terms of the series is 300, find the value of $a$. A geometric series has first term $c$ and common ratio $r$, where $c$ and $r$ are non-zero. An arithmetic series has first term $b$ and common difference $d$, where $b$ and $d$ are non-zero. It is given that the 5th, 8th and 10th terms of the arithmetic series are equal to the 2nd, 3rd and 4th of the geometric series respectively. Show that $r$ satisfies the equation $3r^2 - 5r + 2 = 0$ and hence find the sum to infinity in terms of $c$.

1. Rewrite the terms using the power law of logarithms

The terms are $\log_a 3 = \log_a 3^1$, $\log_a 27 = \log_a 3^3$, $\log_a 243 = \log_a 3^5$, with the $n$th term $\log_a 3^{2n-1}$.

Using $\log_a x^k = k\log_a x$:
$T_n = \log_a 3^{2n-1} = (2n-1)\log_a 3.$
So the terms are $\log_a 3,\; 3\log_a 3,\; 5\log_a 3,\; \ldots$

2. Show the common difference is constant

Consider the difference between consecutive terms:
$T_n - T_{n-1} = (2n-1)\log_a 3 - (2(n-1)-1)\log_a 3 = \big[(2n-1)-(2n-3)\big]\log_a 3 = 2\log_a 3.$
This difference is the same constant for all $n$, so the series is an arithmetic series (with common difference $2\log_a 3$).

1. Identify the first term and common difference of the arithmetic series.

From the given series, the first term is $u_1 = \log_a 3$.
The second term is $u_2 = \log_a 27 = \log_a 3^3 = 3 \log_a 3$.
The common difference is $d = u_2 - u_1 = 3 \log_a 3 - \log_a 3 = 2 \log_a 3$.

2. Form an equation using the sum of the first 30 terms and solve for $a$.

Using the sum formula $S_n = \frac{n}{2}[2u_1 + (n-1)d]$ for $n=30$:
$S_{30} = \frac{30}{2}[2(\log_a 3) + 29(2 \log_a 3)]$
$S_{30} = 15[2 \log_a 3 + 58 \log_a 3] = 15(60 \log_a 3) = 900 \log_a 3$
Equating this to the given sum of 300:
$900 \log_a 3 = 300$
$\log_a 3 = \frac{1}{3}$
Converting from logarithmic form to index form:
$a^{\frac{1}{3}} = 3$
$a = 3^3 = 27$

1. Express the given terms as equations.

Let the 5th, 8th, and 10th terms of the arithmetic series be equal to the 2nd, 3rd, and 4th terms of the geometric series respectively:
$b + 4d = cr$ --- (1)
$b + 7d = cr^2$ --- (2)
$b + 9d = cr^3$ --- (3)

2. Eliminate $b$ to find expressions relating $d$, $c$, and $r$.

Subtract (1) from (2):
$(b + 7d) - (b + 4d) = cr^2 - cr$
$3d = c(r^2 - r)$ --- (4)

Subtract (2) from (3):
$(b + 9d) - (b + 7d) = cr^3 - cr^2$
$2d = c(r^3 - r^2)$ --- (5)

3. Eliminate $d$ to deduce the given quadratic equation in $r$.

From (4), $d = \frac{c(r^2 - r)}{3}$
From (5), $d = \frac{c(r^3 - r^2)}{2}$

Equating the expressions for $d$:
$\frac{c(r^2 - r)}{3} = \frac{c(r^3 - r^2)}{2}$
$2c(r^2 - r) = 3c(r^3 - r^2)$

Since $c \neq 0$, we can divide both sides by $c$:
$2r^2 - 2r = 3r^3 - 3r^2$
$3r^3 - 5r^2 + 2r = 0$

Since $r \neq 0$, we can divide the equation by $r$:
$3r^2 - 5r + 2 = 0$ (Shown)

4. Solve the quadratic equation and reject the invalid value for $r$.

Factorising the quadratic equation:
$(3r - 2)(r - 1) = 0$
$r = \frac{2}{3} \text{ or } r = 1$

If $r = 1$, substituting into equation (4) gives $3d = c(1^2 - 1) = 0$, which means $d = 0$.
However, it is given that $d \neq 0$, so we reject $r = 1$.
Therefore, $r = \frac{2}{3}$.

5. Calculate the sum to infinity of the geometric series.

The sum to infinity $S_\infty$ of the geometric series is given by $\frac{\text{first term}}{1 - r}$:
$S_\infty = \frac{c}{1 - r}$
$S_\infty = \frac{c}{1 - \frac{2}{3}}$
$S_\infty = \frac{c}{\frac{1}{3}} = 3c$