The first $n$ terms of a series are given by $\log_a 3 + \log_a 27 + \log_a 243 + \dots + \log_a 3^{2n-1}$, 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. Express the $n$th term of the series, $u_n$, in a simplified form using the power law of logarithms.

$u_n = \log_a 3^{2n-1}$
$u_n = (2n-1)\log_a 3$

2. Calculate the difference between consecutive terms, $u_n - u_{n-1}$, to show it is a constant independent of $n$.

$u_{n-1} = (2(n-1)-1)\log_a 3 = (2n-3)\log_a 3$
$u_n - u_{n-1} = (2n-1)\log_a 3 - (2n-3)\log_a 3$
$u_n - u_{n-1} = [2n - 1 - (2n - 3)]\log_a 3$
$u_n - u_{n-1} = 2\log_a 3$
Since the difference $u_n - u_{n-1}$ is a constant value ($2\log_a 3$) that is independent of $n$, the series is an arithmetic series.

1. Identify the first term and common difference, and set up the equation for the sum of the first 30 terms.

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

Using the formula for the sum of an arithmetic series $S_n = \frac{n}{2}[2a + (n-1)d]$:
$S_{30} = \frac{30}{2} [2(\log_a 3) + (30 - 1)(2 \log_a 3)]$

2. Simplify the sum expression, equate it to 300, and solve for $a$.

$15 [2 \log_a 3 + 58 \log_a 3] = 300$
$15 [60 \log_a 3] = 300$
$900 \log_a 3 = 300$
$\log_a 3 = \frac{1}{3}$

Converting the logarithmic equation to its exponential form:
$a^{\frac{1}{3}} = 3$
$a = 3^3 = 27$

1. Express the given conditions as a system of equations involving $b, d, c$, and $r$.

Let the arithmetic progression (AP) have terms $T_n = b + (n-1)d$ and the geometric progression (GP) have terms $U_n = cr^{n-1}$.
We are given:
$T_5 = U_2 \implies b + 4d = cr \quad \text{--- (1)}$
$T_8 = U_3 \implies b + 7d = cr^2 \quad \text{--- (2)}$
$T_{10} = U_4 \implies b + 9d = cr^3 \quad \text{--- (3)}$

2. Eliminate $b$ to find expressions for $d$ in terms of $c$ and $r$.

Subtracting (1) from (2) to eliminate $b$:
$(b + 7d) - (b + 4d) = cr^2 - cr$
$3d = c(r^2 - r) \implies d = \frac{c(r^2 - r)}{3}$

Subtracting (2) from (3) to eliminate $b$:
$(b + 9d) - (b + 7d) = cr^3 - cr^2$
$2d = c(r^3 - r^2) \implies d = \frac{c(r^3 - r^2)}{2}$

3. Equate the expressions for $d$ to derive the required quadratic equation.

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

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

Since $r \neq 0$, we can divide by $r$:
$3r^2 - 5r + 2 = 0 \quad \text{(Shown)}$

4. Solve the quadratic equation to find the value of $r$ and reject the invalid solution.

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

If $r = 1$, then $d = \frac{c(1^2 - 1)}{3} = 0$.
Since we are given $d \neq 0$, we reject $r = 1$.
Therefore, $r = \frac{2}{3}$.

5. Calculate the sum to infinity of the geometric series in terms of $c$.

The sum to infinity of a geometric series is $S_\infty = \frac{c}{1 - r}$.
Substituting $r = \frac{2}{3}$:
$S_\infty = \frac{c}{1 - \frac{2}{3}}$
$S_\infty = \frac{c}{\frac{1}{3}} = 3c$