If $\log_3(x^2 - 1)$, $\log_3(2x^2 + 1)$ and $\log_3(6x^2 + 3)$ are the first three terms of an arithmetic progression, then the sum of the next three terms of the progression is

In an arithmetic progression, the difference between consecutive terms is constant. Let's call this common difference $d$.

$d = \log_3(2x^2 + 1) - \log_3(x^2 - 1) = \log_3\left(\dfrac{2x^2 + 1}{x^2 - 1}\right)$

For the second and third terms:

$\log_3(6x^2 + 3) - \log_3(2x^2 + 1) = \log_3\left(\dfrac{6x^2 + 3}{2x^2 + 1}\right)$

For these to form an arithmetic progression, these differences must be equal:

$\log_3\left(\dfrac{2x^2 + 1}{x^2 - 1}\right) = \log_3\left(\dfrac{6x^2 + 3}{2x^2 + 1}\right)$

Since the logarithms are equal, their arguments must be equal:

$\dfrac{2x^2 + 1}{x^2 - 1} = \dfrac{6x^2 + 3}{2x^2 + 1}$

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Cross-multiplying:

$(2x^2 + 1)(2x^2 + 1) = (x^2 - 1)(6x^2 + 3)$

Expanding and rearranging:

$4x^4 + 4x^2 + 1 = 6x^4 - 6x^2 + 3x^2 - 3$

$-2x^4 + 7x^2 + 4 = 0$

Let $y = x^2$ to simplify:

$-2y^2 + 7y + 4 = 0$

$2y^2 - 7y - 4 = 0$

Using the quadratic formula:

$y = \dfrac{7 \pm \sqrt{49 + 32}}{4} = \dfrac{7 \pm \sqrt{81}}{4} = \dfrac{7 \pm 9}{4}$

So $y = 4$ or $y = -\dfrac{1}{2}$

Since $y = x^2$, and $x^2$ cannot be negative, $y = 4$ is our only valid solution.

Therefore, $x^2 = 4$, so $x = \pm 2$.

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Calculating the common difference using $x^2 = 4$:

$d = \log_3\left(\dfrac{2x^2 + 1}{x^2 - 1}\right) = \log_3\left(\dfrac{2(4) + 1}{4 - 1}\right) = \log_3\left(\dfrac{9}{3}\right) = \log_3(3) = 1$

Calculating the first three terms:

$a_1 = \log_3(x^2 - 1) = \log_3(4 - 1) = \log_3(3) = 1$

$a_2 = \log_3(2x^2 + 1) = \log_3(2(4) + 1) = \log_3(9) = 2$

$a_3 = \log_3(6x^2 + 3) = \log_3(6(4) + 3) = \log_3(27) = 3$

We see the pattern: $a_n = n$

Therefore:

$a_4 = 4$

$a_5 = 5$

$a_6 = 6$

The sum of the next three terms = $a_4 + a_5 + a_6 = 4 + 5 + 6 = 15$

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Alternate Solution

Given: $\log _{3}\left(x^{2}-1\right), \log _{3}\left(2 x^{2}+1\right), \log _{3}\left(6 x^{2}+3\right)$ are in AP.

For AP: 2(middle term) $=$ first term + third term

$\begin{aligned} & 2 \log _{3}\left(2 x^{2}+1\right)=\log _{3}\left(x^{2}-1\right)+\log _{3}\left(6 x^{2}+3\right) \\ & \log _{3}\left(2 x^{2}+1\right)^{2}=\log _{3}\left[\left(x^{2}-1\right)\left(6 x^{2}+3\right)\right] \\ & \left(2 x^{2}+1\right)^{2}=\left(x^{2}-1\right)\left(6 x^{2}+3\right) \\ & 4 x^{4}+4 x^{2}+1=6 x^{4}-3 x^{2}-3 \\ & 2 x^{4}-7 x^{2}-4=0 \\ & x^{2}=4\text { or } x^2=-0.5(\text {not possible}) \\ & x^{2}=4, \text { so } x=2 \\ & \text { Terms: } \log _{3}(3)=1, \log _{3}(9)=2, \log _{3}(27)=3 \\ & \text { Common difference }=1 \\ & \text { Next three terms: } 4,5,6 \\ & \text { Sum }=\mathbf{1 5} \end{aligned}$