IPMAT Indore 2025 (SA) - If _3(x^2 - 1), _3(2x^2 + 1) and _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 | PYQs + Solutions

IPMAT Indore 2025

Modern Math

Logarithms

Medium

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

Entered answer:

✅ Correct Answer: 15

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}

Cross-multiplying:

\newline

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

Expanding and rearranging:

\newline

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

\newline

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

Let

y = x^2

to simplify:

\newline

-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}

y = 4

y = -\dfrac{1}{2}

Since

y = x^2

, and

x^2

cannot be negative,

y = 4

is our only valid solution.

\newline

Therefore,

x^2 = 4

, so

x = \pm 2

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:

\newline

a_4 = 4

\newline

a_5 = 5

\newline

a_6 = 6

The sum of the next three terms =

a_4 + a_5 + a_6 = 4 + 5 + 6 = 15