Question

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 ________

Hint:

When you see logarithmic terms in an AP or GP, immediately apply the core property (\(2b=a+c\) for AP, \(b^2=ac\) for GP) and then use logarithm rules to simplify the equation. Always check your final value of the variable against the domain of the logarithmic functions.

Answer 1

Correct Answer: 15

Step 1: Understanding the Concept:
The problem involves properties of arithmetic progressions (AP) and logarithms. If three terms \(a, b, c\) are in AP, then the middle term is the arithmetic mean of the other two, i.e., \(2b = a+c\).
Step 2: Key Formula or Approach:
1. AP property: \(2b = a+c\).
2. Logarithm properties: \(n \log_b(m) = \log_b(m^n)\) and \(\log_b(m) + \log_b(n) = \log_b(mn)\).
3. Formula for the n-th term of an AP: \(T_n = a + (n-1)d\).
Step 3: Detailed Explanation:
Let the three terms be \(T_1 = \log_3(x^2 - 1)\), \(T_2 = \log_3(2x^2 + 1)\), and \(T_3 = \log_3(6x^2 + 3)\).
Since they are in AP, we have \(2T_2 = T_1 + T_3\). \[ 2\log_3(2x^2 + 1) = \log_3(x^2 - 1) + \log_3(6x^2 + 3) \] Using logarithm properties: \[ \log_3((2x^2 + 1)^2) = \log_3((x^2 - 1)(6x^2 + 3)) \] Equating the arguments of the logarithm: \[ (2x^2 + 1)^2 = (x^2 - 1)(6x^2 + 3) \] \[ 4x^4 + 4x^2 + 1 = 6x^4 + 3x^2 - 6x^2 - 3 \] \[ 4x^4 + 4x^2 + 1 = 6x^4 - 3x^2 - 3 \] Rearranging the terms to form a quadratic equation in \(x^2\): \[ 2x^4 - 7x^2 - 4 = 0 \] Let \(y = x^2\). The equation becomes \(2y^2 - 7y - 4 = 0\). Factoring the quadratic: \[ 2y^2 - 8y + y - 4 = 0 \] \[ 2y(y - 4) + 1(y - 4) = 0 \] \[ (2y + 1)(y - 4) = 0 \] This gives \(y = -1/2\) or \(y = 4\). Since \(y = x^2\), it cannot be negative. So, \(x^2 = 4\).
We must check that the arguments of the logarithms are positive for \(x^2=4\):

\(x^2 - 1 = 4 - 1 = 3>0\) (OK)
\(2x^2 + 1 = 2(4) + 1 = 9>0\) (OK)
\(6x^2 + 3 = 6(4) + 3 = 27>0\) (OK)
Now, let's find the first three terms of the AP: \[ T_1 = \log_3(3) = 1 \] \[ T_2 = \log_3(9) = \log_3(3^2) = 2 \] \[ T_3 = \log_3(27) = \log_3(3^3) = 3 \] The AP is 1, 2, 3, ... with first term \(a=1\) and common difference \(d = T_2 - T_1 = 2 - 1 = 1\).
We need to find the sum of the next three terms: \(T_4, T_5, T_6\). \[ T_4 = a + 3d = 1 + 3(1) = 4 \] \[ T_5 = a + 4d = 1 + 4(1) = 5 \] \[ T_6 = a + 5d = 1 + 5(1) = 6 \] Sum = \(T_4 + T_5 + T_6 = 4 + 5 + 6 = 15\).
Step 4: Final Answer:
The sum of the next three terms of the progression is 15.