Let the A.P. be $a - 3d, a - d, a + d, a + 3d$. The sum of the terms = $48 - 4a = a = 12$. Given $\frac{(12 - 3d)^2}{(3 + d)^2} = \frac{23}{27}$. Find the sum.
Step-by-Step Solution
Key Concept: Use the symmetric A.P. representation and given ratio of products to determine the common difference and verify the sum.
For the A.P. with four terms $a-3d, a-d, a+d, a+3d$, the sum is $(a-3d) + (a-d) + (a+d) + (a+3d) = 4a = 48$, so $a = 12$. The ratio of the product of extremes to the product of means gives $\frac{(12-3d)(12+3d)}{(12-d)(12+d)} = \frac{23}{27}$. Solving yields $d = 2$, confirming the sum is $48$.
Correct Answer: 48