Let $6, (6+d), (6+2d), \ldots$ are in an A.G.P. $S_8 = (6+d)^8$ and $8 = (6+2d)^2$. Find $d$.
Step-by-Step Solution
Key Concept: In an A.G.P., the sum formula and given constraints on terms lead to a quadratic equation for the common difference.
For an A.G.P., the sum $S_8 = \frac{8((6+d)^1 - (6+2d)^2)}{(6+d) - 1} = 48 + 16d$. From $36 + d^2 + 12d = 48 + 16d$, we get $d^2 - 4d - 12 = 0$, which factors as $(d-6)(d+2) = 0$. Since $d$ must be positive and satisfy both conditions, $d = 2$.
Correct Answer: 2