NDA 2 2024 Maths, question 43

Solution.

Answer: (d) $6$

1. Let the terms be $a, a + d, a + 2d, a + 3d, a + 4d$. The GP condition $(a + d)^{2} = a(a + 4d)$ gives $d(d - 2a) = 0$, so $d = 2a$.
2. The terms are $a, 3a, 5a, 7a, 9a$ with product $945a^{5} = 229635$, so $a^{5} = 243$ and $a = 3$.
3. So $d = 2a = 6$.

Trap: $d = 0$ also satisfies the GP condition, but then all five terms are equal and 0 is not an option; the question means a non-constant AP.