Suppose four distinct positive numbers $a_1, a_2, a_3, a_4$ are in G.P. Let $b_1 = a_1$, $b_2 = b_1 + a_2$, $b_3 = b_2 + a_3$ and $b_4 = b_3 + a_4$. Statements-I : The numbers $b_1, b_2, b_3, b_4$ are neither in A.P. nor in G.P. and Statements-II : The numbers $b_1, b_2, b_3, b_4$ are in H.P.

Suppose four distinct positive numbers a₁, a₂, a₃, a₄ are in G.P. Let b₁ = a₁, b₂ = b₁ + a₂, b₃ = b₂ + a₃ and b₄ = b₃ + a₄. Statements-I : The numbers b₁, b₂, b₃, b₄ are neither in A.P. nor in G.P. and Statements-II : The numbers b₁, b₂, b₃, b₄ are in H.P. $a_1, a_2, a_3, a_4 \text{ distinct positive, in G.P.}$ $b_1 = a_1, \quad b_k = b_{k-1} + a_k \ \ (k = 2, 3, 4)$ $\text{I: } b_1, b_2, b_3, b_4 \text{ neither A.P. nor G.P.}$ $\text{II: } b_1, b_2, b_3, b_4 \text{ are in H.P.}$ $\text{Choose the correct option}$ $a_1 = A, \ a_2 = Ar, \ a_3 = Ar^2, \ a_4 = Ar^3$ $A > 0, \quad r > 0, \quad r \neq 1$ $b_1 = A$ $b_2 = A + Ar = A(1+r)$ $b_3 = A(1+r) + Ar^2 = A(1+r+r^2)$ $b_4 = A(1+r+r^2) + Ar^3$ $= A(1+r+r^2+r^3)$ $b_2 - b_1 = Ar$ $b_3 - b_2 = Ar^2, \quad b_4 - b_3 = Ar^3$ $Ar = Ar^2 \ \Rightarrow \ r = 1$ $\text{but } r \neq 1 \ \Rightarrow \ \text{not A.P.}$ $b_2^2 = A^2(1+r)^2$ $b_1 b_3 = A^2(1+r+r^2)$ $A^2(1+2r+r^2) \neq A^2(1+r+r^2)$ $\Rightarrow \text{not G.P.}$ $\therefore \text{Statement-I is TRUE}$ $\text{H.P.} \iff \frac{1}{b_1}, \frac{1}{b_2}, \frac{1}{b_3}, \frac{1}{b_4} \text{ in A.P.}$ $\iff \frac{2}{b_2} = \frac{1}{b_1} + \frac{1}{b_3}$ $\frac{2}{A(1+r)} = \frac{1}{A} + \frac{1}{A(1+r+r^2)}$ $\Rightarrow \frac{2}{1+r} = 1 + \frac{1}{1+r+r^2}$ $2(1+r+r^2) = (1+r)(2+r+r^2)$ $2+2r+2r^2 = 2+3r+2r^2+r^3$ $\Rightarrow r + r^3 = 0$ $\Rightarrow r(1+r^2) = 0 \ \Rightarrow \ r = 0$ $\text{impossible, since } r > 0$ $\therefore \text{Statement-II is FALSE}$ $\text{I is True, II is False} \ \Rightarrow \ (3)$