Determine Progression Type for Partial Sums of Geometric Sequence

To determine the correct choice, let us analyze both statements step-by-step.

Since $a_1, a_2, a_3, a_4$ are four distinct positive numbers in Geometric Progression (G.P.), we can express them as:
$a_1 = a, \quad a_2 = ar, \quad a_3 = ar^2, \quad a_4 = ar^3$
where $a > 0$, $r > 0$, and $r \neq 1$ (since the terms are distinct).

The partial sums $b_1, b_2, b_3, b_4$ are defined as:
$b_1 = a_1 = a$
$b_2 = b_1 + a_2 = a + ar = a(1 + r)$
$b_3 = b_2 + a_3 = a(1 + r + r^2)$
$b_4 = b_3 + a_4 = a(1 + r + r^2 + r^3)$

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Analysis of STATEMENT-1:

We need to test if $b_1, b_2, b_3, b_4$ are in Arithmetic Progression (A.P.) or Geometric Progression (G.P.).

1. Testing for A.P.:
   The common differences between consecutive terms are:
   $b_2 - b_1 = a(1 + r) - a = ar$
   $b_3 - b_2 = a(1 + r + r^2) - a(1 + r) = ar^2$
   Since $r \neq 1$ and $a > 0$, we have $ar \neq ar^2$. Thus, $b_2 - b_1 \neq b_3 - b_2$, which implies $b_1, b_2, b_3, b_4$ are not in A.P.

2. Testing for G.P.:
   The ratios between consecutive terms are:
   $\frac{b_2}{b_1} = \frac{a(1 + r)}{a} = 1 + r$
   $\frac{b_3}{b_2} = \frac{a(1 + r + r^2)}{a(1 + r)} = \frac{1 + r + r^2}{1 + r}$
   If $b_1, b_2, b_3, b_4$ were in G.P., then:
   $\frac{b_2}{b_1} = \frac{b_3}{b_2} \implies 1 + r = \frac{1 + r + r^2}{1 + r}$
   $(1 + r)^2 = 1 + r + r^2 \implies 1 + 2r + r^2 = 1 + r + r^2 \implies r = 0$
   However, $r > 0$ because $a_1, a_2, a_3, a_4$ are positive numbers. Thus, $\frac{b_2}{b_1} \neq \frac{b_3}{b_2}$, which implies $b_1, b_2, b_3, b_4$ are not in G.P.

Hence, STATEMENT-1 is True.

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Analysis of STATEMENT-2:

We need to test if $b_1, b_2, b_3, b_4$ are in Harmonic Progression (H.P.).

If $b_1, b_2, b_3, b_4$ are in H.P., then their reciprocals $\frac{1}{b_1}, \frac{1}{b_2}, \frac{1}{b_3}, \frac{1}{b_4}$ must be in A.P., which requires:
$\frac{1}{b_2} - \frac{1}{b_1} = \frac{1}{b_3} - \frac{1}{b_2}$

Evaluating both sides:
$\text{LHS} = \frac{1}{a(1 + r)} - \frac{1}{a} = \frac{1 - (1 + r)}{a(1 + r)} = \frac{-r}{a(1 + r)}$

$\text{RHS} = \frac{1}{a(1 + r + r^2)} - \frac{1}{a(1 + r)} = \frac{(1 + r) - (1 + r + r^2)}{a(1 + r)(1 + r + r^2)} = \frac{-r^2}{a(1 + r)(1 + r + r^2)}$

Equating LHS and RHS:
$\frac{-r}{a(1 + r)} = \frac{-r^2}{a(1 + r)(1 + r + r^2)}$

Since $a > 0$ and $r > 0$, we can cancel $\frac{-r}{a(1 + r)}$ from both sides:
$1 = \frac{r}{1 + r + r^2} \implies 1 + r + r^2 = r \implies 1 + r^2 = 0$

Since $r$ is a real positive number, $1 + r^2 = 0$ has no real solutions.

Hence, $b_1, b_2, b_3, b_4$ are not in H.P., so STATEMENT-2 is False.

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Conclusion:

• STATEMENT-1 is True.
• STATEMENT-2 is False.

This corresponds to Option C.