Understanding the Problem

The problem describes two increasing arithmetic progressions, A and B. We are given the sum of the first n positive terms for each, denoted as P and Q, respectively, and a specific ratio for P and Q in terms of n. We need to find the relationship between their common differences, d for A and D for B.

• Arithmetic Progression A has a first term \(a_1\) and common difference $d$.
• Arithmetic Progression B has a first term \(b_1\) and common difference $D$.
• Since both are increasing, $d > 0$ and $D > 0$.
• The first n terms of both progressions are positive, implying \(a_1 > 0\) and \(b_1 > 0\).
• P is the sum of the first n terms of A, and Q is the sum of the first n terms of B.
• The given ratio is \(\frac{P}{Q} = \frac{5n+4}{9n+6}\).

Formula for the Sum of an Arithmetic Progression

The sum of the first n terms of an arithmetic progression with the first term $a$ and common difference $d$ is given by the formula:

\(\qquad S_n = \frac{n}{2}[2a + (n-1)d]\)

Applying this formula to the given progressions A and B:

• Sum P for progression A: \(P = \frac{n}{2}[2a_1 + (n-1)d]\)
• Sum Q for progression B: \(Q = \frac{n}{2}[2b_1 + (n-1)D]\)

Using the Given Ratio of Sums

We are given the ratio \(\frac{P}{Q}\). Substituting the sum formulas, we get:

\(\qquad \frac{\frac{n}{2}[2a_1 + (n-1)d]}{\frac{n}{2}[2b_1 + (n-1)D]} = \frac{5n+4}{9n+6}\)

Assuming \(n \ge 1\) (since we are summing n terms), we can cancel out \(\frac{n}{2}\) from the numerator and denominator on the left side:

\(\qquad \frac{2a_1 + (n-1)d}{2b_1 + (n-1)D} = \frac{5n+4}{9n+6}\)

We can rewrite the terms in the numerator and denominator on the left side to group the terms involving n:

\(\qquad \frac{dn + (2a_1 - d)}{Dn + (2b_1 - D)} = \frac{5n+4}{9n+6}\)

Establishing the Relationship Between Common Differences

This equation involves rational functions of $n$. For this equality to hold true for all positive integer values of $n$, the coefficients of $n$ in the numerator and denominator on both sides must be proportional, and the constant terms in the numerator and denominator on both sides must also be proportional with the same proportionality constant.

Comparing the coefficients of $n$ from both sides:

\(\qquad \frac{\text{Coefficient of } n \text{ in numerator (left)}}{\text{Coefficient of } n \text{ in denominator (left)}} = \frac{d}{D}\)

\(\qquad \frac{\text{Coefficient of } n \text{ in numerator (right)}}{\text{Coefficient of } n \text{ in denominator (right)}} = \frac{5}{9}\)

For the equality to hold for all $n$, these ratios must be equal:

\(\qquad \frac{d}{D} = \frac{5}{9}\)

This gives us a direct relationship between the common differences $d$ and $D$. Cross-multiplying the equation \(\frac{d}{5} = \frac{D}{9}\) (which is equivalent to \(\frac{d}{D} = \frac{5}{9}\)) yields:

\(\qquad 9d = 5D\)

We can express $D$ in terms of $d$ by dividing by 5:

\(\qquad D = \frac{9}{5}d\)

Comparing D and d based on the Relationship

Our derivation shows that the common difference of arithmetic progression B is \(\frac{9}{5}\) times the common difference of arithmetic progression A. That is, \(D = \frac{9}{5}d\).

Since arithmetic progression A is increasing, its common difference $d$ must be a positive value ($d > 0$).

Now let's compare $D$ and $d$ using the relationship \(D = \frac{9}{5}d\). We can write $D$ as $1.8d$.

We are comparing $1.8d$ with $d$. Since $d > 0$, multiplying the inequality $1.8 > 1$ by $d$ gives $1.8d > d$.

Therefore, based on the derived relationship and the fact that $d$ is positive, we conclude that $D$ is always greater than $d$ in this scenario.

\(\qquad D > d\)

Examining the Options

Let's check which of the given options is consistent with our finding \(D = \frac{9}{5}d\) where $d > 0$:

• Option 1: $D > d$
  As shown above, since $D = 1.8d$ and $d > 0$, it is always true that $D > d$.
• Option 2: $D < d$
  Since $D = 1.8d$ and $d > 0$, $1.8d < d$ would imply $1.8 < 1$, which is false. So, this option is not correct.
• Option 3: $7D > 12d$
  Let's substitute \(D = \frac{9}{5}d\) into this inequality: \(7\left(\frac{9}{5}d\right) > 12d \implies \frac{63}{5}d > 12d \implies 12.6d > 12d\). Since $d > 0$, this inequality is true (as $12.6 > 12$). This statement is also always correct given the problem conditions.
• Option 4: None of the above
  Since Option 1 is always correct, this option is incorrect.

Both Option 1 ($D > d$) and Option 3 ($7D > 12d$) are true consequences of the fundamental relationship \(D = \frac{9}{5}d\). However, Option 1 provides a direct comparison between the magnitudes of the common differences, which is immediately apparent from $D=1.8d$. The question asks which one is "always correct" among the options provided.

Conclusion

The analysis of the sum ratio for the first n positive terms of the two increasing arithmetic progressions leads to the fundamental relationship \(D = \frac{9}{5}d\) between their common differences. Since $d$ is positive, $D$ must also be positive, and specifically, $D$ is 1.8 times $d$. This directly implies that $D$ is greater than $d$.

The relationship that is always correct among the given options is $D > d$.