Understanding the Problem: Ratio of Arithmetic Progressions

The question involves two increasing arithmetic progressions (APs), let's call them A and B. We are given the ratio of the sum of the first 'n' positive terms of A (denoted by P) to the sum of the first 'n' positive terms of B (denoted by Q). We need to find the ratio of the first term of A to the first term of B.

Let the first term of arithmetic progression A be \(a_1\) and its common difference be \(d_A\).

Let the first term of arithmetic progression B be \(b_1\) and its common difference be \(d_B\).

Since the progressions are increasing, \(d_A > 0\) and \(d_B > 0\). The problem also states the first 'n' positive terms, implying \(a_1 > 0\) and \(b_1 > 0\).

Formula for Sum of an Arithmetic Progression

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

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

Using this formula for AP A and AP B:

• Sum of first n terms of A, \( P = \frac{n}{2}[2a_1 + (n-1)d_A] \)
• Sum of first n terms of B, \( Q = \frac{n}{2}[2b_1 + (n-1)d_B] \)

Using the Given Ratio of Sums

We are given the ratio \( P : Q = (5n + 4) : (9n + 6) \). This can be written as:

\( \frac{P}{Q} = \frac{5n + 4}{9n + 6} \)

Substitute the formulas for P and Q:

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

The term \( \frac{n}{2} \) cancels out from the numerator and denominator:

\( \frac{2a_1 + (n-1)d_A}{2b_1 + (n-1)d_B} = \frac{5n + 4}{9n + 6} \)

Relating Sum Ratio to Term Ratio in Arithmetic Progressions

There is a useful property that relates the ratio of the sum of the first 'n' terms of two arithmetic progressions to the ratio of their individual terms. Specifically, the ratio of the sum of the first 'n' terms is equal to the ratio of their \( (\frac{n+1}{2}) \)-th terms.

Let the k-th term of AP A be \( t_k(A) = a_1 + (k-1)d_A \).

Let the k-th term of AP B be \( t_k(B) = b_1 + (k-1)d_B \).

The property states:

\( \frac{S_n(A)}{S_n(B)} = \frac{t_{(n+1)/2}(A)}{t_{(n+1)/2}(B)} \) (Note: This property holds when comparing the sum of n terms to the ratio of the $(n+1)/2$ term. This works by comparing the structure of the sum formula \(\frac{2a+(n-1)d}{2b+(n-1)d}\) with the structure of the term formula \(\frac{a+(k-1)d}{b+(k-1)d}\). If we replace $(n-1)$ with $2(k-1)$, then \(k-1 = \frac{n-1}{2}\), which gives \(k = \frac{n-1}{2} + 1 = \frac{n+1}{2}\).)

So, we have:

\( \frac{t_k(A)}{t_k(B)} = \frac{5n + 4}{9n + 6} \) where \( k = \frac{n+1}{2} \). This means \( n = 2k - 1 \).

Finding the Ratio of the First Terms

We are asked to find the ratio of the first term of A to that of B, which is \( \frac{a_1}{b_1} \). The first term is the term when \( k=1 \).

To find the ratio of the first terms \( \frac{t_1(A)}{t_1(B)} \), we need to substitute \( k=1 \) into the expression for \( \frac{t_k(A)}{t_k(B)} \). According to the relationship \( n = 2k - 1 \), when \( k=1 \), the corresponding value of \( n \) is:

\( n = 2(1) - 1 = 2 - 1 = 1 \)

So, to find the ratio of the first terms (k=1), we need to evaluate the given ratio of sums \( \frac{5n+4}{9n+6} \) at \( n=1 \).

Ratio of first terms \( = \frac{5(1) + 4}{9(1) + 6} \)

\( = \frac{5 + 4}{9 + 6} \)

\( = \frac{9}{15} \)

Simplifying the fraction:

\( \frac{9}{15} = \frac{3 \times 3}{5 \times 3} = \frac{3}{5} \)

Thus, the ratio of the first term of A to that of B is \( 3/5 \).

Summary of Steps

• Write down the sum of first 'n' terms formulas for both APs.
• Set up the equation based on the given ratio of sums.
• Use the property that the ratio of the sum of the first 'n' terms of two APs is equal to the ratio of their \( (\frac{n+1}{2}) \)-th terms.
• Relate the term index 'k' to 'n' using \( k = \frac{n+1}{2} \), which gives \( n = 2k - 1 \).
• To find the ratio of the first terms (k=1), substitute \( n=2(1)-1=1 \) into the given ratio expression.
• Calculate the final ratio.

Conclusion: Ratio of First Terms

Based on the calculations, the ratio of the first term of arithmetic progression A to that of arithmetic progression B is \( 3/5 \).

Revision Table: Arithmetic Progression Ratios

Additional Information: Properties of Arithmetic Progressions

Arithmetic progressions are sequences where the difference between consecutive terms is constant. This constant difference is called the common difference.

• General Term: The \( k \)-th term of an AP is given by \( a_k = a_1 + (k-1)d \), where \( a_1 \) is the first term and \( d \) is the common difference.
• Sum of First n Terms: The sum \( S_n \) can also be calculated as the average of the first and last term multiplied by the number of terms: \( S_n = \frac{n}{2}(a_1 + a_n) \).
• Arithmetic Mean: For any three consecutive terms in an AP, the middle term is the arithmetic mean of the other two. If \( a, b, c \) are in AP, then \( b = \frac{a+c}{2} \).
• Property Used: The property \( \frac{S_n(A)}{S_n(B)} = \frac{t_{(n+1)/2}(A)}{t_{(n+1)/2}(B)} \) is derived from the formulas. Let's look at the sum ratio again: \( \frac{S_n(A)}{S_n(B)} = \frac{2a_1 + (n-1)d_A}{2b_1 + (n-1)d_B} = \frac{a_1 + \frac{(n-1)}{2}d_A}{b_1 + \frac{(n-1)}{2}d_B} \) Compare this with the ratio of the k-th terms: \( \frac{t_k(A)}{t_k(B)} = \frac{a_1 + (k-1)d_A}{b_1 + (k-1)d_B} \) By setting \( (k-1) = \frac{(n-1)}{2} \), we get \( k = \frac{n-1}{2} + 1 = \frac{n+1}{2} \). So, the ratio of sums of n terms equals the ratio of the \( (n+1)/2 \)-th terms. To find the ratio of the first terms (k=1), we set \( k=1 \) in the relationship \( k = \frac{n+1}{2} \), which gives \( 1 = \frac{n+1}{2} \Rightarrow 2 = n+1 \Rightarrow n=1 \). This substitution method works reliably for finding the ratio of any specific term \( t_k \) by substituting \( n=2k-1 \) into the given sum ratio expression.