Let P be the sum of first n positive terms of an increasing arithmetic progression A. Let Q be the sum of first n positive terms of another increasing arithmetic progression B. Let P ∶ Q = (5n + 4) ∶ (9n + 6)
What is the ratio of their 10 th terms ?
Solving Arithmetic Progression Ratio Problems
This problem involves two increasing arithmetic progressions (APs). We are given the ratio of the sums of the first 'n' positive terms of these two APs and asked to find the ratio of their 10th terms.
Let the first arithmetic progression be A, with the first term \(a_1\) and common difference \(d_1\).
Let the second arithmetic progression be B, with the first term \(a_2\) and common difference \(d_2\).
Since the progressions are increasing, we know that \(d_1 > 0\) and \(d_2 > 0\).
Formulas for Arithmetic Progressions
The sum of the first n terms of an arithmetic progression is given by the formula:
\(\qquad S_n = \frac{n}{2}[2a + (n-1)d]\)
where $a$ is the first term and $d$ is the common difference.
The k-th term of an arithmetic progression is given by the formula:
\(\qquad a_k = a + (k-1)d\)
Setting up the Given Information
Let P be the sum of the first n terms of AP A. Using the sum formula:
\(\qquad P = \frac{n}{2}[2a_1 + (n-1)d_1]\)
Let Q be the sum of the first n terms of AP B. Using the sum formula:
\(\qquad Q = \frac{n}{2}[2a_2 + (n-1)d_2]\)
We are given the ratio of P to Q:
\(\qquad \frac{P}{Q} = \frac{5n + 4}{9n + 6}\)
Substituting the formulas for P and Q:
\(\qquad \frac{\frac{n}{2}[2a_1 + (n-1)d_1]}{\frac{n}{2}[2a_2 + (n-1)d_2]} = \frac{5n + 4}{9n + 6}\)
The \(\frac{n}{2}\) terms cancel out:
\(\qquad \frac{2a_1 + (n-1)d_1}{2a_2 + (n-1)d_2} = \frac{5n + 4}{9n + 6}\)
Finding the Ratio of the 10th Terms
We need to find the ratio of the 10th terms of the two APs. Using the formula for the k-th term, the 10th term of AP A is:
\(\qquad a_{10, A} = a_1 + (10-1)d_1 = a_1 + 9d_1\)
The 10th term of AP B is:
\(\qquad a_{10, B} = a_2 + (10-1)d_2 = a_2 + 9d_2\)
We want to find the ratio \(\frac{a_1 + 9d_1}{a_2 + 9d_2}\).
Let's look back at the expression we derived from the sum ratio:
\(\qquad \frac{2a_1 + (n-1)d_1}{2a_2 + (n-1)d_2} = \frac{5n + 4}{9n + 6}\)
We can divide the numerator and denominator of the left side by 2:
\(\qquad \frac{a_1 + \frac{(n-1)}{2}d_1}{a_2 + \frac{(n-1)}{2}d_2} = \frac{5n + 4}{9n + 6}\)
To make the left side match the form of the ratio of the 10th terms (\(\frac{a_1 + 9d_1}{a_2 + 9d_2}\)), we need the coefficient of \(d_1\) and \(d_2\) to be 9.
So, we must have:
\(\qquad \frac{n-1}{2} = 9\)
Solving for n:
\(\qquad n-1 = 9 \times 2\)
\(\qquad n-1 = 18\)
\(\qquad n = 18 + 1\)
\(\qquad n = 19\)
This means that if we substitute $n=19$ into the given ratio of the sums expression, we will get the ratio of the 10th terms.
Calculating the Ratio
Substitute $n=19$ into the expression \(\frac{5n + 4}{9n + 6}\):
Ratio of 10th terms \(= \frac{5(19) + 4}{9(19) + 6}\)
Calculate the numerator:
\(\qquad 5 \times 19 + 4 = 95 + 4 = 99\)
Calculate the denominator:
\(\qquad 9 \times 19 + 6 = 171 + 6 = 177\)
The ratio is \(\frac{99}{177}\).
Simplifying the Ratio
Now, we simplify the fraction \(\frac{99}{177}\). Both numbers are divisible by 3 (since the sum of digits $9+9=18$ and $1+7+7=15$ are divisible by 3).
\(\qquad \frac{99 \div 3}{177 \div 3} = \frac{33}{59}\)
The ratio of the 10th terms of the two arithmetic progressions is 33/59.
| Step | Description | Calculation/Formula |
|---|---|---|
| 1 | Identify formulas for sum and term of AP. | \(S_n = \frac{n}{2}[2a + (n-1)d]\), \(a_k = a + (k-1)d\) |
| 2 | Set up the ratio of sums using formulas. | \(\frac{P}{Q} = \frac{2a_1 + (n-1)d_1}{2a_2 + (n-1)d_2} = \frac{5n + 4}{9n + 6}\) |
| 3 | Express ratio of 10th terms. | \(\frac{a_{10, A}}{a_{10, B}} = \frac{a_1 + 9d_1}{a_2 + 9d_2}\) |
| 4 | Relate sum ratio expression to term ratio expression. | \(\frac{a_1 + \frac{(n-1)}{2}d_1}{a_2 + \frac{(n-1)}{2}d_2} = \frac{a_1 + 9d_1}{a_2 + 9d_2} \implies \frac{n-1}{2} = 9\) |
| 5 | Solve for n. | $n = 19$ |
| 6 | Substitute n=19 into the given sum ratio. | \(\frac{5(19) + 4}{9(19) + 6} = \frac{99}{177}\) |
| 7 | Simplify the resulting ratio. | \(\frac{99}{177} = \frac{33}{59}\) |
Conclusion
The ratio of the 10th terms of the two arithmetic progressions is 33/59.
Revision Table: Arithmetic Progressions
| Concept | Formula | Description |
|---|---|---|
| n-th term (\(a_n\)) | \(a_n = a + (n-1)d\) | Value of the term at position n. |
| Sum of first n terms (\(S_n\)) | \(S_n = \frac{n}{2}[2a + (n-1)d]\) | Sum of the first n terms. |
| Alternative Sum Formula | \(S_n = \frac{n}{2}(a + a_n)\) | Sum using the first and last term. |
| Common Difference (d) | \(d = a_{n} - a_{n-1}\) | Constant difference between consecutive terms. |
Additional Information: Relating Sum and Term Ratios in APs
A common type of problem involving arithmetic progressions gives the ratio of the sums of the first 'n' terms of two different APs and asks for the ratio of their m-th terms.
The sum of the first n terms of an AP with first term $a$ and common difference $d$ is \(S_n = \frac{n}{2}[2a + (n-1)d]\).
The m-th term of an AP is \(a_m = a + (m-1)d\).
If we have two APs, say AP1 (first term \(a_1\), diff \(d_1\)) and AP2 (first term \(a_2\), diff \(d_2\)), the ratio of their sums of n terms is:
\(\qquad \frac{S_{n,1}}{S_{n,2}} = \frac{\frac{n}{2}[2a_1 + (n-1)d_1]}{\frac{n}{2}[2a_2 + (n-1)d_2]} = \frac{2a_1 + (n-1)d_1}{2a_2 + (n-1)d_2} = \frac{a_1 + \frac{(n-1)}{2}d_1}{a_2 + \frac{(n-1)}{2}d_2}\)
The ratio of their m-th terms is:
\(\qquad \frac{a_{m,1}}{a_{m,2}} = \frac{a_1 + (m-1)d_1}{a_2 + (m-1)d_2}\)
To find the ratio of the m-th terms from the ratio of the sums of n terms, we equate the coefficient of the common difference terms:
\(\qquad \frac{n-1}{2} = m-1\)
Solving for n in terms of m gives $n-1 = 2(m-1)$, so $n = 2m - 2 + 1 = 2m - 1$.
This means that the ratio of the m-th terms of two APs is equal to the ratio of the sums of the first $(2m-1)$ terms of those APs.
In this specific problem, we needed the ratio of the 10th terms (so $m=10$). According to the relationship, this ratio is the same as the ratio of the sums of the first $2(10) - 1 = 20 - 1 = 19$ terms. This confirms why setting \(\frac{n-1}{2} = 9\) (where 9 is $m-1$ for $m=10$) and solving for $n=19$ was the correct approach.
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