Ratio of nth Terms from Ratio of Sums in AP: Class 10

Quick answer: If the ratio of the sums of n terms of two APs is given, you can find the ratio of their mth terms by replacing n with 2m − 1 in the given ratio. This works because $\frac{S_n}{S'_n} = \frac{a + \frac{n - 1}{2}d}{a' + \frac{n - 1}{2}d'}$, and $\frac{n - 1}{2} = m - 1$ exactly when $n = 2m - 1$. For example, if $S_n : S'_n = (7n + 1) : (4n + 27)$, the 11th terms are in the ratio $4 : 3$.

The two AP formulas you need

For an AP with first term $a$ and common difference $d$:

Why n = 2m − 1 works

Take two APs: the first with $a$, $d$ and the second with $a'$, $d'$.

Step 1: Write the ratio of their sums of n terms and cancel $\frac{n}{2}$:

$\frac{S_n}{S'_n} = \frac{2a + (n - 1)d}{2a' + (n - 1)d'}$

Step 2: Divide the numerator and denominator by 2:

$\frac{S_n}{S'_n} = \frac{a + \frac{n - 1}{2}d}{a' + \frac{n - 1}{2}d'}$

Step 3: Compare this with the ratio of the mth terms:

$\frac{a_m}{a'_m} = \frac{a + (m - 1)d}{a' + (m - 1)d'}$

Step 4: The two expressions are the same when $\frac{n - 1}{2} = m - 1$, that is $n - 1 = 2m - 2$, so

$n = 2m - 1$

Rule: To get the ratio of the mth terms, substitute $n = 2m - 1$ in the ratio of the sums.

Worked example 1: ratio of 11th terms

The sums of n terms of two APs are in the ratio $(7n + 1) : (4n + 27)$. Find the ratio of their 11th terms.

Step 1: Here $m = 11$, so $n = 2(11) - 1 = 21$.

Step 2: Substitute $n = 21$:

$\frac{a_{11}}{a'_{11}} = \frac{7(21) + 1}{4(21) + 27} = \frac{148}{111}$

Step 3: Simplify. Both numbers are divisible by 37: $148 = 4 \times 37$ and $111 = 3 \times 37$.

Answer: The ratio of the 11th terms is $4 : 3$.

Check: One pair of APs with this sum ratio is $a = 4$, $d = 7$ and $a' = 15.5$, $d' = 4$. Their 11th terms are $4 + 70 = 74$ and $15.5 + 40 = 55.5$, and $74 : 55.5 = 4 : 3$.

Worked example 2: the m² : n² result

The ratio of the sums of m and n terms of an AP is $m^2 : n^2$. Show that the ratio of the mth and nth terms is $(2m - 1) : (2n - 1)$.

Step 1: Write the given ratio using the sum formula:

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

Step 2: Cancel $\frac{m}{n}$ from both sides:

$\frac{2a + (m - 1)d}{2a + (n - 1)d} = \frac{m}{n}$

Step 3: Cross-multiply and expand:

$2an + (mn - n)d = 2am + (mn - m)d$

Step 4: Collect terms:

$2a(n - m) = (n - m)d$

Since $m \ne n$, divide by $(n - m)$ to get $d = 2a$.

Step 5: Substitute $d = 2a$ in the ratio of the terms:

$\frac{a_m}{a_n} = \frac{a + (m - 1)2a}{a + (n - 1)2a} = \frac{2m - 1}{2n - 1}$

Hence proved.

The reverse method: from terms to sums

If the ratio of the mth terms is given and you need the ratio of the sums of n terms, replace m with $\frac{n + 1}{2}$.

Example: The nth terms of two APs are in the ratio $(3n + 1) : (4n - 1)$. Find the ratio of their sums of n terms.

Step 1: Replace n in the term ratio by $\frac{n + 1}{2}$:

$\frac{3 \cdot \frac{n + 1}{2} + 1}{4 \cdot \frac{n + 1}{2} - 1} = \frac{\frac{3n + 5}{2}}{2n + 1}$

Step 2: Multiply the numerator and denominator by 2.

Answer: $S_n : S'_n = (3n + 5) : (4n + 2)$.

Check: The APs $4, 7, 10, \ldots$ and $3, 7, 11, \ldots$ have nth terms $3n + 1$ and $4n - 1$. Their sums are $\frac{n}{2}(3n + 5)$ and $\frac{n}{2}(4n + 2)$, which match.

Summary table

Exam tips

• In a board answer, show the step $\frac{n - 1}{2} = m - 1$, so the examiner sees where $n = 2m - 1$ comes from.
• Always reduce the final ratio to lowest terms; look for common factors such as 37 in 148 and 111.
• Check that you substituted into the given sum ratio, not into a term formula.

Common mistakes

• Putting $n = m$ instead of $n = 2m - 1$.
• Forgetting to cancel $\frac{n}{2}$ before comparing the two expressions.
• Using the shortcut in reverse without changing it: going from terms to sums needs $m = \frac{n + 1}{2}$.
• Leaving the answer as an unsimplified fraction.

Frequently Asked Questions

Why do we replace n with 2m − 1?

Because the ratio of sums contains $\frac{n - 1}{2}d$, and this equals the $(m - 1)d$ in the mth term exactly when $n = 2m - 1$.

What is the ratio of the 11th terms if the sums are in the ratio (7n + 1) : (4n + 27)?

Put n = 21 to get 148 : 111, which simplifies to 4 : 3.

How do you find the ratio of sums from the ratio of nth terms?

Replace m with (n + 1)/2 in the ratio of the terms and simplify.

Does this method work for a single AP?

The substitution is for comparing two APs. For one AP with $S_m : S_n = m^2 : n^2$, you show that d = 2a and then find the term ratio.

Is this topic in the Class 10 syllabus?

Yes. It uses only the nth term and sum formulas from the Arithmetic Progressions chapter, though it is a higher-order question.