m Arithmetic Means Are Inserted Between 1 and 31 So That the Ratio of the 7th and (m−1)th Means Is 5:9. Find the Value of m

A standard arithmetic-means problem, not tied to a specific exam paper.

Problem

$m$ arithmetic means are inserted between $1$ and $31$, so that the ratio of the 7th and $(m-1)$th means is $5:9$. Find the value of $m$.

Key insight. Inserting $m$ means between two numbers creates one long AP of $m+2$ terms total, with $1$ and $31$ as its first and last terms. The common difference $D$ is then completely determined by $m$ alone — from there, “the 7th mean” and “the $(m-1)$th mean” are just specific terms of this single AP, both expressible purely in terms of $m$.

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Approach

Recognise that $1$, followed by $m$ inserted means, followed by $31$, forms an AP of $m+2$ terms. Use the last term to find the common difference $D$ in terms of $m$. The 7th mean is the 8th term of this AP, and the $(m-1)$th mean is the $m$th term — write both using $D$, form their ratio, and solve for $m$.

Solution

Step 1 — Set up the AP and find D

The full sequence is $1, A_1, A_2, \ldots, A_m, 31$ — a total of $m+2$ terms, with first term $1$ and last term $31$:

$$31 = 1 + (m+1)D \implies D = \frac{30}{m+1}$$

Step 2 — Express the 7th mean

The 7th mean $A_7$ is the 8th term of the AP:

$$A_7 = 1+7D = 1+\frac{210}{m+1} = \frac{(m+1)+210}{m+1} = \frac{m+211}{m+1}$$

Step 3 — Express the (m−1)th mean

The $(m-1)$th mean $A_{m-1}$ is the $m$th term of the AP:

$$A_{m-1} = 1+(m-1)D = 1+\frac{30(m-1)}{m+1} = \frac{(m+1)+30(m-1)}{m+1} = \frac{31m-29}{m+1}$$

Step 4 — Form the ratio and set it equal to 5:9

$$\frac{A_7}{A_{m-1}} = \frac{m+211}{31m-29} = \frac{5}{9}$$

Step 5 — Cross-multiply

$$9(m+211) = 5(31m-29)$$

$$9m+1899 = 155m-145$$

Step 6 — Solve for m

$$1899+145 = 155m-9m$$

$$2044 = 146m \implies m = \frac{2044}{146} = 14$$

Answer

Common mistakes

• Miscounting which term number corresponds to “the 7th mean.” Since the sequence starts with $1$ as the first term (not the first mean), the 7th mean is the 8th term of the AP, and the $(m-1)$th mean is the $m$th term — off-by-one errors here silently produce a completely different (and wrong) value of $m$.
• Forgetting that both $D$ and $m$ appear in every expression. Since $D=\dfrac{30}{m+1}$ already depends on $m$, substituting $D$ into the term formulas before simplifying is essential — trying to solve for $m$ and $D$ as if they were independent unknowns leads nowhere.

Practise next

• $n$ arithmetic means are inserted between $2$ and $38$, so that the ratio of the 4th mean to the $(n-3)$th mean is $1:3$. Find $n$, using the same single-AP setup.