Question

Between two numbers whose sum is \(2\frac{1}{6}\), an even number of arithmetic means is inserted between them. The sum of these means exceeds their number by unity. How many means are there?

• A. 12
• B. 6
• C. 24
• D. None

Hint:

Use the rule that the sum of n arithmetic means between two numbers a and b is always \(\frac{n}{2}(a+b)\). Set this equal to n+1 and solve for n.

Answer 1

The Correct Option is A

Take the two given numbers as \(a\) and \(b\), with \(a + b = 2\frac{1}{6} = \frac{13}{6}\). Suppose \(n\) equal arithmetic means \(m_1, m_2, \ldots, m_n\) sit between them, so that \(a, m_1, m_2, \ldots, m_n, b\) is one arithmetic progression with common difference \(d\).

Since there are \(n+2\) terms in total, \(b\) is the \((n+2)\)th term, so

\[ b = a + (n+1)d \]

Now write each mean in terms of \(a\) and \(d\): \(m_k = a + kd\) for \(k = 1, 2, \ldots, n\). Add all n means:

\[ \sum_{k=1}^{n} m_k = na + d\cdot\frac{n(n+1)}{2} \]

From \(b = a+(n+1)d\), we get \(d = \dfrac{b-a}{n+1}\). Put this in:

\[ \sum m_k = na + \frac{n(n+1)}{2}\cdot\frac{b-a}{n+1} = na + \frac{n(b-a)}{2} = \frac{n(a+b)}{2} \]

This confirms the sum of the n means is \(\dfrac{n}{2}(a+b)\). With \(a+b=\frac{13}{6}\), the sum becomes \(\dfrac{13n}{12}\).

The question says this sum is 1 more than n, the count of means:

\[ \frac{13n}{12} = n + 1 \implies 13n = 12n + 12 \implies n = 12 \]

Twelve is indeed even, so it fits the condition "an even number of means" given in the question. Checking the other choices, n=6 gives a sum of 6.5 (should be 7) and n=24 gives a sum of 26 (should be 25), so neither works. Only n=12 satisfies the equation exactly.

\[ \boxed{n = 12} \]