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

Step 1: Set up the two numbers and the means.
Let the two numbers be \(a\) and \(b\) with \(a+b=2\frac{1}{6}=\frac{13}{6}\).
We insert \(n\) arithmetic means between them, where \(n\) is even.
So the full list \(a, m_1, m_2, \ldots, m_n, b\) is one arithmetic progression (AP) with \(n+2\) terms.

Step 2: Write the sum of the means using the AP sum rule.
In any AP, the sum of all terms equals the number of terms times the average of the first and last term.
So the sum of all \(n+2\) terms is \(\dfrac{n+2}{2}(a+b)\).
The sum of only the \(n\) means (drop the two ends \(a\) and \(b\)) is
\[ S = \frac{n+2}{2}(a+b) - (a+b) = \frac{n}{2}(a+b) \]
Put \(a+b=\frac{13}{6}\):
\[ S = \frac{n}{2}\times\frac{13}{6} = \frac{13n}{12} \]

Step 3: Use the condition on the sum of the means.
We are told the sum of the means exceeds their number (\(n\)) by 1, so
\[ S = n+1 \]
\[ \frac{13n}{12} = n+1 \]
Multiply both sides by 12:
\[ 13n = 12n+12 \]
\[ n = 12 \]

Step 4: Check the other options.
\(n=12\) is even, which matches "an even number of means," so it is valid.
Test \(n=6\): \(S=\frac{13\times 6}{12}=6.5\), but \(n+1=7\); these do not match, so option (2) fails.
Test \(n=24\): \(S=\frac{13\times 24}{12}=26\), but \(n+1=25\); these do not match either, so option (3) fails.
Since \(n=12\) is a valid even number that satisfies the condition, option (4), "None," is not needed.

Final Answer:
There are 12 arithmetic means between the two numbers. \[ \boxed{n = 12} \]