Find the sum of the two middle-most terms of the AP: \(-\dfrac{4}{3}, -1, -\dfrac{2}{3},\ldots, 4\dfrac{1}{3}\).
3
Step 1: Identify first term, common difference and last term
The first term \(a = -\dfrac{4}{3}\).
The common difference is \(d = -1 - \left(-\dfrac{4}{3}\right) = -1 + \dfrac{4}{3} = \dfrac{1}{3}\).
The last term is given as \(l = 4\dfrac{1}{3} = \dfrac{13}{3}\).
Step 2: Find the number of terms in the AP
We use the formula for the \(n\)-th term of an AP: \(a_n = a + (n-1) d\).
Here \(a_n = l = \dfrac{13}{3}\).
So, \(\dfrac{13}{3} = -\dfrac{4}{3} + (n-1) \cdot \dfrac{1}{3}\).
Multiply through by 3 to remove denominators: \(13 = -4 + (n-1)\).
So, \(13 + 4 = n - 1 \Rightarrow 17 = n - 1 \Rightarrow n = 18\).
Therefore, there are 18 terms in this AP.
Step 3: Find the middle terms
If \(n\) is even, then there are two middle terms. They are the \(\tfrac{n}{2}\)-th term and the \(\left(\tfrac{n}{2} + 1\right)\)-th term.
Here, \(n = 18\), so the middle terms are the 9th term and the 10th term.
Step 4: Calculate the 9th term
Formula: \(a_n = a + (n-1)d\).
So, \(a_9 = -\dfrac{4}{3} + (9-1) \cdot \dfrac{1}{3}\).
= \(-\dfrac{4}{3} + 8 \cdot \dfrac{1}{3}\).
= \(-\dfrac{4}{3} + \dfrac{8}{3}\).
= \(\dfrac{4}{3}\).
Step 5: Calculate the 10th term
\(a_{10} = -\dfrac{4}{3} + (10-1) \cdot \dfrac{1}{3}\).
= \(-\dfrac{4}{3} + 9 \cdot \dfrac{1}{3}\).
= \(-\dfrac{4}{3} + \dfrac{9}{3}\).
= \(\dfrac{5}{3}\).
Step 6: Add the two middle terms
\(a_9 + a_{10} = \dfrac{4}{3} + \dfrac{5}{3} = \dfrac{9}{3} = 3\).
Final Answer: The sum of the two middle-most terms is 3.