In an AP of 4, 6 1/2, 9, 11 1/2, ..., determine the number of terms needed to make a total of 126.

The given AP is $4 + 6\frac{1}{2} + 9 + 11\frac{1}{2} + \dots$
The first term is $a = 4$.
The common difference $d$ is the difference between consecutive terms:

$$d = 6\frac{1}{2} - 4 = \frac{13}{2} - \frac{8}{2} = \frac{5}{2}$$

We can verify this with the next terms:

$$d = 9 - 6\frac{1}{2} = 9 - \frac{13}{2} = \frac{18}{2} - \frac{13}{2} = \frac{5}{2}$$

The sum $S_n$ is given as 126. The formula for $S_n$ is:

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

Substitute the known values $S_n = 126$, $a = 4$, and $d = \frac{5}{2}$:

$$126 = \frac{n}{2}\left[2(4) + (n-1)\frac{5}{2}\right]$$ $$126 = \frac{n}{2}\left[8 + \frac{5}{2}(n-1)\right]$$

Multiply both sides by 2:

$$2 \times 126 = n\left[8 + \frac{5}{2}(n-1)\right]$$ $$252 = n\left[8 + \frac{5n}{2} - \frac{5}{2}\right]$$

Combine the constant terms inside the bracket:

$$252 = n\left[\frac{16}{2} + \frac{5n}{2} - \frac{5}{2}\right]$$ $$252 = n\left[\frac{11 + 5n}{2}\right]$$

Multiply both sides by 2 again:

$$2 \times 252 = n(11 + 5n)$$ $$504 = 11n + 5n^2$$

Rearrange into the standard quadratic form $An^2 + Bn + C = 0$:

$$5n^2 + 11n - 504 = 0$$

Using the quadratic formula $n = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}$:
Here, $A=5$, $B=11$, $C=-504$.

$$n = \frac{-11 \pm \sqrt{11^2 - 4(5)(-504)}}{2(5)}$$ $$n = \frac{-11 \pm \sqrt{121 + 10080}}{10}$$ $$n = \frac{-11 \pm \sqrt{10201}}{10}$$

Calculate the square root: $\sqrt{10201} = 101$.

$$n = \frac{-11 \pm 101}{10}$$

Two possible solutions for $n$:

$$n_1 = \frac{-11 + 101}{10} = \frac{90}{10} = 9$$ $$n_2 = \frac{-11 - 101}{10} = \frac{-112}{10} = -11.2$$

Since the number of terms cannot be negative or a fraction, we take the positive integer value.

The number of terms needed is 9.