The number of terms in an A.P. is even, the sum of the odd terms in it is 24 and that of the even terms is 30. If the last term exceeds the first term by $$10\frac{1}{2}$$, then the number of terms in the A.P. is:

Let the total number of terms in the arithmetic progression (A.P.) be $$2n$$, since it is even. This means there are $$n$$ terms in odd positions and $$n$$ terms in even positions.

Assume the first term is $$a$$ and the common difference is $$d$$. The sequence is:

Term 1: $$a$$, Term 2: $$a + d$$, Term 3: $$a + 2d$$, Term 4: $$a + 3d$$, ..., Term $$2n$$: $$a + (2n-1)d$$.

The odd-positioned terms are: $$a$$, $$a + 2d$$, $$a + 4d$$, ..., up to $$n$$ terms. This forms an A.P. with first term $$a$$, common difference $$2d$$, and $$n$$ terms. The sum of these odd terms is given as 24:

Sum = $$\frac{n}{2} \times [2a + (n-1) \cdot 2d] = n[a + (n-1)d] = 24$$. ...(1)

The even-positioned terms are: $$a + d$$, $$a + 3d$$, $$a + 5d$$, ..., up to $$n$$ terms. This forms an A.P. with first term $$a + d$$, common difference $$2d$$, and $$n$$ terms. The sum of these even terms is given as 30:

Sum = $$\frac{n}{2} \times [2(a + d) + (n-1) \cdot 2d] = n[a + d + (n-1)d] = n[a + nd] = 30$$. ...(2)

Subtract equation (1) from equation (2):

$$n[a + nd] - n[a + (n-1)d] = 30 - 24$$

$$n[a + nd - a - (n-1)d] = 6$$

$$n[nd - (n-1)d] = 6$$

$$n[d] = 6$$

$$nd = 6$$. ...(3)

It is given that the last term exceeds the first term by $$10\frac{1}{2} = \frac{21}{2}$$:

Last term = $$a + (2n-1)d$$, First term = $$a$$, so:

$$a + (2n-1)d - a = \frac{21}{2}$$

$$(2n-1)d = \frac{21}{2}$$. ...(4)

Substitute $$nd = 6$$ from equation (3) into equation (4):

$$(2n-1)d = \frac{21}{2}$$

$$2n \cdot d - d = \frac{21}{2}$$

$$2 \times 6 - d = \frac{21}{2}$$

$$12 - d = \frac{21}{2}$$

Solve for $$d$$:

$$d = 12 - \frac{21}{2} = \frac{24}{2} - \frac{21}{2} = \frac{3}{2}$$.

Now substitute $$d = \frac{3}{2}$$ into equation (3):

$$n \times \frac{3}{2} = 6$$

$$n = 6 \times \frac{2}{3} = 4$$.

Therefore, the total number of terms is $$2n = 2 \times 4 = 8$$.

Verification:

Using equation (1): $$n[a + (n-1)d] = 4 \left[ a + (4-1) \cdot \frac{3}{2} \right] = 4 \left[ a + \frac{9}{2} \right] = 24$$

So, $$a + \frac{9}{2} = 6$$, thus $$a = 6 - \frac{9}{2} = \frac{12}{2} - \frac{9}{2} = \frac{3}{2}$$.

Using equation (2): $$n[a + nd] = 4 \left[ \frac{3}{2} + 4 \cdot \frac{3}{2} \right] = 4 \left[ \frac{3}{2} + 6 \right] = 4 \left[ \frac{15}{2} \right] = 4 \times 7.5 = 30$$, which matches.

Last term: $$a + (2n-1)d = \frac{3}{2} + (8-1) \cdot \frac{3}{2} = \frac{3}{2} + \frac{21}{2} = \frac{24}{2} = 12$$

First term: $$a = \frac{3}{2}$$

Difference: $$12 - \frac{3}{2} = \frac{24}{2} - \frac{3}{2} = \frac{21}{2} = 10.5$$, which matches.

Hence, the correct answer is Option B.