If $$n$$ arithmetic means are inserted between a and 100 such that the ratio of the first mean to the last mean is 1 : 7 and $$a + n = 33$$, then the value of $$n$$ is

Let $$n$$ arithmetic means be inserted between $$a$$ and $$100$$.

Express the common difference and means:

The resulting AP has first term $$a$$, last term $$100$$, and $$(n + 2)$$ terms total.

$$d = \frac{100 - a}{n + 1}$$

First mean $$= a + d$$, Last mean $$= a + nd$$

Apply the ratio condition:

$$\frac{a + d}{a + nd} = \frac{1}{7}$$

$$7(a + d) = a + nd$$

$$7a + 7d = a + nd$$

$$6a = d(n - 7) \quad \cdots (1)$$

Substitute $$d$$ and $$a = 33 - n$$:

From $$a + n = 33$$, we get $$a = 33 - n$$.

$$d = \frac{100 - (33 - n)}{n + 1} = \frac{67 + n}{n + 1}$$

Substituting into equation (1):

$$6(33 - n) = \frac{(67 + n)(n - 7)}{n + 1}$$

$$6(33 - n)(n + 1) = (67 + n)(n - 7)$$

Expand and solve:

Left side: $$6(33n + 33 - n^2 - n) = 6(32n + 33 - n^2) = -6n^2 + 192n + 198$$

Right side: $$67n - 469 + n^2 - 7n = n^2 + 60n - 469$$

Setting them equal:

$$-6n^2 + 192n + 198 = n^2 + 60n - 469$$

$$7n^2 - 132n - 667 = 0$$

Using the quadratic formula:

$$n = \frac{132 \pm \sqrt{132^2 + 4 \times 7 \times 667}}{2 \times 7} = \frac{132 \pm \sqrt{17424 + 18676}}{14} = \frac{132 \pm \sqrt{36100}}{14} = \frac{132 \pm 190}{14}$$

Taking the positive root: $$n = \frac{322}{14} = 23$$

The correct answer is Option C: $$23$$.