1. Set up the arithmetic progression

If $n$ arithmetic means are inserted between $a$ and $100$, then the full A.P. is:

$a,\ A_1,\ A_2,\ \dots,\ A_n,\ 100$

So the total number of terms is $n+2$.

Let the common difference be $d$.

Then:

• First mean: $A_1 = a+d$
• Last mean: $A_n = a+nd$
• Final term: $a+(n+1)d = 100$

2. Use the given ratio

Given that the ratio of the first mean to the last mean is $1:7$:

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

Cross-multiplying:

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

$7a+7d=a+nd$

$6a=(n-7)d \quad \text{(1)}$

3. Use the last term condition

Since the last term is $100$:

$a+(n+1)d=100 \quad \text{(2)}$

From (1),

$d=\frac{6a}{n-7}$

Substitute into (2):

$a+(n+1)\cdot \frac{6a}{n-7}=100$

$a\left(1+\frac{6(n+1)}{n-7}\right)=100$

$a\left(\frac{n-7+6n+6}{n-7}\right)=100$

$a\left(\frac{7n-1}{n-7}\right)=100$

Thus,

$a=100\cdot \frac{n-7}{7n-1}$

4. Use the condition $a+n=33$

Given:

$a+n=33$

So,

$100\cdot \frac{n-7}{7n-1}+n=33$

Multiply by $7n-1$:

$100(n-7)+n(7n-1)=33(7n-1)$

$100n-700+7n^2-n=231n-33$

$7n^2+99n-700=231n-33$

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

5. Solve the quadratic

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

Discriminant:

$\Delta = (-132)^2-4(7)(-667)$

$\Delta = 17424+18676=36100$

$\sqrt{\Delta}=190$

Therefore,

$n=\frac{132\pm 190}{14}$

So,

$n=\frac{322}{14}=23$

or

$n=\frac{-58}{14}<0$

Reject the negative value.

Hence,

$\boxed{n=23}$

6. Check with options

Option C is $23$, so the correct answer is:

$\boxed{\text{C}}$