Formula for compounding interest with yearly increasing additional payments

Suppose we have a constant per period interest rate $i$ and have the following investment schedule:

$$ \text{period } 0: \text{ I invest } P,\\ \text{period } 1: \text{ I invest } A_1,\\ \vdots \\ \text{period } n: \text{ I invest } A_n.\\ $$

Then the future value of our investment as of period $n$ will be

$$P(1+i)^n+\sum_{k=1}^nA_k(1+i)^{n-k}.$$

The above is general, but now let's look at the special case where the additional payments grow at a constant rate $g$, i.e. $$A_k=A_1(1+g)^{k-1},k\geq1 .$$ Then the above formula is a finite geometric series:

$$P(1+i)^n+\sum_{k=1}^nA_1(1+g)^{k-1}(1+i)^{n-k}\\ =P(1+i)^n+A_1(1+i)^{n-1}\sum_{k=0}^{n-1}\left(\frac{1+g}{1+i}\right)^{k}\\ =P(1+i)^n+A_1(1+i)^{n-1}\frac{\left(\frac{1+g}{1+i}\right)^{n}-1}{\frac{1+g}{1+i}-1}\\ =P(1+i)^n+A_1\frac{(1+i)^n-(1+g)^n}{i-g}, $$ which is valid if $i\neq g$. Note this boils down to the additional payments formula you mention when there is no growth in the additional payments: $g=0$.

Your problem would use this formula, with $P=75,000*10\%,A_1=75,000*10\%*(1.05),g=5\%,i=12\%,n=30.$

As an addendum, in the case $i=g$, you can work out the future value is $P(1+i)^n+nA_1(1+i)^{n-1}.$