A loan is amortized over 5 years with monthly payments at nominal interest rate i=9% convertible monthly. First payment of $1000 is to be paid 1 month before loan date. Each succeeding monthly payments will decrease by 2% after the first payment. Calculate the outstanding loan balance after the 40th payment has been made.
My textbook only showed geometric progression that is increasing which is given by,
$$ \frac{1-(\frac{1+k}{1+i})^n}{i-k} $$
but I am not sure how I can evaluate a geometric progression that is decreasing.
Recall the present value for an annuity-immediate $$\require{enclose} a_{\enclose{actuarial}{n} i} = v + v^2 + \cdots + v^n = \frac{1 - v^n}{i}$$ where $v = (1+i)^{-1}$ is the periodic present value discount factor. When the payments are in geometric sequence, e.g. $$v + (1+k) v^2 + (1+k)^2 v^3 + \cdots + (1+k)^{n-1} v^n,$$ this is simply $1/(1+k)$ times the same annuity-immediate with a different present value discount factor $$v' = (1+k)v,$$ or $$i' = \frac{1+i}{1+k} - 1 = \frac{i-k}{1+k}.$$ Thus it has present value $$\frac{1}{1+k} a_{\enclose{actuarial}{n} i'} = \frac{1 - (v')^n}{i'(1+k)} = \frac{1 - \left(\frac{1+k}{1+i}\right)^n}{i-k},$$ as claimed.
The above derivation does not stipulate that $k > 0$. The only requirement is that $k \ne i$ for this formula to work. If $-1 < k < 0$, the payments are geometrically decreasing since then $0 < 1+k < 1$. So if payments decrease by $2\%$ per month, then $k = -0.02$ in your formula.
You may also want to refer back to this earlier question you asked and I answered. In that answer, I used $r$ instead of $1+k$.