Finding rooms of Summation

Question

Hi I was wondering how do I Solve this question. I have to solve for $a$. I can solve for it when there's one summation but now there are three. My guess is factoring out the $A$. Divide $s$ by the $3$ summations $X$ $3$. The constants are given:
$$i = 0.03$$ $$s = 100000$$ $$n = 12$$ $$\sum_{k=1}^{8}a\cdot(1+i)^{\frac{-k}{n}}+\sum_{k=13}^{20}a\cdot(1+i)^{\frac{-k}{n}}+\sum_{k=25}^{32}a\cdot(1+i)^{\frac{-k}{n}}=s$$

Would this be equivalent? $$ a = s / (\sum_1^8\cdot(1+i)^{\frac{-k}{n}} + \sum_{13}^{20}\cdot(1+i)^{\frac{-k}{n}} +$$ $$\sum_{25}^{32}\cdot(1+i)^{\frac{-k}{n}}) $$

Answer 1

Note that the number of terms in each sum is the same, so you can write $$\sum_{k=1}^{8}a\cdot(1+i)^{\frac{-k}{n}}+\sum_{k=13}^{20}a\cdot(1+i)^{\frac{-k}{n}}+\sum_{k=25}^{32}a\cdot(1+i)^{\frac{-k}{n}}=\\ \left(1+(1+i)^{\frac{-12}n}+(1+i)^{\frac{-24}n}\right)\sum_{k=1}^{8}a\cdot(1+i)^{\frac{-k}{n}}$$ and the part in the big parentheses is a geometric sum as well. It looks like you are paying off a loan over three years with eight payments per year and four months of no payments.

Answer 2

Factor out $a$ and insert the known values $i$, $s$, $n$ to obtain:

$$a \left( \underbrace{\sum_{k=1}^{8} (\frac{1}{1.03})^{\frac{k}{12}} + \sum_{k=13}^{20} (\frac{1}{1.03})^{\frac{k}{12}} + \sum_{k=25}^{32} (\frac{1}{1.03})^{\frac{k}{12}}}_{3\,geometric\,series} \right) = 100000$$

I assume you know how to calculate sums of geometric series. Divide 100000 by their sum and you have found $a$.