# Finding rooms of Summation

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

• It's in the link he pasted. Commented Feb 26, 2014 at 15:20
• Just saw the link and deleted my "Which equation?" comment. Sorry
– Keba
Commented Feb 26, 2014 at 15:20
• A tutorial for writing equations on this site is here People don't like to click through, and it makes typing an answer difficult. Commented Feb 26, 2014 at 15:23
• Cheers bud, I'll definitely apply that the next question I got. Commented Feb 26, 2014 at 15:25
• You can use the geometric sum formula I gave in my answer to your previous question to convert each sum into a formula without summation. Commented Feb 26, 2014 at 15:25

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.
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 $a$.