# Daniel’s Sum of Geometric Series

Continuing with the search of young Mathematicians, Daniel James Agsaullo Molina from Saint Louis College, La Union wants to share one of his favorite problems. He is an incoming freshman student. He was one of the national finalists in the individual category of 2014 Metrobank MTAP-Dep-Ed Math Challenge representing region 1. Daniel at the center

Problem:

Given a geometric sequence whose sum of the first 20 terms is 7. And whose sum from the 21st to the 60th term is 84, find the sum from the 61st to the 120th term.

Solution:

The sum of a geometric sequence is defined as: $S_n=\displaystyle\frac{a(r^n-1)}{r-1}$

Where n=number of terms, a=first term, and r = common ratio

Expressing the sum of the first 20 terms, $S_{20}=\displaystyle\frac{a(r^{20}-1)}{r-1}=7$     …#1

Another given is the sum of the terms from the 21st to the 60th which is 84.

It is expressed as $S_{21\to 60}=S_{60}-S_{20}=84$

Therefore, $S_{60}=S_{21\to 60}+S_{20}=84+7=91$

Expressing the sum $S_{60}$,

We get $S_{60}=\displaystyle\frac{a(r^{60}-1)}{r-1}=91$   …#2

From  #2 ÷ #1 and simplifying, $r^{40}+r^{20}+1=13\to r^{40}+r^{20}-12=0$ $(r^{20}+4)(r^{20}-3)=0$

Since $r^{20}\rangle 0$ as r ranges to all possible common ratio. $r^{20}-3=0 \to r^{20}=3$ …#3

We are ask to find the sum of 61st to 120th $S_{61\to 120}=S_{120}-S_{60}$  ***main equation

Where $S_{60}=91$ $S_{120}=\displaystyle\frac{a(r^{120}-1)}{r-1}=\displaystyle\frac{a(r^{60}-1)}{r-1}\cdot ( r^{60}+1)$

But $\displaystyle\frac{a(r^{60}-1)}{r-1}=S_{60}=91$ and from #3, $r^{60}=(r^{20})^3=3^3=27$ $S_{120}=91(27+1)=91(28)$

Substituting to our main equation, $S_{61\to 120}=91(28)-91=91(27)=2457$

### Dan

Blogger and a Math enthusiast. Has no interest in Mathematics until MMC came. Aside from doing math, he also loves to travel and watch movies.