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Geometric Series


If the first term of a geometric progression is $t_1=a$t1=a and the common ratio is $r$r, then the sequence becomes:


Suppose we wish to add the first $n$n terms of this sequence. Technically, this is known as a series, or alternatively the partial sum of the sequence. We could write the sum as:


If we multiply both sides of this equation by the common ratio $r$r we see that:


Then, by carefully subtracting $rS_n$rSn from $S_n$Sn term by term, we see that all of the middle terms disappear:


This means that $S_n-rS_n=a-ar^n$SnrSn=aarn and when common factors are taken out on both sides of this equation, we find $S_n\left(1-r\right)=a\left(1-r^n\right)$Sn(1r)=a(1rn). Finally, by dividing both sides by $\left(1-r\right)$(1r) (excluding the trivial case of $r=1$r=1) we reveal the geometric sum formula:


An extra step, multiplying the numerator and denominator by $-1$1, reveals a slightly different form for $S_n$Sn:


This form is usually easier to use when $\left|r\right|>1$|r|>1.


As an example, adding up the first $10$10 terms of the geometric progression that begins $96+48+12+...$96+48+12+... is straightforward. Since $a=96$a=96 and $r=\frac{1}{2}$r=12 we have:


To add up the first $20$20 terms of the sequence $2,6,18,...$2,6,18,... we might use the second version of the formula to reveal:



You might be wondering that the two forms of the sum formula excludes the case for $r=1$r=1. This is not an issue, for if $r=1$r=1, then the series becomes:

$S_n=a+a+a+...+a=a\times n$Sn=a+a+a+...+a=a×n


Practice questions

Question 1

Consider the series $5+10+20$5+10+20 ...

Find the sum of the first $12$12 terms.



Consider the number $0.252525$0.252525$\ldots$

  1. Which option represents it as an infinite geometric series?






  2. Therefore express $0.252525$0.252525$\ldots$ as a fraction.


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