If the first term of a geometric sequence is $a$a and the common ratio is $r$r, then the sequence is given by:
$a,ar,ar^2,ar^3,ar^4,\ldots$a,ar,ar2,ar3,ar4,…
Suppose we wish to add the first $n$n terms of this sequence. This will form a geometric series. We could write the sum as:
$S_n=a+ar+ar^2+ar^3+...+ar^{n-1}$Sn=a+ar+ar2+ar3+...+arn−1
We saw a nifty trick for finding the formula for an arithmetic series by adding two sums together and pairing up terms. We will use a similar method here. If we multiply both sides of our geometric sum by the common ratio $r$r we see that:
$rS_n=ar+ar^2+ar^3+...+ar^{n-1}+ar^n$rSn=ar+ar2+ar3+...+arn−1+arn
Then, by carefully subtracting $rS_n$rSn from $S_n$Sn term by term, we see that all of the middle terms disappear:
$S_n-rS_n=a+\left(ar-ar\right)+\left(ar^2-ar^2\right)+...+\left(ar^{n-1}-ar^{n-1}\right)-ar^n$Sn−rSn=a+(ar−ar)+(ar2−ar2)+...+(arn−1−arn−1)−arn
This means that:
$S_n-rS_n=a-ar^n$Sn−rSn=a−arn
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(1−r)=a(1−rn)
Finally, by dividing both sides by $\left(1-r\right)$(1−r) (excluding the trivial case of $r=1$r=1) we reveal the geometric sum formula:
$S_n=\frac{a\left(1-r^n\right)}{1-r}$Sn=a(1−rn)1−r
An extra step, multiplying the numerator and denominator by $-1$−1, reveals a different form for $S_n$Sn. Both formulas will work in any situation, particularly when using a calculator. This form is generally easier to manage when the common ratio is greater than $r=1$r=1:
$S_n=\frac{a\left(r^n-1\right)}{r-1}$Sn=a(rn−1)r−1
For any geometric sequence with starting value $a$a and common ratio $r$r, we can find the sum of the first $n$n terms, using:
$S_n=\frac{a\left(1-r^n\right)}{1-r}$Sn=a(1−rn)1−r, for $r<1$r<1 or $S_n=\frac{a\left(r^n-1\right)}{r-1}$Sn=a(rn−1)r−1, convenient if $r>1$r>1
The formulas for $S_n$Sn exclude the case for $r=1$r=1. For the case where $r=1$r=1, then the sequence becomes $a,a,a,a,\ldots$a,a,a,a,….
Is this a geometric sequence with $r=1$r=1 or an arithmetic sequence with $d=0$d=0? Either way, every term is clearly identical. Hence, the sum of the first $n$n terms is:
$S_n$Sn | $=$= | $a+a+a+...+a$a+a+a+...+a | ($n$n times) |
$S_n$Sn | $=$= | $na$na |
If the sum for the first $n$n terms of the geometric sequence $5,10,20,\ldots$5,10,20,… is $5115$5115, find $n$n.
Think: We have an increasing geometric sequence. State $a$a, $r$r and $S_n$Sn, then substitute into the formula $S_n=\frac{a\left(r^n-1\right)}{r-1}$Sn=a(rn−1)r−1 and rearrange.
Do: $a=5$a=5, $r=2$r=2 and $S_n=5115$Sn=5115, so we have:
$S_n$Sn | $=$= | $\frac{a\left(r^n-1\right)}{r-1}$a(rn−1)r−1 | |
$5115$5115 | $=$= | $\frac{5\left(2^n-1\right)}{2-1}$5(2n−1)2−1 |
Substitute values into formula
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Hence,$5\left(2^n-1\right)$5(2n−1) | $=$= | $5115$5115 |
Simplify fraction and bring unknown to left-hand side
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$2^n-1$2n−1 | $=$= | $1023$1023 |
Divide both sides by $5$5
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$2^n$2n | $=$= | $1024$1024 |
Add $1$1 to both sides
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$\therefore n$∴n | $=$= | $10$10 |
Solve for $n$n, using guess and check, technology or logarithms.
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Consider the series $4+12+36$4+12+36 ...
Find the sum of the first $12$12 terms.
Consider the series $5+\frac{5}{2}+\frac{5}{4}$5+52+54 ...
Find the common ratio, $r$r.
Find the sum of the first $9$9 terms, rounding your answer to one decimal place.
Consider the series $4-12+36-\text{. . . }-708588$4−12+36−. . . −708588.
Solve for $n$n, the number of terms in the series.
Find the sum of the series.
Average annual salaries are expected to increase by $4$4 percent each year. If the average annual salary this year is found to be $\$40000$$40000:
Calculate the expected average annual salary in $6$6 years, correct to the nearest cent.
This year, Vincent starts at a new job in which he will receive the average annual salary for each year of his employment. Over the coming $6$6 years (including this year) he plans to save half of each year’s annual salary.
What will be his total savings over these $6$6 years? Give your answer correct to the nearest cent.