Geometric sequences and seriesEdexcel A-Level Maths: Revision notes

A geometric sequence has a constant common ratio $r$: $u_{n+1}=ru_n$, so $r=\frac{u_{n+1}}{u_n}$. With first term $a$,

$u_n=ar^{n-1}.$

For $a=64$, $r=\frac34$: $u_5=64\left(\frac34\right)^4=\frac{81}{4}$. If $r>1$ the terms grow; if $0<r<1$ they shrink towards zero; if $r<0$ the signs alternate. Given two terms, divide them: if $u_2=12$ and $u_5=324$ then $r^3=27$, so $r=3$ and $a=4$.

For $r\ne1$,

$S_n=\frac{a(1-r^n)}{1-r}=\frac{a(r^n-1)}{r-1}.$

Use the first form when $|r|<1$ and the second when $r>1$ to keep numbers positive. Example: $a=4$, $r=3$, $n=8$: $S_8=\frac{4(3^8-1)}{2}=13120$. If $r=1$ every term equals $a$ and $S_n=na$.

You must be able to prove $S_n$:

$S_n=a+ar+ar^2+\cdots+ar^{n-1}$

$rS_n=ar+ar^2+\cdots+ar^{n-1}+ar^{n}$

Subtracting, the middle terms cancel: $S_n-rS_n=a-ar^n$. Factorising, $S_n(1-r)=a(1-r^n)$, and since $r\ne1$ you may divide by $1-r$.

A geometric series converges when the terms shrink fast enough, which happens exactly when $|r|<1$, i.e. $-1<r<1$. The modulus $|r|$ is the size of $r$ ignoring its sign. As $n\to\infty$, $r^n\to0$, so

$S_\infty=\frac{a}{1-r},\qquad |r|<1.$

For $a=64$, $r=\frac34$: $S_\infty=\frac{64}{1/4}=256$. If $|r|\ge1$ the series does not converge. A series such as $1+3x+9x^2+\cdots$ has $r=3x$, so it converges when $|3x|<1$, i.e. $|x|<\frac13$.

When $n$ is in an index, take logarithms. To find the least $n$ with $64\left(\frac34\right)^{n-1}<1$: $\left(\frac34\right)^{n-1}<\frac1{64}$, so $(n-1)\ln\frac34<\ln\frac1{64}$. Because $\ln\frac34<0$, dividing reverses the inequality: $n-1>14.46$, so $n=16$.

For sums: $S_n=8000(1-0.9^n)>7500$ gives $0.9^n<\frac1{16}$, so $n>26.3$ and $n=27$. Always give the answer as a whole number and check by substitution.

The terms from the $m$th onwards form a geometric series with first term $ar^{m-1}$ and the same ratio, so for $|r|<1$ the tail sum is $\frac{ar^{m-1}}{1-r}$. For $a=800$, $r=0.9$ the tail from the $10$th term is $8000\times0.9^9$. Equivalently it is $S_\infty-S_{m-1}$.

When two pieces of information give $a$ and $r$, eliminate one variable: from $\frac{a}{1-r}=48$ and $ar=9$ you get $16r^2-16r+3=0$, so $r=\frac14$ or $\frac34$; both are valid because $|r|<1$.