Geometric seriesAQA A-Level Maths: Revision notes

In a geometric sequence each term is the previous term multiplied by a constant common ratio $r$. With first term $a$: $u_n=ar^{n-1}.$ For $a=2$, $r=3$: $2,6,18,54,162,\dots$ and $u_5=2\times3^4=162$. Find $r$ by dividing consecutive terms, $r=\frac{u_{n+1}}{u_n}$. From two terms, divide to eliminate $a$: $ar=12$ and $ar^4=1.5$ give $r^3=\frac18$, so $r=\frac12$ and $a=24$.

The sum of the first $n$ terms is $S_n=\frac{a(1-r^n)}{1-r}=\frac{a(r^n-1)}{r-1},\qquad r\ne1.$ Use the first form when $|r|<1$ and the second when $r>1$ (to avoid negatives). For $a=2$, $r=3$, $n=6$: $S_6=\frac{2(3^6-1)}{2}=728$. To find $n$ from a sum, rearrange to isolate $r^n$ and take logarithms: $3^n-1>10^6\Rightarrow n>\frac{\log(10^6+1)}{\log3}=12.58$, so $n=13$.

If $|r|<1$ the terms shrink towards $0$ and the series converges to a finite sum to infinity: $S_\infty=\frac{a}{1-r},\qquad|r|<1.$ Here $|r|$ is the modulus of $r$ (its size, ignoring sign), so $|r|<1$ means $-1<r<1$. For $a=40$, $r=0.6$: $S_\infty=\frac{40}{0.4}=100$. If $|r|\ge1$ there is no sum to infinity (the series diverges). The gap $S_\infty-S_n=\frac{ar^n}{1-r}$ shows how close the sum is after $n$ terms: $100(0.6)^n<0.001$ needs $n\ge23$.

The terms from the $k$th onwards form a new geometric series with first term $ar^{k-1}$ and the same $r$. For $a=24$, $r=\frac12$, the terms from the $5$th onwards sum to $\frac{1.5}{1-\frac12}=3$. Alternatively subtract: $S_\infty-S_4=48-45=3$. Questions often give a relationship and ask for $r$. If $S_\infty$ is $4$ times the second term: $\frac{a}{1-r}=4ar\Rightarrow1=4r(1-r)\Rightarrow(2r-1)^2=0$, so $r=\frac12$. Remember to check that your $r$ satisfies $|r|<1$ whenever the question mentions a sum to infinity.

Where the ratio contains $x$, use modulus notation. For $3+3(2x-1)+3(2x-1)^2+\dots$ the ratio is $r=2x-1$, so convergence needs $|2x-1|<1$: $-1<2x-1<1\ \Rightarrow\ 0<x<1.$ Then $S_\infty=\frac{3}{1-(2x-1)}=\frac{3}{2-2x}$. If $S_\infty=5$, then $2-2x=0.6$ and $x=0.7$, which lies inside $0<x<1$, so it is valid. Always check a solution against the convergence range.