Advanced Higher Maths
Sequences and Series

Let \(a\) represent the first term and \(d\) represent the common difference. We will need to find these first.

The 2nd term \(u_2=a+d\) and the 5th term \(u_5=a+4d.\)

So we have a pair of simultaneous equations:

$$ \begin{gather} \:a+d=7\\[4pt] \:a+4d=19 \end{gather} $$

Subtracting the equations, \(3d=12\) so the common difference \(d=4\small.\)

Substituting, the first term \(a=7-4=3\small.\)

Now we can use the formula for the sum of the first \(n\) terms, with \(n=50.\)

$$ \begin{eqnarray} S_n &=& \small\frac12\normalsize n\left[2a+(n\!-\!1)d\right] \\[9pt] S_{50} &=& \small\frac{50}{2}\normalsize\left[2(3)+(50\!-\!1)(4)\right] \\[9pt] &=& 25\left[6+49\!\times\!4\right] \\[6pt] &=& 25(202) \\[6pt] &=& 5050 \end{eqnarray} $$

Let \(a\) represent the first term and \(r\) represent the common ratio. We will need to find these first.

The 2nd term \(u_2=ar\) and the 5th term \(u_5=ar^4.\)

So, like the previous example, we have a pair of simultaneous equations:

$$ \begin{gather} ar=24\\[3pt] ar^4=3 \end{gather} $$

Dividing the equations, \(r^3=\large\frac{3}{24}\normalsize =\large\frac{1}{8}\) so the common ratio \(r=\sqrt[\leftroot{-1}\uproot{2}\scriptstyle 3]{\large\frac{1}{8}\normalsize}=\large\frac{1}{2}\small.\)

Substituting, \(\large\frac{1}{2}\normalsize a=24\) so the first term \(a=48\small.\)

Now we can use the formula for the sum of the first \(n\) terms, with \(n=10\small.\)

$$ \begin{eqnarray} S_n &=& \frac{a(1-r^n)}{1-r} \\[9pt] S_{10} &=& \frac{48\bigl(1-\left(\frac{1}{2}\right)^{10}\bigr)}{1-\frac{1}{2}} \\[9pt] &=& \frac{3069}{32}\:\,\textsf{or}\:\,95.90625 \end{eqnarray} $$

Let \(a\) represent the first term and \(r\) represent the common ratio.

$$ \begin{gather} S_\infty=\frac{a}{1\!-\!r}\\[9pt] 18=\frac{6}{1\!-\!r}\\[9pt] 18(1-r)=6\\[8pt] 1-r=\frac{1}{3}\\[9pt] r=\frac{2}{3} \end{gather} $$

So now we can find the fourth term, \(u_4.\)

$$ \begin{eqnarray} u_4 &=& ar^{4-1}\\[6pt] &=& 6\,\left(\frac{2}{3}\right)^{3}\\[6pt] &=& 6\left(\frac{8}{27}\right)\\[6pt] &=& 2\left(\frac{8}{9}\right)\\[6pt] &=& \frac{16}{9} \end{eqnarray} $$

As usual, let \(a\) represent the first term and \(r\) represent the common ratio.

This solution relies upon the fact that any term in a geometric sequence or series divided by the preceeding term is, by definition, equal to the common ratio.

$$ \begin{eqnarray} \frac{u_2}{u_1} &=& \frac{u_3}{u_2}\\[9pt] \frac{x+2}{x+6} &=& \frac{x-1}{x+2}\\[9pt] (x+2)(x+2) &=& (x+6)(x-1)\\[6pt] x^2+4x+4 &=& x^2+5x-6\\[6pt] 4x+4 &=& 5x-6\\[6pt] x &=& 10\\[6pt] \end{eqnarray} $$

So \(u_1=10\!+\!6=16,\,\) \(u_2=10\!+\!2=12\,\) and \(\,u_3=10\!-\!1=9\small.\)

The common ratio \(r=\large\frac{12}{16}\normalsize=\large\frac{3}{4}\small.\) The series converges because \(-1\lt r\lt 1\small.\) So the sum to infinity exists.

$$ \begin{eqnarray} S_\infty &=& \frac{a}{1\!-\!r}\\[9pt] &=& \frac{16}{1-\frac{3}{4}}\\[9pt] &=& 16\!\times\!4\\[9pt] &=& 64 \end{eqnarray} $$

The first term \(a=5\) and the common difference \(d=3\small.\)

So we set up our inequality for the sum of the first \(n\) terms exceeding \(500\):

$$ \begin{gather} S_n \gt 500 \\[9pt] \small\frac12\normalsize n\,\bigl[2a+(n\!-\!1)d\bigr]\gt 500 \\[9pt] \small\frac12\normalsize n\,\bigl[2(5)+3(n\!-\!1)\bigr]\gt 500 \\[9pt] n(10+3n-3) \gt 1000 \\[9pt] n(3n+7) \gt 1000 \\[9pt] 3n^2+7n-1000 \gt 0 \\[9pt] \end{gather} $$

The critical values are \(n=\large\frac{-7\,\pm\,\sqrt{7^2-4(3)(-1000)}}{6}\normalsize\) which gives \(n\approx 17.1\) or \(n\approx -19.5\small.\)

The negative value can clearly be disregarded, so \(S_{17}\lt 500\) but \(S_{18}\gt 500.\) The answer is \(n=18\small.\)

The first term \(a=7\) and the common difference \(d=4\small.\)

First we need to find the number of terms, \(n\small.\)

$$ \begin{gather} u_n = a+(n-1)d \\[6pt] 163 = 7+4(n-1) \\[6pt] 163 = 7+4n-4 \\[6pt] 163 = 4n+3 \\[6pt] 4n = 160 \\[6pt] n = 40 \end{gather} $$

Now we can find the sum of the first \(40\) terms:

$$ \begin{eqnarray} S_n &=& \small\frac12\normalsize n\bigl[2a+(n\!-\!1)d\bigr] \\[9pt] S_{40} &=& \small\frac{40}{2}\normalsize\bigl[2(7)+(40\!-\!1)(4)\bigr] \\[9pt] &=& 20\left(14+39\!\times\!4\right) \\[8pt] &=& 20\!\times\!170 \\[8pt] &=& 3400 \end{eqnarray} $$

The first term \(a=65\) and the common difference \(d=-4\small.\)

First we must find the number of terms, \(n\small.\)

$$ \begin{gather} S_n = \small\frac12\normalsize n\,\bigl[2a+(n-1)d\bigr] \\[9pt] 96 = \small\frac{n}{2}\normalsize \bigl[2(65)-4(n-1)\bigr] \\[9pt] 96 = \small\frac{n}{2}\normalsize \bigl(130-4n+4\bigr) \\[9pt] 96 = \small\frac{n}{2}\normalsize \bigl(134-4n\bigr) \\[9pt] 96 = 67n-2n^2 \\[9pt] 2n^2-67n+96 = 0 \\[9pt] (2n-3)(n-32)= 0 \\[9pt] 2n-3=0\:\:\:\small\textsf{or}\normalsize\:\:\:n-32=0 \\[6pt] n=\small\frac{3}{2}\normalsize\:\:\:\small\textsf{or}\normalsize\:\:\:n=32 \\[6pt] \end{gather} $$

Clearly only positive integer values of \(n\) make sense, so \(n=32\small.\)

Now we can find \(L\small,\) because we know that it is the \(32^{\text{nd}}\) term.

$$ \begin{eqnarray} u_n &=& a+(n-1)d \\[6pt] u_{32} &=& 65+(32-1)(-4) \\[6pt] &=& 65-4\!\times\!31 \\[6pt] &=& 65-124 \\[6pt] &=& -\!59 \end{eqnarray} $$

So our answer is that \(L=-59\small.\)

This example requires the summation formulae, which are given on your formulae list .

$$ \begin{eqnarray} \sum^{\normalsize {n}}_{\normalsize {r=1}}\,\left(r^3\!-\!2r\right) &=& \frac{n^2(n\!\small+\normalsize\!1)^2}{4} - 2\left[\frac{n(n\!\small+\normalsize\!1)}{2}\right]\\[9pt] &=& \frac{n^2(n\!\small+\normalsize\!1)^2}{4} - \frac{4n(n\!\small+\normalsize\!1)}{4}\\[9pt] &=& \frac{n(n\!\small+\normalsize\!1)}{4}\biggl\{\normalsize n(n+1)-4\biggr\}\\[9pt] &=& \frac{n(n\!\small+\normalsize\!1)}{4}\bigl(n^2+n-4\large\bigr)\\[9pt] &=& \frac{n(n\!\small+\normalsize\!1)(n^2\!+\!n\!-\!4)}{4} \end{eqnarray} $$

The given summation formula  starts counting at the first term, so we sum from 1 to 50 and subtract 1 to 19:

$$ \begin{eqnarray} \sum^{50}_{r=20}\,3r^2 &=& \sum^{50}_{r=1}\,3r^2 - \sum^{19}_{r=1}\,3r^2 \\[9pt] &=& 3\left\{\frac{50(51)(101)}{6} - \frac{19(20)(39)}{6}\right\}\\[9pt] &=& \frac{257\,550}{2} - \frac{14\,820}{2}\\[9pt] &=& 121\,365\\[9pt] \end{eqnarray} $$

We have been given two facts about this sequence, so we use each of them to obtain an equation involving the first term \(a\) and the common difference \(d\small.\) Then we solve the two equations simultaneously and evaluate \(S_{10}\small.\)

$$ \begin{gather} S_{20} = 320 \\[7pt] \small\frac12\normalsize (20)\large[\normalsize 2a+(20\!-\!1)d\large]\normalsize = 320 \\[7pt] 10(2a+19d) = 320 \\[7pt] 2a+19d = 32\:\:① \\[18pt] u_{21} = 37 \\[7pt] a+(21-1)d = 37 \\[7pt] a+20d = 37\:\:② \\[7pt] \end{gather} $$

By doubling equation \(②\) we can eliminate \(a\) and find \(d\small.\)

$$ \begin{gather} 2a+40d=74\:\:②\!\times\!2\\[7pt] 2a+19d = 32\:\:① \end{gather} $$

Subtracting gives us \(21d=42\) so \(d=2\small.\)

Substituting into equation \(②\) gives us \(a+40=37\) so \(a=-3\small.\)

Now we can find the sum of the first ten terms:

$$ \begin{eqnarray} S_{10} &=& \small\frac12\normalsize (10)\bigl[2a+(10\!-\!1)d\bigr]\\[9pt] &=& \small\frac12\normalsize (10)\bigl[2(-3)+(10\!-\!1)(2)\bigr]\\[9pt] &=& 5(-6+9\!\times\!2)\\[9pt] &=& 5\times 12\\[9pt] &=& 60\\[9pt] \end{eqnarray} $$

(a)  Using the given summation formulae :

$$ \begin{eqnarray} &\phantom{=}& \sum^{\normalsize {n}}_{\normalsize {r=1}}\,\left(r^2\!+\!3r\right)\\[9pt] &=& \sum^{\normalsize {n}}_{\normalsize {r=1}}\,r^2 + 3\sum^{\normalsize {n}}_{\normalsize {r=1}}\,r\\[9pt] &=& \frac{n(n\!\small+\normalsize\!1)(2n\!\small+\normalsize\!1)}{6} + 3\left\{\frac{n(n\!\small+\normalsize\!1)}{2}\right\}\\[9pt] &=& \frac{n(n\!\small+\normalsize\!1)(2n\!\small+\normalsize\!1)}{6} + \frac{9n(n\!\small+\normalsize\!1)}{6}\\[9pt] &=& \small\frac{1}{6}\normalsize n(n\!\small+\normalsize\!1)\bigl[(2n\!\small+\normalsize\!1)+9\bigr]\\[9pt] &=& \small\frac{1}{6}\normalsize n(n\!\small+\normalsize\!1)\left(2n\!\small+\normalsize\!10\right)\\[9pt] &=& \small\frac{1}{3}\normalsize n(n\!\small+\normalsize\!1)\left(n\!\small+\normalsize\!5\right) \end{eqnarray} $$

(b)  This part is similar to example 9 above. We sum the first 20 terms and subtract the sum of the first 10 terms.

$$ \begin{eqnarray} &\phantom{=}& \sum^{\normalsize {20}}_{\normalsize {r=11}}\,\left(r^2\!+\!3r\right)\\[9pt] &=& \sum^{\normalsize {20}}_{\normalsize {r=1}}\,\left(r^2\!+\!3r\right) - \sum^{\normalsize {10}}_{\normalsize {r=1}}\,\left(r^2\!+\!3r\right) \\[9pt] &=& \small\frac{20}{3}\normalsize(20\!\small+\normalsize\!1)\left(20\!\small+\normalsize\!5\right) - \small\frac{10}{3}\normalsize(10\!\small+\normalsize\!1)\left(10\!\small+\normalsize\!5\right)\\[9pt] &=& \small\frac{20\times 21\times 25}{3}\normalsize - \small\frac{10\times 11\times 15}{3}\normalsize\\[9pt] &=& 3500 - 550\\[9pt] &=& 2950\\[9pt] \end{eqnarray} $$

Let \(a\) represent the first term and \(r\) represent the common ratio.

(a) (i)  The 4th term \(u_4=ar^3\) and the 7th term \(u_7=ar^6.\) So:

$$ \begin{eqnarray} ar^3 &=& 9\\[6pt] ar^6 &=& 243 \end{eqnarray} $$

Dividing, \(r^3=\large\frac{243}{9}\normalsize =27\) so \(r=\sqrt[\leftroot{-1}\uproot{2}\scriptstyle 3]{27}=3\small.\)

(a) (ii)  Substituting:

$$ \begin{eqnarray} ar^3 &=& 9\\[6pt] 27a &=& 9\\[6pt] a &=& \small\frac{9}{27}\\[6pt] a &=& \small\frac{1}{3} \end{eqnarray} $$

(b)  Now we can use the formula for the sum of the first \(n\) terms.

$$ \begin{eqnarray} S_n &=& \frac{a\small\,\normalsize(1\!-\!r^n)}{1\!-\!r} \\[9pt] &=& \frac{\frac{1}{3}(1\!-\!3^n)}{1\!-\!3} \\[9pt] &=& -\!\small\frac{1}{6}\normalsize (1\!-\!3^n) \\[9pt] \end{eqnarray} $$

Replacing the \(n\) by \(2n\small,\) we can see that:

$$ S_{2n}= -\small\frac{1}{6}\normalsize (1\!-\!3^{2n}) $$

Now we are ready to prove the given statement. Starting with the left hand side and making use of the difference of two squares structure of the numerator:

$$ \begin{eqnarray} \frac{S_{2n}}{S_{n}} &=& \large\frac{\small-\frac{1}{6}\normalsize (1\!-\!3^{2n})}{\small-\frac{1}{6}\normalsize (1\!-\!3^n)}\\[6pt] &=& \frac{1\!-\!3^{2n}}{1\!-\!3^n}\\[6pt] &=& \frac{1^2\!-\!(3^{n})^2}{1\!-\!3^n}\\[6pt] &=& \frac{(1\!+\!3^n)(1\!-\!3^n)}{1\!-\!3^n}\\[6pt] &=& 1+3^n \small\textsf{ as required.} \end{eqnarray} $$

(a)  Let \(a\) represent the first term and \(r\) represent the common ratio.

The third term \(u_3=ar^2\) and the fifth term \(u_5=ar^4.\) So:

$$ \begin{eqnarray} ar^2 &=& 36\\[3pt] ar^4 &=& 16 \end{eqnarray} $$

Dividing, \(r^2=\large\frac{16}{36}\normalsize=\large\frac{4}{9}\normalsize\) so \( r=\sqrt{\large\frac{4}{9}\normalsize}=\large\frac{2}{3}\small.\)

(b)  Substituting, \(\large\frac{4}{9}\normalsize a=36\) so \(a=\large\frac{9}{4}\normalsize\times 36=81\small.\)

(c)  The associated geometric series has a sum to infinity because \(-1\lt \large\frac{2}{3}\normalsize\lt 1\small.\)

(d)  This is just a simple substitution into the summation formula:

$$ \begin{eqnarray} S_\infty &=& \frac{a}{1\!-\!r}\\[9pt] &=& \frac{81}{1\!-\!\frac{2}{3}}\\[9pt] &=& 81\times 3\\[9pt] &=& 243 \end{eqnarray} $$

(a)  Using \(a\) for the first term and \(r\) for the common ratio, the third term is \(a+2d = -3+2d\small.\)

(b)  The eighth term is \(a+7d = -3+7d\small.\)

The 8th term is five times the 3rd term, so:

$$ \begin{gather} -3+7d=5(-3+2d)\\[6pt] -3+7d=-15+10d\\[6pt] 7d-10d=-15+3\\[6pt] -3d=-12\\[6pt] d=4\\[6pt] \end{gather} $$

(c)  We use the summation formula for an arithmetic series with \(a=-3\) and \(d=4\small.\)

$$ \begin{gather} S_n \gt 500 \\[9pt] \small\frac12\normalsize n\bigl[2a+(n\!-\!1)d\bigr]\gt 500 \\[9pt] \small\frac12\normalsize n\bigl[2(-3)+4(n\!-\!1)\bigr]\gt 500 \\[9pt] n(-6+4n-4) \gt 1000 \\[9pt] n(4n-10) \gt 1000 \\[9pt] 4n^2-10n-1000 \gt 0 \\[9pt] 2n^2-5n-500 \gt 0 \\[9pt] \end{gather} $$

The critical values are \(n=\large\frac{5\,\pm\,\sqrt{(-5)^2-4(2)(-500)}}{4}\normalsize\) which gives \(n\approx 17.1\) or \(n\approx -14.6\small.\)

The negative value can clearly be disregarded, so \(S_{17}\lt 500\) but \(S_{18}\gt 500\) and the answer is \(18\small.\)

This exam question is very similar to example 8 above, although the factorisation takes a slightly different form.

Using the standard summations from the formulae sheet  gives:

$$ \begin{eqnarray} \sum^{n}_{r=1}\,\left(r^3\!-\!3r\right) &=& \frac{n^2(n\!\small+\normalsize\!1)^2}{4} - 3\left\{\frac{n(n\!\small+\normalsize\!1)}{2}\right\}\\[9pt] &=& \frac{n^2(n\!\small+\normalsize\!1)^2}{4} - \frac{6n(n\!\small+\normalsize\!1)}{4}\\[9pt] &=& \small\frac14\normalsize n(n\!\small+\normalsize\!1)\bigl\{\normalsize n(n\!\small+\normalsize\!1)\small-\normalsize6\bigr\}\\[9pt] &=& \small\frac14\normalsize n(n\!\small+\normalsize\!1)\large\!\left(\normalsize n^2\!\small+\normalsize\!n\small-\normalsize\!6\large\right)\normalsize\\[9pt] &=& \small\frac14\normalsize n(n\!\small+\normalsize\!1)(n\!\small-\normalsize\!2)(n\!\small+\normalsize\!3)\\[9pt] \end{eqnarray} $$