Advanced Higher Maths
Sequences and Series

Let \(a\) represent the first term and \(d\) represent the common difference.

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

$$ \begin{eqnarray} a &+& d &=& 7\\[3pt] a &+& 4d &=& 19 \end{eqnarray} $$

Subtracting, \(3d=12\) so \(d=4.\)

Substituting, \(a=7\!-\!4=3.\)

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

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

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

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

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

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

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

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

$$ \begin{eqnarray} S_n &=& \frac{a(1\!-\!r^n)}{1\!-\!r} \\[9pt] S_{10} &=& \frac{48\left(1\!-\!(\frac{1}{2})^{10}\right)}{1\!-\!\frac{1}{2}} \\[9pt] &=& \frac{3069}{32} \end{eqnarray} $$

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

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

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.

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

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

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

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

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

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

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 \(d=4.\)

First we must 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:

$$ \begin{eqnarray} S_n &=& \small\frac12\normalsize n\large[\normalsize 2a+(n\!-\!1)d\large]\normalsize \\[9pt] S_{40} &=& \small\frac{40}{2}\large[\normalsize 2(7)+(40\!-\!1)(4)\large]\normalsize \\[9pt] &=& 20\large[\normalsize 14+39\times 4\large]\normalsize \\[6pt] &=& 20(170) \\[6pt] &=& 3400 \end{eqnarray} $$

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

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

$$ \begin{gather} S_n = \small\frac12\normalsize n\large[\normalsize 2a+(n\!-\!1)d\large]\normalsize \\[9pt] 96 = \small\frac{n}{2}\large[\normalsize 2(65)-4(n\!-\!1)\large]\normalsize \\[9pt] 96\times 2 = n(130-4n+4) \\[9pt] 96\times 2 = n(-4n+134) \\[9pt] 96 = n(-2n+67) \\[9pt] 96 = -2n^2+67n \\[9pt] 2n^2-67n+96 = 0 \\[9pt] (2n-3)(n-32)= 0 \\[9pt] \end{gather} $$

$$ \begin{eqnarray} 2n-3=0 &\:\ \small\textsf{or}\normalsize\ & n-32=0 \\[6pt] n=\small\frac{3}{2}\normalsize &\:\ \small\textsf{or}\normalsize\ & n=32 \\[6pt] \end{eqnarray} $$

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 \(32^{nd}\) term.

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

So \(L=-\!59\small.\)

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

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

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

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

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:

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

(a)  Using the given summation formulae:

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

(a)  This part is similar to the previous example.

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

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

(a) (i)  The 4nd 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 \(r=\sqrt[\leftroot{-1}\uproot{2}\scriptstyle 3]{27}=3.\)

(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.

$$ \begin{eqnarray} S_n &=& \frac{a(1\!-\!r^n)}{1\!-\!r} \\[9pt] &=& \frac{\frac{1}{3}(1\!-\!3^n)}{1\!-\!3} \\[9pt] &=& -\!\small\frac{2}{3}\normalsize (1\!-\!3^n) \\[9pt] \end{eqnarray} $$

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

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

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:

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