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.

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

(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:

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