1. series and sequences

prospectors are drilling for oil. The cost of drilling to a depth of 50m is £500. To drill a further 50m costs £640 and, hence, the total cost of drilling to a depth of 100m is £1140. Each subsequent extra depth of 50m costs £140 more to drill than the previous 50m.
the total sum of money available for drilling is £76000. Find, to the nearest 50m, the freatest depth that can be drilled.
a=500 d=140 i'm not sure what n=....

2. Hello, girlpower1991!

Prospectors are drilling for oil.
The cost of drilling to a depth of 50m is $500. To drill a further 50m costs$640
and, hence, the total cost of drilling to a depth of 100m is $1140. Each subsequent extra depth of 50m costs$140 more to drill than the previous 50m.
The total sum of money available for drilling is \$76,000.
Find, to the nearest 50m, the greatest depth that can be drilled.
I had to baby-talk mny way through it . . .

$\begin{array}{cccc}\text{Level} & \text{Cost per level} & \text{Total cost} \\ \hline 1 & 500 \\ & & 500 \\ 2 & 640 \\ & & 1140 \\ 3 & 780 \\ & & 1920 \\ 4 & 920 \\ & & 2840 \end{array}$
. . $\begin{array}{ccccc} 5 & \qquad\;\; 1060 \\ & & \qquad\quad\; 3900 \end{array}$

We have a Cost sequence: . $500, 1140, 1920, 2840, 3900, \hdots$

Take the difference of consecutive terms,
. . then take differences of the differences, etc.

$\begin{array}{cccccccccc} \text{Sequence} & 500 && 1140 && 1920 && 2840 && 3900 \\ \text{1st diff.} & & 640 & & 780 & & 920 & & 1060 \\ \text{2nd diff.} & & & 140 & & 140 & & 140 \end{array}$

The second differences are constant.
. . Hence, the function is of the second degree . . . a quadratic.

The general quadratic function is: . $f(n) \;=\;an^2 + bn + c$

Plug in the first three values of the sequence:

. . $\begin{array}{ccccc} f(1) = 500\!: & a + b + c &=& 500 & [1] \\
f(2) = 1140\!: & 4a + 2b + c &=& 1140 & [2] \\
f(3) = 1920\!: & 9a + 3b + c &=& 1920 & [3] \end{array}$

$\begin{array}{ccccc}\text{Subtract [2] - [1]:} & 3a + b &=& 640 & [4] \\
\text{Subtract [3] - [2]:} & 5a + b &=& 780 & [5] \end{array}$

$\begin{array}{cccccccc}\text{Subtract [5] - [4]:} & 2a \:=\: 140 & \Rightarrow &\boxed{a \:=\:70}\end{array}$

$\begin{array}{ccccccc}\text{Substitute into [4]:} & 3(70) + b \:=\:640 & \Rightarrow &\boxed{b \:=\:430} \end{array}$

$\begin{array}{cccccc}\text{Substitute into [1]:} & 70 + 430 + c &=& 500 & \Rightarrow &\boxed{c \:=\:0} \end{array}$

Hence: . $\boxed{f(n) \;=\;70n^2 + 430n}$

We have: . $70n^2 + 430n \;=\;76,\!000 \quad\Rightarrow\quad 7n^2 + 43n - 7600 \;=\;0$

Quadratic Formula: . $n \;=\;\frac{-43\pm\sqrt{43^2 + 4(7)(7600)}}{14} \;=\;
\frac{-43 \pm\sqrt{214,649}}{14}$

The positive root is: . $n \;=\;30.02159122 \;\approx\;30$ levels

The greatest depth is: . $30 \times 50 \;=\;1500$ m.