Question 1.12: (a) The following equation gives the price (€P) of a concert......

(a)(i) To make Q the subject of the formula, solve for Q:

P = 12 000 − 4Q

4Q = 12 000 − P

Q = 3000 − P/4           divide both sides by 4

(b)(i) Transpose the formula S ={\frac{n}{2}}\left(2a+(n-1)d\right) to make d the subject of the formula:

S_{n}={\frac{n}{2}}\left(2a+(n-1)d\right)

S_{n} × 2 =\cancel{2}\times{\frac{n}{\cancel{2}}}\left(2a+(n-1)d\right)      multiply both sides by 2: the twos on the RHS cancel, hence

\frac {2S_n}{n} = \frac {\cancel{n}(2a  +  (n  –  1)d)}{\cancel{n}}  divide both sides by n and simplify the RHS

\frac {2S_n}{n} = 2a + (n  −  1)d  now subtract 2a from each side

\frac {2S_n}{n}  –  2a = (n  –  1)d      bring the 2a to the other side

(\frac {2S_n}{n}  –  2a) × \frac {1}{n  –  1} = (n  –  1)d × \frac {1}{n  –  1}    multiplying both sides by  × \frac {1}{n  –  1}

That is,

d=\left(\frac{2S_{n}}{n} – 2a\right)\times{\frac{1}{n – 1}}

Hence d is the subject of the formula.

One would normally simplify this expression if possible. In this case we could add the terms inside the bracket and factor out the ‘2’

d=\left(\frac{2S_{n}}{n} – \frac{2a}{1}\right)\frac{1}{(n – 1)}=\left(\frac{2S_{n} –  2a n}{n}\right)\frac{1}{(n –  1)}=\frac{2(S_{n} –  2a n)}{n}\frac{1}{(n – 1)}

={\frac{2(S_{n} –  a n)}{n(n  –  1)}}

Hence, in its most simplified form, the formula for d is

d = ={\frac{2(S_{n} –  a n)}{n(n  –  1)}}

(ii)       Next, substitute n = 35, a = 200 and S_n = S_{35} = 5512.5:

d=\frac{2(5512.5  –  200\times35)}{35(35  –  1)}=\frac{2(5512.5  –  7000)}{35(34)} =\frac{2(-1487.5)}{1190}=-\frac{2975}{1190}=-2.5

If you look back at Worked Example 1.11, where S was evaluated, you will see that this problem is the same except d is evaluated given the other three variables n = 35, a = 200 and S_n = 5512.5.