Question about sum of arithmetic series.

Question

A store begins to stock a new range of DVD players and achieves sales of £1500 of these products during the first month.

In a model it is assumed that sales will decrease by x£ in each subsequent month, so that sales of £(1500-x) and £(1500-2x) will be achieved in the second and third month respectively.

Given that sales total £8100 during the first six months, use the model to

(a) find the value of x, (b) find the expected value of sales in the eight month, (c) show that the expected total of sales in pounds during the first n months is given by kn(51-n), where k is an integer to be found. (d) Explain why this model cannot be valid over a long period of time.

I can’t solve the part c). So far I think x is 60. But when I do the sum formula I can’t get that it is equal to kn(51-n).

Can someone help me please?

Answer 1

Sales during the first $n$ months are given by $$S_n=1500n-\frac{n(n-1)x}{2}$$.This can be derived by using the summation formula for first $n-1$ natural numbers. Substitute $x=60$ and you will get $S_n=30n(51-n)$.

Answer 2

(i)$n/2(2a+(n-1)d$)=total sum

$n=6 a= 1500 d=?$

$8100=6/2((1500 \times 2)+(6-1)d)$

$d=\frac{(8100/3)-(1500 \times 2)}{5}$

(ii)$\frac{n}{2}((2\times 1500)+(n-1)\times(-60))$

$\frac{n}{2}(3000-60n+60)$

$\frac{n}{2}(3600-60n)$

$n((1530-30n))$ take $30$ outside

$30n(51-n)$