# Math Help - arithmetic progression

1. ## arithmetic progression

I need help solving this word problem..

Q1 )There was a need for one table at the catering company's first event. Two table were needed for the second event, three table the third event and so on. one table seats 10 and two table seats 18. If the table are always set up end to end and a total of 74, 800 guests have attended the company's events, how many events has the catering company handled? (Assume every seat has been filled at each even)...

2. Hello, Vedicmaths!

Q1) There was a need for one table at the catering company's first event.
Two table were needed for the second event, three tables for the third event and so on.
One table seats 10, and two table seat 18.
If the table are always set up end to end and a total of 74,800 guests have attended,
how many events has the catering company handled?
(Assume every seat has been filled at each event.)

We assume that: each additional table seats 8 more guests.

The number of tables and seats look like this:

. . $\begin{array}{|c|c|c|} \text{Event} & \text{Tables} & \text{Seats} \\ \hline 1 & 1 & 10 \\ 2 & 2 & 18 \\ 3 & 3 & 26 \\ 4 & 4 & 34 \\ \vdots & \vdots & \vdots \\ n & n & 8n+2 \end{array}$

$\text{The total number of seats (guests) is: }\:\underbrace{10 + 18 + 26 + 34 + \cdots + (8n+2)}_{n\text{ terms}} \;=\;74,800$

The sum of an arithmetic series is: . $S_n \;=\;\frac{n}{2}\bigg[2a + (n-1)d\bigg]$
. . where: . $n = \text{no. of terms},\;\;a = \text{first term},\;\;d = \text{common difference}$

We have: . $a = 10,\;\;d = 8$

Hence: . $\frac{n}{2}\bigg[2(10) + (n-1)8\bigg] \:=\:74,800$

. . which simplifies to: . $2n^2 + 3n - 37,400 \:=\:0$

. . which factors: . $(n - 136)(2n + 275) \:=\:0$

. . and has the positive root: . $n \:=\:136$

Therefore, the catering company handled 136 events.

3. wow!

thanks for the help!