Arthmetic Sequences and Series

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lovely_nancy

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Hi, could someone guide me through this question please:

As part of a fund - raising campaign, I have been given some books of raffle tickets to sell. Each book has the same number of tickets and all the tickets I have been given are numbered in sequence. The Number of the ticket on the front of the 5th book is 205 and that on the front of 19th book is 373.

(i) By writing the number of the ticket on the front of the first book as "a" and the number of the ticket in each book as "d" write down two equations involving "a" and "d"

(ii) From these equations find how many tickets ae in each book and the number on the front of the book I have been given.

(iii) The last ticket I have been given is numbered 492.
How many books have I been given?
 
I'll start you off. For an arithmetic sequence, finding d is the same as finding the slope of a line. In this case

\(\displaystyle d=\frac{373-205}{19-5}\)
 
lovely_nancy said:
Hi, could someone guide me through this question please:

As part of a fund - raising campaign, I have been given some books of raffle tickets to sell. Each book has the same number of tickets and all the tickets I have been given are numbered in sequence. The Number of the ticket on the front of the 5th book is 205 and that on the front of 19th book is 373.

(i) By writing the number of the ticket on the front of the first book as "a" and the number of the ticket in each book as "d" write down two equations involving "a" and "d"

(ii) From these equations find how many tickets ae in each book and the number on the front of the book I have been given.

(iii) The last ticket I have been given is numbered 492.
How many books have I been given?
Click to expand...

the number of the ticket on the front of the first book as "a"

the number of the ticket in each book as "d"

What is the number of the ticket on the back (last ticket) of the first book?

Please share your work with us, indicating exactly where you are stuck - so that we may know where to begin to help you.
 
DrSteve said:
I'll start you off. For an arithmetic sequence, finding d is the same as finding the slope of a line. In this case

\(\displaystyle d=\frac{373-205}{19-5}\)
Click to expand...

in which case I get "d" as 12
 
Correct. Now remember the general term for an arithmetic sequence is \(\displaystyle a_n=a_1+(n-1)d\). We just found \(\displaystyle d\), and you were given \(\displaystyle a_5\) . So you should be able to find \(\displaystyle a=a_1\)
 
Hi, could someone guide me through this question please:

As part of a fund - raising campaign, I have been given some books of raffle tickets to sell. Each book has the same number of tickets and all the tickets I have been given are numbered in sequence. The Number of the ticket on the front of the 5th book is 205 and that on the front of 19th book is 373.

(i) By writing the number of the ticket on the front of the first book as "a" and the number of the ticket in each book as "d" write down two equations involving "a" and "d"

(ii) From these equations find how many tickets ae in each book and the number on the front of the book I have been given.

(iii) The last ticket I have been given is numbered 492.
How many books have I been given?

Let N = first ticket number on a book of tickets, "a" = the number of the ticket on the front of the first book, "d" = the number of tickets in each book.

From N = a + (n-1)d

373 = a + (18)d and
205 = a + (4)d

Subtracting, 168 = 14d making d = 12, the number of tickets in each book

There must have been 204/12 = 17 up front books of tickets available, of which 12 went to other people. you received book 5 on up.

Since the first ticket on the 5th book that you received was 205, the 1st book's first ticket number must be 204 - 4(12) = 156.

As for ticket #492, 492 - 372 = 120/12 = 10 additional books above the 19th making 41 books of tickets all told.

You were given 41 - 12 = 29 books of tickets, having the following properties:
Actual book number....13............14.............31.............41
Book #-Your books......1..............5.............19.............29 (Note-12 0f the series of 41 12 ticket books were given to other people.)
First ticket #..........156............205...........373...........480
Last ticket #..........168............217...........385...........492 (Note-there were probably additional tickets disposed of)
 
I think final answer is incorrect

The final answer (c) is wrong

The 14th book has tickets 157 - 168, inclusive of 157. The person has given books with 13 tickets.

492 - 11 = 481 (this is the first ticket of nth book)

481=157 +12(n-1)
324=12(n-1)
n= 28

The following statement is incorrect:
s for ticket #492, 492 - 372 = 120/12 = 10 additional books above the 19th making 41 books of tickets all told.

372 is final ticket in the 18th book so there 10+18 = 28 books which the person has. Though correcrtly there are 41 books in total.
 
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