Can you help me with this arithmetic sequence and series word problem?

Question

I really don't know what to do with this word problem...please help, 10 pts for best!

A company responsible for laying the foundation of a new high-rise office building must pay $5000 a day for the first five days that the completion of the foundation is late. On the sixth day and each day thereafter, the penalty is increase by $200 a day. If the company was penalized $65,6000, how many days late were they laying the foundation?

sweet2...can you show your work and explain how you did it?

Joe...but doesn't it says they were penalized $65,600 and not the remaining like you said which is $40,600?

Can someone clarify the answer?

Basic questions:

an = a1 + (n -1) d

Sn = n(a1 + an)
-------------
2

Sn = n[(2*a1) + (n-1) d]
------------------------
2

$65,600 penalized......sorry!

Answer 1

Well, we know that they were penalized $5,000 a day for the first 5 days. $5,000 x 5 = $25,000

For each day after, they get penalized $200 more a day than the day before.

If you subtract 25,000 from 65,600, you're left with 40,600.

On the first "late" day, the company gets charged 5200, then 5400, then 5600, then 5800, then 6000, then 6200, then 6400.

If you add these up, it equals the remaniing 40,600, so they were 7 days late.

Answer 2

Arithmetic Sequence Word Problems

Answer 3

They were 12 days late, altogether.

If you were meant to do this using an arithmetic series approach, rather than by a brute force "Let's keep adding increasing sums until we get there" "method" [ !! ], here's how to do it:

They paid a total penalty of $20,000 for the first FOUR days they were late. That means that they paid $45,600 for the remaining days. That was paid in sums of $5,000, $5,200, $5,400 ... etc.

[Note that starting with $5000 --- rather than $5,200 --- makes the arithmetic series we need to sum just that little bit more aesthetic to work with.]

Let there be ' n ' days in this arithmetic series. The sum of those n terms (the standard (1/2) n [2 a + (n - 1) d] sum for an arithmetic series) is:

(1/2) n [ 10000 + 200 (n - 1)] = 45,600,

so that, dividing both sides by 100 and multiplying out by the "1/2" on the LHS,

n [49 + n] = 456, or n^2 + 49 n - 456 = 0.

This factorizes into (n - 8) (n + 57) = 0.

Here, n = - 57 is unphysical; so n = 8 is our required additional number of days (additional to the FOUR days we started with).

Therefore the total number of days that the penalty is for is 4 + 8 = 12 days.

Live long and prosper.

Answer 4

5000(5) = 25,000 for 1st 5 days.

arithmetic series for remaining days:
5200 + 5400 + ... + [5200 + (n-1)(200)] has sum of 65,600 - 25,000 = 40,600.

sum = n/2 ( 5200 + 5200 + (n-1)(200)) = 40,600
n/2 [ 10,400 + 200n - 200] = 40,600
5200n + 100n² - 100n = 40,600
5100n + 100n² = 40600
n² + 51n - 406 = 0
(n+58)(n-7) = 0
n = 7

add the 1st 5 days and get total 12 days.

seems like it would have been simpler to set up a column total in Excel and extend the series until the total was right.

Answer 5

is it 65,600 or 656,000? For 65,600 the answer is twelve days.

I just punched in the numbers.

5*(5000)+5200+5400+5600+5800+6000+6200+6400 which comes to 12 days. I hope that helps. good luck.

Answer 6

308 days