Bob wants to read an 80 page book. On the first day he reads 5 pages, and for each subsequent day he decides to read \(\frac{1}{2}\) page more than the previous day. How many days does Bob need in order to finish the book?

Logic Oct 9, 2019

#1**+1 **

This is an arithemetic sequence. Use the arithmetic sequence formula to solve for N:

Solve for N:

N/2*[2*5 + 1/2*(N - 1) ]= 80, solve for N

1/2 N ((N - 1)/2 + 10) = 80

Multiply both sides by 2:

N ((N - 1)/2 + 10) = 160

Expand out terms of the left hand side:

N^2/2 + (19 N)/2 = 160

Multiply both sides by 2:

N^2 + 19 N = 320

Add 361/4 to both sides:

N^2 + 19 N + 361/4 = 1641/4

Write the left hand side as a square:

(N + 19/2)^2 = 1641/4

Take the square root of both sides:

N + 19/2 = sqrt(1641)/2 or N + 19/2 = -sqrt(1641)/2

Subtract 19/2 from both sides:

N = sqrt(1641)/2 - 19/2 or N + 19/2 = -sqrt(1641)/2

Subtract 19/2 from both sides:

**N = sqrt(1641)/2 - 19/2 = 10.75, or about ~11 days to read the book **

Guest Oct 9, 2019

#2**0 **

For a geometric series S_{n} = a_{1} (1-r^{n) }/ (1-r)

The common ratio is 1.5 (1.5 more pages each day) a_{1 }= 5

80 = 5 (1- 1.5^{n) }/ (1-1.5)

-40 = 5 (1-1.5^{n} )

-8 = (1-1.5^{n})

-9 = -1.5^{n}

9 = 1.5^{n }Take log of both sides

log9/log1.5 = n = 5.41 days ~~~ 6 days to read the 80 page book.

ElectricPavlov Oct 9, 2019

#4**+2 **

The arthmetic "formula" for the number of pages read on Day N is given by :

5 + (N-1)(.5) =

5 + .5N - .5 =

4.5 + .5N

The first term = 5

And the last term is 4.5 + .5N

So....we want to solve this :

(N/2) [ first term + last term ] = 80

(N/2)[ 5 + 4.5 + .5N ] = 80

(N/2) [ 9.5 + .5N ] = 80 mutiply through by 2

N [ 9.5 + .5N ] = 160

.5N^2 + 9.5N - 160 = 0 multiply through by 2 again

N^2 + 19N - 320 = 0

The graph here (letting N = x) shows two possible solutions : https://www.desmos.com/calculator/ze8gobvp4n

Taking the positive one...it will take about 10.75 days = 11 days

CPhill Oct 9, 2019