+0

0
305
4
+1200

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?

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

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

Oct 9, 2019
#2
+26006
0

For a geometric series    Sn = a1 (1-rn)  / (1-r)

The common ratio is 1.5    (1.5 more pages each day)     a1 = 5

80 = 5 (1- 1.5n) / (1-1.5)

-40 = 5 (1-1.5n )

-8 = (1-1.5n)

-9 = -1.5n

9 = 1.5n       Take log of both sides

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

Oct 9, 2019
#3
+1200
0

Thanks Guys!!!

Oct 9, 2019
#4
+111438
+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

Oct 9, 2019