Math Questions (2unit) - help (1 Viewer)

[JSBboag]

I need help with a couple of Q's

1.) m, n, 8 are in armithmetic sequence and m, n, 9 are in geometric sequence. Find the two sets of values for m,n

--> I know the numbers are 6 and 4 with the difference being 2 and the ratio being 3/2 but i don't know how to work it out (worth 5 marks).

2.) By expressing the recurring decimal 0.48 ([48] being recurring) as the sum of an infinite number of terms of a series, find its value as a fraction in simplest form.

 

[Just.Snaz]

1) m, n, 8
n – m = 8 – n
2n – m = 8 ---> [1]

m, n , 9
n/m = 9/n
n^2 = 9m
m = (n^2)/9 ---> [2]

Sub [2] into [1]

2n - (n^2)/9 = 8

x by -9

n^2 - 18n + 72 = 0

(n - 12)(n - 6) = 0

n = 12, n = 6

Sub into [1]
n = 12:
2x12 - m = 8
m = 16

n = 6:
2x6 - m = 8
m = 4

therefore n = 12, m = 16 OR n = 6, m = 4

 

[Just.Snaz]

2) 0.48 (recurring) =
0.48 + 0.0048 + 0.000048 + 00000048 + etc...

S = a/(1- r)

a = 0.48

r = 0.0048/0.48
=0.01

S = 0.48/(1-0.01)
= 0.48/0.99
=48/99

 

[JSBboag]

2) 0.48 (recurring) =
0.48 + 0.0048 + 0.000048 + 00000048 + etc...


S = a/(1- r)

a = 0.48

r = 0.0048/0.48
=0.01

S = 0.48/(1-0.01)
= 0.48/0.99
=48/99

The recurring demcimal is 0.48484848484848
Also, thanks for your help.

 

[Just.Snaz]

The recurring demcimal is 0.48484848484848
Also, thanks for your help.

huh? lol

and yeah, no problems