Sequence's P1

Well, this maths question is quite hard to do (for me anyway) - i understand whats happening i just cant prove it. I just mess up, anyway:

A salesman is paid commission of £10 per week for each life insurance policy which he has sold. Each week he sells one new policy so that he is paid £10 commission in the first week, £20 in the second week, £30 in the third etc.
In the second year the commission increases to £11 per week on new policies sold, although it remains at £10 per week for policies sold in the first year. He continues to see one policy a week.
[Q] show that he is paid £542 in the second week of his second year.
ref: Heinman mod maths, P1 , page 185.

if anyone can help me out that would be great thanks :smile:

Reply 1

john !!

14

a = 10
d = 10

a + 51d = 10 + 510 = 520

this is what he is paid on the final week of the first year.

two weeks in, he is paid 520 + 11*2
because d changes to 11

which makes 542.

Reply 2

Revelation

OP

1

ah ye, why did i not recognise it carried on from previous year - thanks.

Reply 3

john !!

14

Pretty handy I had the book in front of me already. I did the whole question while I was there. :biggrin:

wait until you get to question 35 !

Reply 4

Revelation

OP

1

hehe i did 18,24,32,38 for hwk :smile:

my maths class is ez and slow - 18 is quite hard too

mm q35 looks fun :P

Reply 5

elpaw

15

i sense a challenge... care to post q35?

Reply 6

john !!

14

the first three terms in a series are (m-4), (m+2), and (3m +1)

given that the series is arithmetic, find m

given that the series is geometric, find the two possible values of m

...

find the sum to infinity of the convergant of the two geometric series'.

summed up.

I've finished the first one and I got a fraction. I'll do the rest after dinner. I think it looks more scary than it is.

Reply 7

elpaw

15

mik1a

the first three terms in a series are (m-4), (m+2), and (3m +1)

given that the series is arithmetic, find m

given that the series is geometric, find the two possible values of m

...

find the sum to infinity of the convergant of the two geometric series'.

summed up.

I've finished the first one and I got a fraction. I'll do the rest after dinner. I think it looks more scary than it is.

a) (m-4)+(3m+1) = 2(m+2)

=> 2m = 7

=> m=7/2

b) (m-4)(3m+1)=(m+2)²

=> 2m² -15m -8 = 0

=> m = 8 or m = -1/2

c) [for] r = (10)/(4) = 5/2 >1 => divergent

[for] r = (3/2)/(-9/2) = -1/3 > -1 => convergent

sum to infinity = a/(1-r) = (-9/2)/(1+1/3)

= (-9/2)/(4/3) = -27/8

Reply 8

Revelation

OP

1

wd, i couldnt even find m :smile:

Reply 9

JamesF

1

Common difference = m+2 - (m-4) = 6, so
(3m+1) - (m+2) = 6
....m = 3.5

Common ratio = m+2/m-4 = 3m+1/m+2
Multiply out etc, m = 8, -1/2

Sum = (m-4)/1-(1/2) = (6-4)/(1/2)
= 4.

Reply 10

JamesF

1

You beat me to it Eplaw. We seem to have a difference of opinion for part b, but i got the same equation as u, check ur solutions.

Reply 11

Fermat

8

(m-4), (m+2), (3m+1)

Arithmetic:
===========
(m+2)- (m-4)= (3m+1) - (m+2)
6 = 2m-1
m=3.5
=====

Geometric
=========
(m+2)/(m-4)=(3m+1)/(m+2)
m^2 + 4m + 4 = 3m^2 - 9m - 4
2m^2 - 15m - 8 = 0
(2m+1)(m-8)=0
m=-1/2, m=8
===========

Sum to infinity
===============
take r=(m+2)/(m-4) to ensure the series converges, (|r|<1)
(m=-0.5)
r=1.5/-4.5
=-1/3
=====

a0=m-4=-4.5

the sum to infinity is S=a0(1/(1-r))

S=a0/(1-r)
S=-4.5/(1+1/3)
S=-4.5/(4/3)
S=-27/8
=====

made a mistake earlier. I think that's it now.