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C2 Help on Geometric Series/M1 Friction

I've come across questions from C2 and M1 that I have become stuck on:

Geometric Series

A geometric series has the first term 1200. Its sum to infinity is 960

a) show that the common ratio of the series is -1/4

b) find, to 3 decimal places, the difference between the ninth and tenth terms of the series

c) write down an expression for the sum of the first n terms of the series

Given that n is odd,
d) prove that the sum of the first n terms of the series is 960 (1 + 0.25&#8319 :wink:

Mechanics - Friction

A body of mass 2kg is held in limiting equilibrium on a rough plane inclined at 20˚ to the horizontal by a horizontal force X. The coefficient of friction between the body and the plane is 0.2. Modelling the body as a particle find X when the body is on the point of slipping

a) Up the Plane

Reply 1

steve2005

17

lilydelarocks

I've come across questions from C2 and M1 that I have become stuck on:

Geometric Series

A geometric series has the first term 1200. Its sum to infinity is 960

a) show that the common ratio of the series is -1/4

Here is the first part.

Picture 86.png12.9KB

Reply 2

nota bene

15

lilydelarocks

Geometric Series

A geometric series has the first term 1200. Its sum to infinity is 960

a) show that the common ratio of the series is -1/4

b) find, to 3 decimal places, the difference between the ninth and tenth terms of the series

c) write down an expression for the sum of the first n terms of the series

Given that n is odd,
d) prove that the sum of the first n terms of the series is 960 (1 + 0.25&#8319 :wink:

a)

960=\frac{1200}{1-r}

solve for r

b)

u_n=1200(\frac{1}{4})^{n-1}

sub for n=10 and subtract n=9

c)

S_n=\frac{1200(\frac{1}{4}^n-1)}{\frac{1}{4}-1}

d) see steve's solution it is correct

Reply 3

steve2005

17

lilydelarocks

I've come across questions from C2 and M1 that I have become stuck on:

Geometric Series

A geometric series has the first term 1200. Its sum to infinity is 960

a) show that the common ratio of the series is -1/4

b) find, to 3 decimal places, the difference between the ninth and tenth terms of the series

My solution to part c looks rather too complicated??

Picture 87.png38.4KB

Reply 4

drizzle

OP

1

hmm I'm not sure about c as you showed that the sum of the first n terms of the series was an expression which was the same as the one you had to prove for d)

The first 2 parts I could do, I just wanted to check the answers as I haven't actually got the answers myself.

Reply 5

steve2005

17

lilydelarocks

hmm I'm not sure about c as you showed that the sum of the first n terms of the series was an expression which was the same as the one you had to prove for d)

The first 2 parts I could do, I just wanted to check the answers as I haven't actually got the answers myself.

Where does the question come from, I may have the answer.

Reply 6

nota bene

15

lilydelarocks

hmm I'm not sure about c as you showed that the sum of the first n terms of the series was an expression which was the same as the one you had to prove for d)

The first 2 parts I could do, I just wanted to check the answers as I haven't actually got the answers myself.

I think he is correct, it is just a straightforward application of the formula

S_n=\frac{u_1(r^n-1)}{r-1}=\frac{u_1(1-r^n)}{1-r}

Reply 7

DFranklin

18

lilydelarocks

hmm I'm not sure about c as you showed that the sum of the first n terms of the series was an expression which was the same as the one you had to prove for d)It's fine. The only point about (d) is that because you know n is odd, you can write -(-0.25)^n = 0.25^n.

Reply 8

drizzle

OP

1

OH RIGHT! I see!! Thanks guys for the help - Any ideas about the mechanics question on Friction?

Reply 9

steve2005

17

lilydelarocks

OH RIGHT! I see!! Thanks guys for the help - Any ideas about the mechanics question on Friction?

I hope this is correct , but I have not checked it. Do you have the answer for X

I have corrected the 30 degree angles ..they should have been 20

Picture 90.png58.1KB

Reply 10

drizzle

OP

1

In the back of the text book the answer is X is 11.9N...

EDIT: just had a look at your answer steve2005 and Ive noticed that where it says:

2(9.8)sin20 - 2(0.2)(9.8)cos20 divided by cos20 - 0.2sin20 it should say: 2(9.8)sin20 + 2(0.2)(9.8)cos20 divided by cos20 - 0.2sin20

..well i think so anyways. When you change the sign to a plus the answer will be 11.92164109 which is 11.9N which is the answer in the back of the text book.

Reply 11

steve2005

17

lilydelarocks

In the back of the text book the answer is X is 11.9N...

EDIT: just had a look at your answer steve2005 and Ive noticed that where it says:

2(9.8)sin20 - 2(0.2)(9.8)cos20 divided by cos20 - 0.2sin20 it should say: 2(9.8)sin20 + 2(0.2)(9.8)cos20 divided by cos20 - 0.2sin20

..well i think so anyways. When you change the sign to a plus the answer will be 11.92164109 which is 11.9N which is the answer in the back of the text book.

Sorry for the careless mistake.

Picture 92.png73.2KB

C2 Help on Geometric Series/M1 Friction

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