Question

geometric and arithmetic progression exam question help (the mark scheme isn't being useful by itself)

Answer 1

This would be the geometric first four terms\[\Large a, ar, ar^2, ar^3\]
the first, second, fourth terms form an arithmetic sequence, so these three \[\Large a, ar, ar^3\]where there must be a common difference between terms.

Answer 2

ar = a+d
and
ar^3 = a+2d

Answer 3

Which means the difference between 1st and 2nd = diff. between 2nd and 3rd of the arithmetic
\[\Large ar^3- ar = ar-a\]

Answer 4

Simplify that and you have part a :)

Answer 5

divide by a

r^3 - r = r -1

r^3 - 2r + 1 = 0

Answer 6

its part b and c im really stuck on though

Answer 7

You have to solve for r. The "converges" bit tells you 0<r<1. The value when summed to infinity helps you get a because when 0<r<1, the sum equals \[\sum_{i=1}^{\infty}r=a/(1-r)\] where a is the first term, all subsequent terms multiplied by powers of r.

Answer 8

Hmm, if it converges that means |r| < 1, and the sum S= a/(1-r)

Maybe we need to use the facts from part (i), with the geometric/arithmetic series...

Answer 9

tricky.

okay heres my plan. using

r^3 - r = r -1

we can make r = 1 to make it all zero thereby proving that r = 1 is a root. however the solution cannot be 1 because then all terms would be the same.


r^3 - 2r + 1 = 0

divided by (r-1)

becomes (r^2 + r - 1)(r-1)=0

then use quadratic formulae

does that look right or is it horribly flawed?

Answer 10

Hmm, that might be right.

http://www.wolframalpha.com/input/?i=%28r%5E2+%2B+r+-+1%29%28r-1%29%3D0

So r looks like \[\Large r = \frac{ 1 }{2 } ( \sqrt 5 - 1)\]

Now finding a should be easy from \[\Large S= \frac{ a }{ 1-r }\]

Answer 11

does it become + root 5 or minus root 5?

Answer 12

It should be \[\Large3 +\sqrt 5= \frac{ a }{ 1- \frac{ 1 }{2 } ( \sqrt 5 - 1)}\]

Answer 13

im still confused why r = 0.5(root5-1)
i thought it was r = 0.5(root5-1) or r = 0.5(-root5-1)

Answer 14

Only one of those satisfies |r| < 1...

Answer 15

oh. oh thats clever

Answer 16

a = 2