Thread: Sequence and Series

I'm having problems getting started on this question. I think once this parts cracked I can continue with the other parts.

The 1st, 2nd & 3rd terms of an AP are p , q & p^2 resepctively, where p is negative.

The 1st, 2nd & 3rd terms of a GP are p , p^2 & q resectively.

Show that p = -1/2 and find the value of q.

So, as this is an AP I know that U_2 - U_1 = U_3 - U_2 = d

Therefore, (q - (-p)) = (p^2 - q)
rearrange and I get q = (p^2 - p)/2 = d

I try subbing this in for U_2: a + 2d = q
as I have just found q and d to be (p^2 - p)/2

I dont end up with -1/2 though. Am I approaching this completely wrong or have I made a schoolboy error.

Your help is much appreciated - I have a test tomorrow!!!

thanks, D

Knowing that p is negative, there's no need to write it as such. That is...

From the AP: q - p = p^2 - q

From the GP: p^2/p = q/p^2

Solve the simultaneous equations, finding two values for p, only one of which is correct with the given condition. Then use either one of the equations to find q.

Great! So I use substitution to solve the simultaneous eq. and get

p (-2p^2 + p + 1) = 0

when p = 0 , p = 1 , p = -1/2

p cannot equal 0 as it would be convergent, p cannot equal 1 because the GP would not be defined and therefore it must be p = -1/2

is this right??

D

I've looked at this again and I see that p must be -1/2 because this is defined in the question - It is negative!

I get 3/8 for q