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Question on AP and Gp
- Thread starter emrade
- Start date
- An AP and GP have the same first term. If the sum of the corresponding first three terms of the two sequences form the sequence 10, 37, 114, find the two possible progressions.
Question isn't quite clear to me. I assume AP is arithmetic progression and GP is geometric progression but what does 'sum of the corresponding first three terms of the two sequences' mean? Especially since you gave three terms. Letting ai be the AP and gi be the GP, do you mean
a1 + a2 + a3 = 10
g1 + g2 + g3 = 37
or just what do you mean? I had thought that maybe you might mean
a1 + a2 + a3 = 10
a2 + a3 + a4 = 37
a3+ a4 + a5 = 114
but that is contradictory.
HallsofIvy
Elite Member
- Joined
- Jan 27, 2012
- Messages
- 7,763
I will take it that "AP" and "GP" do stand for "arithmetic progression" and "geometric progression! The first three terms of an arithmetic progression will be of the form a, a+ p, a+ 2p while the first three terms of a geometric progression, having the same first term,, will be of the form a, ar, ar^2.
I interpret "the sum of the corresponding first three terms of the two sequences" differently, I think, than Ishuda. I take that to be 2a, (a+ p)+ ar, and (a+ 2p)+ ar^2.
So we have 2a= 10, a+ p+ ar= 37, and a+ 2p+ ar^2= 114, three equations to solve for three unknown. For example, we can write the second equation as a(1+ r)+ p= 37. if we divide both sides of the first equation by 2 then multiply by 1+ r, we have a(1+ r)= 5(1+ r). Now, subtract that equation from the second: a(1+ r)+ p- a(1+ r)= p= 37- 5(1+ r), eliminating a. We can write the third equation as a(1+ r^2)+ 2p= 114. If we divide the first equation by 2 and multiply by 1+ r^2, we will have a(1+ r^2)= 5(1+ r^2). Subtracting that from a(1+ r^2+ 2p= 114 again eliminates a: 2p= 114- 5(1+ r^2).
So we have the equations p= 37- 5(1+ r) and 2p= 114- 5(1+ r^2). Multiplying the first of those equations by 2 and subtracting from the other eliminates p giving a single quadratic equation to solve for r.
I interpret "the sum of the corresponding first three terms of the two sequences" differently, I think, than Ishuda. I take that to be 2a, (a+ p)+ ar, and (a+ 2p)+ ar^2.
So we have 2a= 10, a+ p+ ar= 37, and a+ 2p+ ar^2= 114, three equations to solve for three unknown. For example, we can write the second equation as a(1+ r)+ p= 37. if we divide both sides of the first equation by 2 then multiply by 1+ r, we have a(1+ r)= 5(1+ r). Now, subtract that equation from the second: a(1+ r)+ p- a(1+ r)= p= 37- 5(1+ r), eliminating a. We can write the third equation as a(1+ r^2)+ 2p= 114. If we divide the first equation by 2 and multiply by 1+ r^2, we will have a(1+ r^2)= 5(1+ r^2). Subtracting that from a(1+ r^2+ 2p= 114 again eliminates a: 2p= 114- 5(1+ r^2).
So we have the equations p= 37- 5(1+ r) and 2p= 114- 5(1+ r^2). Multiplying the first of those equations by 2 and subtracting from the other eliminates p giving a single quadratic equation to solve for r.
Last edited:
Thanks guys for your replies.. well i tried solving the question then i got stuck
taking "a" as the first term, "d" as common difference and "r" as common ratio
if both thier first times are equal tat means a=a, then since the sum of the corresponding first term is 10 and a=a, that mean a=5,
then, a+d+ar=37 ie.. both corresponding second term from AP and GP
also a+2d+2r^2=114....
with that out of the way, can you guys help me futher?
taking "a" as the first term, "d" as common difference and "r" as common ratio
if both thier first times are equal tat means a=a, then since the sum of the corresponding first term is 10 and a=a, that mean a=5,
then, a+d+ar=37 ie.. both corresponding second term from AP and GP
also a+2d+2r^2=114....
with that out of the way, can you guys help me futher?