Question

Sequences and series. Warning: very challenging. Help would be very much appreciated!

Answer 1

The sum of the first 4 terms in an arithmetic sequence is 50. If the 3rd term is taken out, it forms a geometric sequence. What is the ratio of the geometric sequence?

Answer 2

\[a_{1}+a_{2}+a_{3}+a_{4}=50\]

Answer 3

This forms a geometric sequence: \[a_{1}+a_{2}+a_{4}\]
I need to figure out the ratio of the geometric sequence.

Answer 4

It's kinda simple. Let the progression be\[12.5-3d, 12.5 - d, 12.5+d, 12.5+3d\]The above is a general arithmetic series with sum = 50. Now\[12.5-3d, 12.5 -d, 12.5+3d\]That's a geometric sequence. So just apply the condition:\[(12.5-d)^2 = (12.5-3d)(12.5+3d)\]

Answer 5

Can you please explain this, I don't quite understand the above example :(

Answer 6

Yes, sure.
All you really have to do here is assume a general form of an arithmetic sequence, and then make equations from the given restrictions. You can let it be \(a, ~a+d,~ a+2d,~a+3d\) if you want to. What are you comfortable with?

Answer 7

I prefer the
a, a + d, a + 2d, a + 3d...

Answer 8

Alright, so the sum of that is\[a+a+d+a+2d+a+3d = 4a + 6d\]So one equation is\[4a+6d=50\]The other equation is the condition for three numbers to be in geometric progression.\[(a+d)^2 = a(a+3d)\]Just solve these two.

Answer 9

Okay thanks!

Answer 10

@ParthKohli I don't get (a+d)^2 =a(a+3d) part. Please, explain

Answer 11

If three numbers \(x, y, z\) are in geometric progression, then it is required that \(y^2 = xz\). This is obtained from the fact that ratio of subsequent terms is the same:\[\frac{y}x = \frac{z}y \Rightarrow y^2 = xz\]

Answer 12

Also, once I solve for a, I will also need to solve for d to get the ratio right?

Answer 13

like: a, a*r, a*r^2, hence the middle term square, (a*r)^2 = the first term*the last term, right?

Answer 14

Got you. Thanks for explanation.

Answer 15

@steve816 To figure the whole arithmetic sequence out, you'll need to find \(a\) and \(d\).

Yes, @Loser66!