Q4 - Arithmetic and Geometric series

[ExamFever]

Homework Statement

Let a₁, a₂, a₃ denote the first three terms of a geometrical sequence, for which a₁ + a₂ + a₃ = 26.

a₁ + 3, a₂ + 4, a₃ - 3 are the first three terms of an arithmetical sequence.

Find the first term and the common quotient (ratio) of the geometrical sequence!

Homework Equations

Arithmetical sequences:

a_n = a₁ + (n-1)d
S_n = (n/2)(a₁ + a_n)

Geometric sequences:

a_n = a₁r^n-1
S_n = a₁ * ((r-1n)/(r-1))

The Attempt at a Solution

By taking the sum of the given arithmetical sequence:

(a₁ + a₂ + a₃) + 3 + 4 -3 = 26 +4 = 30

We know that:

S_n = (n/2)(a₁ + a_n)
30 = (3/2)(a₁ + 3 + a₃ - 3) > (3/2)(a₁ + a₃)
30 = (3/2)(20)

Thus a₁ + a₃ = 20

Because of the given

a₁ + a₂ + a₃ = 26

This implies

(a₁ + a₃) + a₂ = 26
20 + a₂ = 26
a₂ = 6

Now, I found the second term of each sequence, however, I do not know how to go forward from this point. The four functions given for arithmetic and geometric sequences do not seem to help me any further because there are always two unknown variables!

 

[ehild]

You do not need the sum of the arithmetic sequence, but it might be useful that the middle term is the arithmetic mean of the neighbours.

ehild

 

[ExamFever]

I might be missing some important understandig about the mean, but I cannot resolve the problem any further with the help of this hint. I have been breaking my head over it a while longer, but do not see where this fits in.

Could you perhaps provide me with another hint? Thanks!

 

[eumyang]

In other words, the middle term of the arithmetic sequence, a₂ + 4, is the average of the two other terms a₁ + 3 and a₃ - 3.


69

 

[ehild]

You know that the first three terms of the geometric sequence is a₁, a₁r, a₁r², and their sum is 26:

a₁+a₁r+a₁r²=26. *

There is the arithmetical sequence, with first term

b₁=a₁+3,

second term

b₂=a₂+4=a₁r+4,

and third term

b₃=a₃-3=a₁r²-3.

By definition,

b₂-b₁=b₃-b₂, which is equivalent by

b₃+b₁=2b₂

so

a₁+3+a₁r²-3=2(a₁r+4). **

You have two unknowns (a₁ and r) and two equations. Solve.

 

[ExamFever]

Now I got the following problem. I tried to use the logical steps you provided in the previous problem but I think I am missing something.

Problem
The sum of the first three terms of a geometrical sequence is 26 and their product is 216. Find the sum of the first four terms!

so we have

a + ar + ar² = 26
a * ar * ar² = 216

where a is the first term

we can get r in terms of a by putting

a = 26 * ((1-r)/1-r²))

Now it is possible to solve the equation

a³r³ = 216 by putting the value we found for a

However, this results in a very long and tiresome caclulation ending up with the following:

2170r⁶ - 6591r⁵ + 6510r⁴ - 2197r³ - 81r² + 27 = 0

I suppose I could solve this equation with synthetic division and then checking for which values the original equations holld, but I feel I got on the wrong track real soon here. Help greatly appreciated.

Cheers,

 

[ehild]

so we have

a + ar + ar² = 26
a * ar * ar² = 216

You made a mistake: a = 26 * ((1-r)/1-r^3)). But it is better to keep the first equation in the original form and start with the second one.

The problem is not that terrible, as you can take the cubic root of the second equation: ar=6

ehild