Friday, July 5, 2024

Algebra+ 09 - Progressions: UNIZOR.COM - Math+ & Problems - Algebra

Notes to a video lecture on http://www.unizor.com

Algebra+ 09 - Progression

Problem A

Solve the following equation
X3+7·X2−21·X−27=0
if it's given that its three solutions form a geometric progression.

Solution A

Let's analyze the relationship between the solutions to an equation of the 3rd power
X3+P·X2+Q·X+R=0
and its coefficients.
If a, b and c are the solutions of an equation of the 3rd power with a coefficient 1 at X3 then
(X−a)·(X−b)·(X−c) = 0
Opening the parenthesis, we obtain
X3−(a+b+c)·X2+
+(ab+bc+ca)·X−abc = 0

Notice that for this type of equation the product of its solutions equals to a free coefficient R with minus sign, while a sum of its solutions equals to a coefficient P at X2 with a minus sign.
Therefore, in case of a given equation
a·b·c = −R = 27 and
a + b + c = −P = −7

Since it's given that three solutions of this equation form a geometric progression,
b = a·r and c = a·r²
where r is a progression's common ratio.
This allows us to represent two equations above in terms of a and r as follows
a·(a·r)·(a·r²) = 27
a + (a·r) + (a·r²) = −7
From the first equation follows:
(a·r)³ = 27
a·r = 3
which, incidentally, is one of the solutions: b=3.

From this and the above equations connecting solutions to coefficients of an equation, we conclude
a·3·c = 27 and
a + 3 + c = −7
or
a·c = 9 and
a + c = −10
Therefore expressing c in terms of a from the second equation and substituting into the first, we obtain
c = −10 − a and
a·(−10−a)=9
or
a² + 10·a + 9 = 0
Solutions to this simple quadratic equations are
−1 and −9.
Putting all solutions into a geometric sequence,
a=−1, b=3, c=−9
or
a=−9, b=3, c=−1
The common ratio of these progressions are r=−3 for the first and r=−1/3 for the second sequence, but the set of solutions is the same.

Solutions to a given equation are:
−1, 3, −9
They form a geometric progression with the first member a=−1 and the common ratio r=−3 or a geometric progression with the first member a=9 and the common ratio r=−1/3.

Problem B

Solve the following equation
X3−15·X2+26·X+120=0
if it's given that its three solutions form an arithmetic progression.

Solution B

Let's use the same relationship between the solutions to an equation of the 3rd power as above
a·b·c = −R = −120 and
a + b + c = −P = 15

Since it's given that three solutions of this equation form an arithmetic progression,
b = a+d and c = a+2d²
where d is a progression's common difference.
This allows us to represent two equations above in terms of a and d as follows
a·(a+d)·(a+2d) = −120
a + (a+d) + (a+2d) = 15
From the second equation follows:
3a+3d = 15
a+d = 5
which, incidentally, is one of the solutions: b=5.

From this and the above equations connecting solutions to coefficients of an equation, we conclude
a·5·c = −120 and
a + 5 + c = 15
or
a·c = −24 and
a + c = 10
Therefore expressing c in terms of a from the second equation and substituting into the first, we obtain
c = 10 − a and
a·(10−a) = −24
or
a² − 10·a − 24 = 0
Solutions to this simple quadratic equations are
−2 and 12.
Putting all solutions into an arithmetic sequence,
a=−2, b=5, c=12
The common difference of this progression is d=7.

Alternatively, using obvious relationships
a = b − d and c = b + d,
we can express the equation
a·5·c = −120
as
(5−d)·5·(5+d) = −120
or
25 − d2 = −24
from which we derive
d = ± 7,
which, in turn, leads to either
a = −2, c = 12
or
a = 12, c = −2
which is the same set of solutions in a different order.

Solutions to a given equation are:
−2, 5, 12
They form an arithmetic progression with the first member a=−2 and the common difference d=7 or an arithmetic progression with the first member a=122 and the common difference d=−7.

Problem C
Given two progressions, arithmetic and geometric, that satisfy the following conditions:
(a) the first members of these progressions are the same and are not equal to zero;
(b) the sums of the first three members of these progressions are the same;
(c) the sum of the first two members of arithmetic progression is greater than the sum of the first two members of geometric progression by the triple value of the first member of arithmetic progression.
What is the common ratio of the geometric progression?

Solution C

Arithmetic progression can be represented as
{a+k·d}
where
a is the first member,
d is the common difference,
k∈[0,N] is a sequence number (N is greater than 3).

Geometric progression can be represented as
{b·rk}
where
b is the first member,
r is the common ratio,
k∈[0,M] is a sequence number (M is greater than 3).

Conditions specified in the problem can be represented as
(a) a = b
(b) a + (a+d) + (a+2d) =
= b + b·r + b·r2

(c) a + (a+d) = 3a + b + b·r

Condition (a) can be used to replace unknown b in conditions (b) and (c) to simplify them as follows:
(b) a + (a+d) + (a+2d) =
= a + a·r + a·r2

(c) a + (a+d) = 3a + a + a·r
which can be simplified further as
(b) 2a + 3d = a·r·(1+r)
(c) d = 2a + a·r

Now we can use condition (c) to substitute d in condition (b):
(b) 2a + 3·(2a+a·r) = a·r·(1+r)

We can reduce this equation by a getting a quadratic equation for unknown ratio r:
2 + 3·(2+·r) = r·(1+r)
or
r2 − 2r −8 = 0
with solutions
r1=4; r2=−2