It is given that aa, bb, cc are the first three terms of a geometric progression. It is also given that aa, cc, bb are the first three terms of an arithmetic progression.

  1. (i)

    Show that

    b2=acb^2 = ac and c=a+b2c = \dfrac{a+b}{2}.

  2. (ii)

    Hence show that

    2(ba)2−(ba)−1=0.2\left(\dfrac{b}{a}\right)^2 - \left(\dfrac{b}{a}\right) - 1 = 0.

  3. (iii)

    Given that the sum to infinity of the geometric progression is SS, find SS in terms of aa.

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Worked solution

Q6(i)

1. Derive the relationship from the geometric progression.

Since aa, bb, cc are the first three terms of a geometric progression, the ratio between consecutive terms is constant:
ba=cb\dfrac{b}{a} = \dfrac{c}{b}
b2=ac(Shown)b^2 = ac \quad \text{(Shown)}

2. Derive the relationship from the arithmetic progression.

Since aa, cc, bb are the first three terms of an arithmetic progression, the difference between consecutive terms is constant:
c−a=b−cc - a = b - c
2c=a+b2c = a + b
c=a+b2(Shown)c = \dfrac{a + b}{2} \quad \text{(Shown)}

Answer

Shown

Q6(ii)

1. Substitute the expression for $c$ from the arithmetic progression into the geometric progression equation.

Substitute c=a+b2c = \dfrac{a+b}{2} into b2=acb^2 = ac:
b2=a(a+b2)b^2 = a\left(\frac{a+b}{2}\right)

2. Rearrange the equation and divide by $a^2$ to obtain the required quadratic form in terms of $(b/a)$.

2b2=a2+ab2b^2 = a^2 + ab
2b2−ab−a2=02b^2 - ab - a^2 = 0
Divide the entire equation by a2a^2 (since for a valid geometric progression, a≠0a \neq 0):
2b2a2−aba2−a2a2=0\frac{2b^2}{a^2} - \frac{ab}{a^2} - \frac{a^2}{a^2} = 0
2(ba)2−(ba)−1=0(shown)2\left(\frac{b}{a}\right)^2 - \left(\frac{b}{a}\right) - 1 = 0 \quad \text{(shown)}

Answer

Shown.

Q6(iii)

1. Find the possible values of the common ratio $r = \frac{b}{a}$.

Let r=bar = \frac{b}{a} be the common ratio of the geometric progression.
From (ii), we have the equation 2r2−r−1=02r^2 - r - 1 = 0.
Factorising the quadratic equation gives:
(2r+1)(r−1)=0(2r + 1)(r - 1) = 0
r=−12r = -\frac{1}{2} or r=1r = 1

2. Select the valid common ratio based on the condition for sum to infinity.

For an infinite geometric progression to converge and have a sum to infinity, its common ratio must satisfy ∣r∣<1|r| < 1.
Therefore, we reject r=1r = 1 and accept r=−12r = -\frac{1}{2}.

3. Calculate the sum to infinity in terms of $a$.

The sum to infinity is given by S=a1−rS = \frac{a}{1 - r}.
Substitute r=−12r = -\frac{1}{2} into the formula:
S=a1−(−12)S = \frac{a}{1 - \left(-\frac{1}{2}\right)}
S=a32S = \frac{a}{\frac{3}{2}}
S=2a3S = \frac{2a}{3}

Answer

\frac{2a}{3}

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