The first, second and forth terms of a convergent geometric progression are consecutive terms of an arithmetic progression. Prove that the common ratio of the geometric progression is $\frac{-1+\sqrt{5}}{2}$ .[5]
The first term of the geometric progression is positive. Show that the sum of the first 5 terms of this progression is greater than nine tenths of the sum of infinity.[3]

The first, second and forth terms of a convergent geometric progression are consecutive terms of an arithmetic progression. Prove that the common ratio of the geometric progression is $\frac{-1+\sqrt{5}}{2}$ .

The first term of the geometric progression is positive. Show that the sum of the first 5 terms of this progression is greater than nine tenths of the sum of infinity.

Recurrence Relations and Summation

International Baccalaureate Higher Level Mathematics

IB HL Math

N2003 P2 Q4

GP terms: $a, ar, ar^{2}, ar^{3}, \ldots$

1st, 2nd, 4th terms are $a, ar, ar^{3}$ — these form an AP.

Common difference: $ar - a = ar^{3} - ar$

$a(r-1) = a(r^{3}-r)$

Since $a > 0$, divide by $a$:

$r-1 = r^{3}-r$

$r^{3}-2r+1 = 0$

Since $r = 1$ is a root, $(r-1)$ is a factor:

$r^{3}-2r+1 = (r-1)(r^{2}+r-1) = 0$

Since $|r| < 1$ (convergent GP), $r \neq 1$. So $r^{2}+r-1 = 0$:

$r = \dfrac{-1 \pm \sqrt{5}}{2}$

$r = \dfrac{-1+\sqrt{5}}{2} \approx 0.618$ \;\; or \;\; $r = \dfrac{-1-\sqrt{5}}{2} \approx -1.618$

Since $|r| < 1$, reject $r = \dfrac{-1-\sqrt{5}}{2}$.

$\therefore r = \dfrac{-1+\sqrt{5}}{2}$ (proven)

$a > 0$, $r = \dfrac{-1+\sqrt{5}}{2}$

Need to show: $S_{5} > \dfrac{9}{10}S_{\infty}$

$S_{5} = \dfrac{a(1-r^{5})}{1-r}$ ... (1), \;\; $S_{\infty} = \dfrac{a}{1-r}$ ... (2)

Substitute (2) into (1): $S_{5} = S_{\infty}(1-r^{5})$

$S_{5} = S_{\infty}\left[1-\left(\dfrac{-1+\sqrt{5}}{2}\right)^{5}\right]$

$S_{5} = 0.90983\ldots \times S_{\infty} > 0.9\,S_{\infty}$

$\therefore S_{5} > \dfrac{9}{10}S_{\infty}$ (shown)

N2003 P2 Q4

N2003 P2 Q4

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