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.

Default solution

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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Default solution