Solving problems involving APs and GPs

Three terms in arithmetic or geometric progression, the arithmetic and geometric means, simultaneous equations from two given terms, and using logarithms to find when a GP passes a value.

Three numbers in AP or GP

$\dfrac{a + b}{2}$ is called the arithmetic mean of $a$ and $b$. When $a$ and $b$ are both positive, $\sqrt{ab}$ is their geometric mean. In a GP the middle term could also be $-\sqrt{ab}$.

Example 1●●

Find $x$ if $x + 1$, $x + 5$ and $3x + 1$ are in GP.

$(x + 5)^2 = (x + 1)(3x + 1)$

$x^2 + 10x + 25 = 3x^2 + 4x + 1$, so $2x^2 - 6x - 24 = 0$, which is $x^2 - 3x - 12 = 0$.

$x = \dfrac{3 \pm \sqrt{57}}{2}$.

Example 2●

Find the arithmetic mean and the geometric mean of $4$ and $25$.

Arithmetic mean $= \dfrac{4 + 25}{2} = 14.5$. Geometric mean $= \sqrt{100} = 10$.

Two terms given

Example 3: a GP●●●

A GP has $T_3 = 18$ and $T_6 = -486$. Find $a$ and $r$.

$ar^2 = 18$ and $ar^5 = -486$.

$r^3 = -27$, so $r = -3$.

Divide the equations: a cancels.

Then $a = \dfrac{18}{9} = 2$.

When does a GP pass a value?

Example 4●●●

The GP $4000, 3200, 2560, \ldots$ has ratio $0.8$. Which is the first term less than $1$?

$4000 \times 0.8^{n-1} < 1$, so $0.8^{n-1} < 0.00025$.

$(n - 1)\ln 0.8 < \ln 0.00025$ gives $n - 1 > \dfrac{\ln 0.00025}{\ln 0.8} \approx 37.17$.

ln 0.8 is negative, so dividing by it reverses the inequality.

So $n > 38.17$, and the first such term is $T_{39}$.

Flashcards

$a, x, b$ in AP

$x = \dfrac{a + b}{2}$

$a, x, b$ in GP

$x^2 = ab$

Two terms of a GP given

Divide the equations to remove $a$.

Practice

Find the arithmetic mean of $4$ and $25$.
Find the geometric mean of $4$ and $25$.
A GP has $T_3 = 18$ and $T_6 = -486$. Find $r$.
Which term of $4000, 3200, 2560, \ldots$ is the first to be less than $1$?
$x - 2$, $x + 4$ and $3x$ are in AP. Then:
An AP has $T_4 = 25$ and $T_{10} = 67$. Find $d$.
An AP has $T_4 = 25$ and $T_{10} = 67$. Find $a$.
Solving $0.9^{n} < 0.1$ with logarithms gives:
Find the positive $x$ if $4, x, 9$ are in GP.