# Thread: Arithmetic and geometric sequences.

1. ## Arithmetic and geometric sequences.

(1) The lengths of the sides of a right triangle are three consecutive terms of an increasing arithmetic sequence. Show that the lengths are in the ratio 3:4:5.

(2) If the pth and qth terms of an arithmetic sequence are q and p respectively, find the (p+q)th term.

(3) Find X and y if the sequence 2y, 2xy, x, xy/2 .... is geometric.

Guys i really need you help
Thank you

2. Hello, vvc531!

(1) The lengths of the sides of a right triangle are three consecutive terms of
an increasing arithmetic sequence. Show that the lengths are in the ratio 3:4:5.

The first three terms of an arithmetic sequence are: .$\displaystyle a,\:a+d,\:a+2d$

If they are the sides of a righ triangle: .$\displaystyle a^2 + (a+d)^2 \:=\:(a+2d)^2$

. . which simplifies to: .$\displaystyle a^2 - 2ad - 3d^2 \:=\:0$

. . which factors: .$\displaystyle (a -3d)(a +d) \:=\:0$

. . and has the positive root: .$\displaystyle a \:=\:3d$

Hence, the three sides are: .$\displaystyle \begin{Bmatrix}a &=& 3d \\a+d &=& 4d \\ a+2d &=&5d \end{Bmatrix}$

And their ratio is: .$\displaystyle 3d:4d:5d \;=\;3:4:5$

(3) Find $\displaystyle x$ and $\displaystyle y$ if the sequence $\displaystyle 2y, 2xy, x, \tfrac{xy}{2}$ is geometric.

The first four terms of a geomeric sequence are: .$\displaystyle a,\:ar,\:ar^2,\:ar^3$

So we have: .$\displaystyle \begin{array}{cccc}a &=& 2y & [1] \\ ar &=& 2xy & [2] \\ ar^2 &=& x & [3] \\ ar^3 &=& \frac{xy}{2} & [4] \end{array}$

$\displaystyle \begin{array}{ccccccc}\text{Divide }[2] \div [1] & \dfrac{ar}{a} \:=\:\dfrac{2xy}{2y} & \Rightarrow & r \:=\:x & [5] \\ \\[-4mm] \text{Divide }[3] \div [2] & \dfrac{ar^2}{ar} \:=\:\dfrac{x}{2xy} & \Rightarrow & r \:=\:\dfrac{1}{2y} & [6] \\ \\[-4mm] \text{Divide }[4] \div[3] & \dfrac{ar^3}{ar^2} \:=\:\dfrac{\frac{xy}{2}}{x} & \Rightarrow & r \:=\:\dfrac{y}{2} & [7]\end{array}$

Equate [7] and [6]: .$\displaystyle \frac{y}{2}\:=\:\frac{1}{2y}\quad\Rightarrow\quad y^2 \:=\:1\quad\Rightarrow\quad y \:=\:\pm1$

Substiute into [7]: .$\displaystyle r \:=\:\pm\frac{1}{2}$

Substitute into [5]: .$\displaystyle x \:=\:\pm\frac{1}{2}$

Therefore: .$\displaystyle (x,y) \;=\;\left(\tfrac{1}{2},\tfrac{1}{2}\right),\;\lef t(\text{-}\tfrac{1}{2},\text{-}\tfrac{1}{2}\right)$

3. ## Thank you so much

Thank you so much
can you please take a look at my other post
really hard pre-cal questions (4) ~(5) ?
You are great!