# Thread: [SOLVED] Arithmetic and Geometric Progressions Question

1. ## [SOLVED] Arithmetic and Geometric Progressions Question

Question:
The 1st term of an arithmetic progression is $\displaystyle a$ and the common difference is $\displaystyle d$, where $\displaystyle d \neq 0$.

(i) Write down expressions, in terms of $\displaystyle a$ and $\displaystyle d$, for the 5th term and 15th term.

The 1st term, the 5th term and the 15th term of the arithmetic progression are the first terms of a geometric progression.

(ii) Show that $\displaystyle 3a = 8d$.
(iii) Find the common ratio of the geometric progression.

Attempt:

(i) $\displaystyle U_5 = a + 4d$
$\displaystyle U_{15} = a + 14d$

(ii) Arithmetic Progression
$\displaystyle U_5 = a + 4d$
$\displaystyle U_{15} = a + 14d$

Geometric Progression
$\displaystyle U_2 = a.r$
$\displaystyle U_3 = a.r^2$

I don't know what to do next... Need help!

2. Hello,

What you did was good

Let's rename the geometric series into S :

$\displaystyle S_2 = a.r$
$\displaystyle S_3=a.r^2$

Now the text says that $\displaystyle S_2=U_5$ and $\displaystyle S_3=U_{15}$

Hence you have to solve this :

$\displaystyle \begin{bmatrix} a+4d=ar \\ a+14d=ar^2 \end{bmatrix}$

Can you try to get 3a=8d ?

3. Originally Posted by looi76
Question:
The 1st term of an arithmetic progression is $\displaystyle a$ and the common difference is $\displaystyle d$, where $\displaystyle d \neq 0$.

(i) Write down expressions, in terms of $\displaystyle a$ and $\displaystyle d$, for the 5th term and 15th term.

The 1st term, the 5th term and the 15th term of the arithmetic progression are the first terms of a geometric progression.

(ii) Show that $\displaystyle 3a = 8d$.
(iii) Find the common ratio of the geometric progression.

Attempt:

(i) $\displaystyle U_5 = a + 4d$
$\displaystyle U_{15} = a + 14d$

(ii) Arithmetic Progression
$\displaystyle U_5 = a + 4d$
$\displaystyle U_{15} = a + 14d$

Geometric Progression
$\displaystyle U_2 = a.r$
$\displaystyle U_3 = a.r^2$

I don't know what to do next... Need help!
$\displaystyle a + 4d = ar$
$\displaystyle a + 14d = ar^2$

$\displaystyle ar + 4dr = ar^2$
$\displaystyle a + 4d + 4dr = ar^2$
$\displaystyle a + 4d + 4dr = a + 14d$
$\displaystyle 4d + 4dr = 14d$
$\displaystyle 4dr = 10d$
$\displaystyle r = \frac{10d}{4d} = \frac{5}{2}$

$\displaystyle a + 4d = \frac{5}{2}a$
$\displaystyle 4d = \frac{3}{2}a$
$\displaystyle 8d = 3a$

4. Originally Posted by icemanfan
$\displaystyle a + 4d = ar$
$\displaystyle a + 14d = ar^2$

$\displaystyle ar + 4dr = ar^2$
$\displaystyle \color{red}{a + 4d + 4dr = ar^2}$
$\displaystyle a + 4d + 4dr = a + 14d$
$\displaystyle 4d + 4dr = 14d$
$\displaystyle 4dr = 10d$
$\displaystyle r = \frac{10d}{4d} = \frac{5}{2}$

$\displaystyle a + 4d = \frac{5}{2}a$
$\displaystyle 4d = \frac{3}{2}a$
$\displaystyle 8d = 3a$
How did you get the equation in red?

5. ar=a+4d

6. Originally Posted by Moo
ar=a+4d
Tell now I didn't get it. How did u get that equation?!

7. Originally Posted by looi76
Tell now I didn't get it. How did u get that equation?!

$\displaystyle \color{blue}{a + 4d = ar}$
$\displaystyle a + 14d = ar^2$

$\displaystyle {\color{red}{ar}} + 4dr = ar^2$
But from the blue equation above this $\displaystyle ar = a+4d$, so replace the red
$\displaystyle \color{red}ar$ with $\displaystyle \color{blue}a+4d$

$\displaystyle {\color{blue}{a + 4d}} + 4dr = ar^2$

Got it ?