# Math Help - [SOLVED] Arithmetic and Geometric Progressions Question

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

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

(i) Write down expressions, in terms of $a$ and $d$, 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 $3a = 8d$.
(iii) Find the common ratio of the geometric progression.

Attempt:

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

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

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

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 :

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

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

Hence you have to solve this :

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

Can you try to get 3a=8d ?

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

(i) Write down expressions, in terms of $a$ and $d$, 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 $3a = 8d$.
(iii) Find the common ratio of the geometric progression.

Attempt:

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

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

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

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

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

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

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

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

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

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

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

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

Got it ?