# Thread: Geometric progression

1. ## Geometric progression

A roll of thread with 90pi cm long is cut into 5 parts to make 5 circles.The radii of the circles increase by 1 cm consecutively.Calculate

a)the radius of the smallest circle
b)the circumference of the 6 th circle
c)the number of circles obtained if the original length of the thread is 660pi cm

2. Hi there mastermin346

Originally Posted by mastermin346
A roll of thread with 90pi cm long is cut into 5 parts to make 5 circles.The radii of the circles increase by 1 cm consecutively.Calculate

a)the radius of the smallest circle
Well i'm reading this as the length of string is 90cm?

Therefore

90 = length of string #1 + length of string #2+ .... +length of string #5

And the difference in each length, shaped as a circle is, adding 1 to the next radius so

$90 = 2\pi r+2\pi (r+1)+2\pi (r+2)+2\pi (r+3)+2\pi (r+4)$

Factoring out and then dividing $2\pi$

$90 = 2\pi (r+ (r+1)+ (r+2)+ (r+3)+ (r+4))$

$\frac{90}{2\pi} = (r+ (r+1)+ (r+2)+ (r+3)+ (r+4))$

Grouping like terms on the RHS

$\frac{90}{2\pi} = 5r+10$

$\frac{45}{\pi} = 5r+10$

$\frac{45}{\pi} -10= 5r$

$r= \frac{\frac{45}{\pi} -10}{5}$

Originally Posted by mastermin346
b)the circumference of the 6 th circle
c)the number of circles obtained if the original length of the thread is 660pi cm
You should be able to use the same logic to finish the rest.

3. Hello mastermin346
Originally Posted by mastermin346
A roll of thread with 90pi cm long is cut into 5 parts to make 5 circles.The radii of the circles increase by 1 cm consecutively.Calculate

a)the radius of the smallest circle
b)the circumference of the 6 th circle
c)the number of circles obtained if the original length of the thread is 660pi cm
A number of points:

1 There's no reason to suppose there was a typo in the original question: the length of the thread could quite well be $90\pi$ cm; in which case, the radius of the smallest circle is $7$ cm.

2 This is an example of an arithmetic progression, not a geometric progression.

3 For part (c), if the initial radius is $7$ cm and there are $n$ circles altogether, then the total length of thread used is:
$2\pi\big(7+(7+1) + (7+2) + ... + (7+[n-1])\big)$
$=2\pi\times \tfrac12n(13+n)$, using the AP formula $S = \tfrac12n(a+l)$

$=660\pi$
This gives a quadratic equation for $n$ with one positive root.
Spoiler:
n=20

Grandad