Eton Education Centre Two approaches to solving a problem on geometric progression By Wee WS (wenshih.wordpress.com) Consider the following problem which has been worded like questions that appeared in the 2011 H2 Mathematics Examination (refer to P1/Q9 and P2/Q10): A circle with radius 10 cm is cut into 10 sectors of decreasing areas. The first sector, which is the largest, has an area of a cm 2 . Each subsequent sector has an area that is r of the area of the previous one. It is given that the total area of the odd-numbered sectors exceeds that of the remaining sectors by 10π cm 2 . Obtain an equation involving r, and hence find the area of the smallest sector. Analysis: We are given three pieces of information, i.e., • the total area of 10 sectors refers to the area of a circle with radius 10 cm; • the sector areas follow a decreasing geometric progression (GP) with first term a and common ratio r; • the relationship between the total area of the odd-numbered sectors and the total area of the even-numbered sectors. The objective is clear – to obtain an equation involving r using the information provided and to find the value of ar 9 , the area of the smallest sector.

Approach: Based on the first and second pieces of information, we can form this relationship easily:

(

) =π

a r10 − 1

(10 ) 2

r −1 ⇒ a r10 − 1 = 100π ( r − 1) --- (1).

(

)

The third piece of information gives us this equation: a + ar 2 + ar 4 + ar 6 + ar 8 = 10π + ar + ar 3 + ar 5 + ar 7 + ar 9 . Using the GP sum formula, we may condense it to: 5 5 a  r 2 − 1 ar  r 2 − 1   = 10π +   2 2 r −1 r −1 10 2 ⇒ a r − 1 = 10π r − 1 + ar r10 − 1 --- (2).

( ) (

( )

)

(

Page 1 of 2

)

(

)

To obtain the desired equation, we may consider two ways:

(

)

replace the terms a r10 − 1 in (2), or

replace the term 10π r 2 − 1 = 10π ( r − 1)( r + 1) in (2).

(

)

If we choose the first way, we obtain:

(

)

100π ( r − 1) = 10π r 2 − 1 + r 100π ( r − 1)  . Upon simplification (you are encouraged to do it!), we arrive at a nice quadratic equation: 11r 2 − 20r + 9 = 0 . Factorisation leads to: giving r =

( r − 1)(11r − 9 ) = 0 ,

9 , since the sectors must become smaller progressively. 11

With the second way we get:

(

)

a r −1 = 10

(

) ( r + 1) + ar

a r10 − 1 10

(r

10

)

−1 .

Simplifying (again, you are encouraged to get your hands dirty!), we arrive at: 11r11 − 9r10 − 11r + 9 = 0 , which looks more intimidating than the equation of the first way but still remains easy for us to factorise:

(r

With r =

10

)

− 1 (11r − 9 ) = 0 .

9 , a can be found from (1) and the value of ar 9 can then be established. 11

From this example, we see the need to reduce our problem-solving effort to arrive at a solution as efficiently as possible. It is essential that we make mathematics work favourably for us, rather than have it stump us unwittingly.

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