好easy的geometric questions@.@ ( 5點分)~

question) (a) Find m in terms of a:

answer:
( a + 3m ) / ( a + m ) = ( a + 11m ) / ( a + 3m )
( a + 3m ) ( a + 3m) = (a + 11m ) ( a + m )
( a + 3m) ² = a² + ma + 11ma + 11m²
a² + 2(a)(3m) + (3m)² = a² + 12ma + 11m²
6ma + 9m² = 12ma + 11m²
0 = 12ma - 6ma + 11m² - 9m²
0 = 6ma + 2m²
-6ma = 2m²
m(-6a) = m(2m) *(抽出因子m並且約掉)
-6a = 2m
-6a / 2 = m
-3a = m
∴ m = -3a
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question) (b) Hence find the common ratio of the geometric squence:

let common ratio be r,
r = ( a + 3m ) / ( a + m ) ------------------------(1)

m = -3a (proved) ------------------------------------(2)

pur (2) into (1)
r = [ a + 3(-3a) ] / [ a + (-3a)]
= (a - 9a) / ( a - 3a )
= -8a / -2a
= -8 / -2
= 4

∴The common ratio is 4

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a.
(a+3m)/(a+m)=(a+11m)/(a+3m) (common ratio)
(a+3m)^2=(a+m)(a+11m)
a^2+6am+9m^2=a^2+12am+11m^2
2m^2+6am=0
m(m+3a)=0
m=0(rej.)or m= -3a

b. the sequence becomes -2a, -8a, -32a
common ratio = (-8a)/(-2a)=4