20分,,,麻煩晒

i) Find the value of a.
The common difference is : T(3) – T(2) = T(2) – T(1)
( 3a – 1 ) – ( a – 1 ) = ( a – 1 ) – a
2a = –1
a = –1/2

ii) Find the common difference, d, of the A.P.
d = T(2) – T(1) = ( a – 1 ) – a = –1

iii) Find the general term of the A.P.
T(n) = a + ( n – 1 ) d = –1/2 + ( n – 1 ) ( –1 )
= –1/2 – n + 1
= 1/2 – n

iv) Find the sum of the first 10 terms of the arithmetic series.
Sum of the first 10 terms of the arithmetic series is :
(1/2) n [ 2a + ( n – 1 ) d ]
= (1/2) x 10 x [ 2 (–1/2) + ( 10 – 1 ) ( –1 ) ]
= 5 (–1 – 9 )
= –50

v) Find the value(s) of a such that the above A.P. becomes a geometric progression G.P
The common ratio is : T(3) / T(2) = T(2) / T(1)
( 3a – 1 ) / ( a – 1 ) = ( a – 1 ) / a
a ( 3a – 1 ) = ( a – 1 )^2
3a^2 – a = a^2 – 2a + 1
2a^2 +a – 1 = 0
( 2a – 1 ) ( a + 1 ) = 0
2a – 1 = 0 or a + 1 = 0
a = 1 / 2 or a = –1

Let k be the 1st term and d is the common difference.
So k = a............(1)
k + d = a - 1...............(2)
k + 2d = 3a - 1.............(3)
(3) - (2) we get d = 3a - 1 - (a - 1) = 3a - 1 - a + 1 = 2a.
Sub into (2)
k + 2a = a - 1
k = -1 -a
Sub into (10
-1 - a = a
2a = -1
a = -1/2.
d = 2(-1/2) = -1.
and k = - 1 - a = -1 - (-1/2) = - 1 + 1/2 = -1/2
so general term T(n) = k + (n -1)d = -1/2 + (n -1)(-1) = -1/2 - n + 1
= 1/2 - n.
Sum of first 10 terms = 10[2(-1/2) + 9(-1)]/2 = 5[-1 - 9] = -50.



















2009-07-08 18:11:17 補充：
For a G.P. T(2)/T(1) = T(3)/T(2) = common ratio. Therefore, (3a -1)/(a - 1) = (a -1)/a
solving the equation we get a = 1/2 or -1.