急 F6 Geometric Sequences 02q11

17a) Let a and d be the first term and common difference of the sequence respectively.
T(3) = -22
a + d(3-1) = -22
a + 2d = -22 ................(1)

T(5) + T(9) = -4
[a + (5-1)d] + [a + (9-1)] =-4
2a + 12d =-4
a + 6d = -2........(2)
(2) -(1) , 4d = 20
d = 5
put d =5 into (1), a + 2(5) =-22
a =-32

Therefore , T(n) = -32 + (n-1)(5)
= 5n -37

b) T(m) >0
5m -37 >0
m > 7.4
Therefore, the required term = 5 (8) -37 = 3


18a) T(1) = 1
T(2) = 4 = 2^(2)
T(3) = 9 = 3^(2)
T(4) = 16 = 4^(2)

Therefore, T(n) = n^(2)

b) U(n) = T(n+1) - T(n)
= (n+1)^(2) - n^(2)
= n^(2) + 2n +1 - n^(2)
= 2n +1

c) U(1) = 2(1) + 1 = 3
U(2) = 2(2) + 1 = 5
U(3) = 2(3) + 1 = 7

Since U(3) - U(2) = U(2) - U(1) =2
Therefore it is a arithmetic sq.