Maths problem: how to do question 35, thanks.?

a_n = 1.5(-2)^(n-1)
a_n < 0 iff. n even

n = 2k, a_n = 1.5(-2)^(2k-1) = -3(4)^(k-1)
a_(2k) > -6000 iff. 4^(k-1) < (-6000)/(-3) = 2000
   iff. (k-1)log(4) < log(2000)
   iff. k-1 < (3.301030)/0.602060 
   iff. k-1 ≦ 5
   iff. k ≦ 6

共有 6 項是負的, 且值比 -6000 大, 即
    a_2 = -3
    a_4 = -12
    a_6 = -48
    a_8 = -192
    a_10 = -768
    a_12 = -3072

35. 
r = 6/-3 = -2
nth term, aₙ = arⁿ⁻¹ > -6000
(1.5)(-2)ⁿ⁻¹ > -6000
(-2)ⁿ⁻¹ > -4000
(-2)ⁿ < 8000 - - - - -(*)

Since negative terms occur only when n is even,
∴ (*) becomes: 
2ⁿ < 8000
log 2ⁿ < log 8000
n log 2 < log 8000
n < log 8000 / log 2
n < 12.97 ( to 2 dec. pl.)
∴ n = 12

So, 12 -ve terms in the sequence are > -6000