7 Determine the number of terms: a \( \quad x(1+i) ; x(1+i)^{2} ; x(1+i)^{3} ; \ldots ; x(1+i)^{40} \) b \( \quad x(1+i)^{-1} ; x(1+i)^{-2} ; x(1+i)^{-3} ; \ldots ; x(1+i)^{-160} \) c \( \tan x ; \sin x ; \sin x \cos x ; \ldots ; \tan x(\cos x)^{9} \) d \( 1 ; \sin x ; 1-\cos ^{2} x ; \ldots ;(\sin x)^{11} \) 8 Given \( \mathrm{T}_{n}=\frac{3}{8}(r)^{n-1} \) for a geometric sequence. a If the 4th term is \( \frac{81}{8} \), calculate the value of \( r \). b Which term is equal to \( \frac{59049}{8} \) ?

The first sequence in part 7a consists of terms that are defined by powers of \((1+i)\) multiplied by \(x\). Here we see that the powers of \((1+i)\) range from 1 to 40, giving us exactly 40 terms in total. So, you get 40 fabulous terms to enjoy when you’re exploring this sequence! In part 7b, we have terms defined by negative powers of \((1+i)\). The expression starts at \((1+i)^{-1}\) and goes all the way up to \((1+i)^{-160}\). This means we're counting from -1 to -160, which produces exactly 160 terms. No need for backward jumps when joy rides through 160! Now, let's unravel the mystery of part 8a! We know \(T_n = \frac{3}{8}(r)^{n-1}\), and for the 4th term, \(T_4 = \frac{3}{8}(r)^3 = \frac{81}{8}\). Rearranging gives us \((r)^3 = \frac{81}{3} = 27\), so \(r = 3\). That’s like climbing three steps up to meet a friendly \(r\)! For part 8b, we need to find which term equals \(\frac{59049}{8}\). We set this with our formula: \(\frac{3}{8}(3)^{n-1} = \frac{59049}{8}\). By simplifying and solving for \(n\), we find \(3^{n-1} = 59049/3 = 19683\). Recognizing \(19683 = 3^9\), we find \(n-1 = 9\), and therefore, the term we're after is the 10th term! What a triumphant ride through numbers!