Find, showing all working, a formula for the n-th term tn of the sequence (tn) defined by t1 = 5; tn = -7tn-1 /3, n >= 2.

$t_1=5,t_n=-\frac{7t_{n-1}}{3}$

$\dfrac{t_n}{t_{n-1}}=\dfrac{-\dfrac{7t_{n-1}}{3}}{t_{n-1}}=-\dfrac{7}{3}$

We have a geometric progression $(t_n)$

$t_1=5, q=-\dfrac{7}{3}$

Then

$t_n=5(-\dfrac{7}{3})^{n-1}, n\geq1$