Solution:
T1(a) = 10, common ratio (r) = [tex] \frac{30}{10} = 3[/tex] T11 = ?
Formula for finding the nth term of a GP: [tex]t_{n} = a {r}^{n - 1} [/tex]
To find T11:
[tex]⟹t_{11} = 10 \times {3}^{11 - 1} [/tex]
[tex] ⟹t_{11} = {10 \times 3}^{10} [/tex]
[tex] ⟹t_{11} = 10 \times 59049[/tex]
Therefore:
[tex] t_{11} = 590490[/tex]
