Answer :
Step-by-step explanation:
we know,
[tex]tn= a {(r)}^{n - 1} \\ T5 = a {(2)}^{5 - 1} \\ t5 = a \times {2}^{4} \\ = 16a[/tex]
Therefore, nth term of a given gp is
16a
Step-by-step explanation:
we know,
[tex]tn= a {(r)}^{n - 1} \\ T5 = a {(2)}^{5 - 1} \\ t5 = a \times {2}^{4} \\ = 16a[/tex]
Therefore, nth term of a given gp is
16a