Considering that the amount of certain terms in a G.P is \[\mathbf{315}.\]

Leave the quantity of terms alone n.

We realize that, amount of terms is

Considering that the initial term an is \[\mathbf{5}\]  and normal proportion r is \[\mathbf{2}.\]

Subsequently, the last term of the \[\mathbf{G}.\mathbf{P}\text{ }=\text{ }\mathbf{sixth}\text{ }\mathbf{term}\text{ }=\text{ }\mathbf{ar6}\text{ }\text{ }\mathbf{1}\text{ }=\text{ }\left( \mathbf{5} \right)\left( \mathbf{2} \right)\mathbf{5}\text{ }=\text{ }\left( \mathbf{5} \right)\left( \mathbf{32} \right)\text{ }=\text{ }\mathbf{160}\]

In this manner, the last term of the G.P. is \[\mathbf{160}.\]