Answer:
The 4th term of the geometric sequence with = 5 and ratio (multiplier) = -3 is -135
Solution:
Given that, first term a of a G.P = 5 and common ratio ( r ) = -3 for an geometric progression.
We have to find the 4th term of the above given geometric progression
We know that, nth term of an G.P is given by
![t_(n)=a \cdot r^(n-1)](https://img.qammunity.org/2020/formulas/mathematics/high-school/woftk26pffsyh7z6n5gj6l1wszi2bi42e3.png)
So, now, 4th term is
![\begin{aligned} t_(4) &=5 *(-3)^(4-1) \\ t_(4) &=5 *(-3)^(3) \\ t_(4) &=5 *(-27) \end{aligned}](https://img.qammunity.org/2020/formulas/mathematics/high-school/5ibkaorr8wza7yqfdevticsyd4njcxfquf.png)
![t_(4)=-135](https://img.qammunity.org/2020/formulas/mathematics/high-school/zz9qy42upez3cjfj1ye5qcz1fftdzuc6td.png)
hence, the 4th term of the given G.P is -135