Answer:
Option (b) is correct.
The three terms of the GP are
with common ratio
![√(2)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/t462m14cxkj26cw9cmocfpgj44y1v8li5n.png)
Explanation:
Consider the given geometric sequence
![3√(2),6,6√(2)......](https://img.qammunity.org/2020/formulas/mathematics/high-school/qojlbl4gudif5gvti08ibte65p69dko131.png)
Geometric sequence is a sequence of numbers where each term is find by multiplying the previous one by a fixed number called the common ratio (r).
![a,\ ar,\ ar^(2),\ ar^(3),\ ar^(4),\ \ldots](https://img.qammunity.org/2020/formulas/mathematics/high-school/sllvwuqujgcy6zxqqj9mfsz9vbglb6d2qt.png)
Consider the first two terms of the given GP.
thus r can be find by dividing ar by a,
thus the common ratio is
![r=(6)/(3√(2))=√(2)](https://img.qammunity.org/2020/formulas/mathematics/high-school/urga60yefc7xvwkuoyg3yhzuu8w1rfx7v4.png)
Now we have to find the next three terms of GP . so multiply r in given last term to obtain next three terms , we get ,
![6√(2) * √(2)=6* 2= 12 \\\\\\12 * √(2)= 12√(2)\\\\\\\12√(2) *√(2)=24](https://img.qammunity.org/2020/formulas/mathematics/high-school/41f3e2vipiotpipca5rfjjpmoa0s8xa7z2.png)
Thus, the three terms of the GP are
with common ratio
.
Option (b) is correct.