Answer:
6a. r^5
6b. 18(4+√2)
Step-by-step explanation:
6a. The first term of the sequence is √r, and the common ratio is √r. Hence the 10th term will be ...
a1·r^(n-1) . . . for a1=√r and n=10.
√r·(√r)^(10-1) = (√r)^10 = r^5
_____
6b. The sum of n terms is given by ...
Sn = a1·(R^n -1)/(R -1)
For a1=√8 and the common ratio R = √8, the sum of 4 terms is ...
[tex]\sqrt{8}\dfrac{(\sqrt{8})^4-1}{\sqrt{8}-1}=\sqrt{8}\dfrac{(64-1)(\sqrt{8}+1)}{(\sqrt(8))^2-(1)^2}\\\\=\sqrt{8}\dfrac{63}{7}(\sqrt{8}+1)=9(8+\sqrt{8})\\\\=18(4+\sqrt{2})[/tex]
Or, you could add up the 4 terms:
[tex]2\sqrt{2}+8+16\sqrt{2}+64=72+18\sqrt{2}=18(4+\sqrt{2})[/tex]
