Question

Can someone help me please

Answer 1

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 ...

$\sqrt{8}\dfrac{(\sqrt{8})^4-1}{\sqrt{8}-1}=\sqrt{8}\dfrac{(64-1)(\sqrt{8}+1)}{(\sqrt(8))^2-(1)^2}\\\\=\sqrt{8}\dfrac{63}{7}(\sqrt{8}+1)=9(8+\sqrt{8})\\\\=18(4+\sqrt{2})$

Or, you could add up the 4 terms:

$2\sqrt{2}+8+16\sqrt{2}+64=72+18\sqrt{2}=18(4+\sqrt{2})$