Question

Can someone help me please

Answer 1

6a. r^5

6b. 18(4+√2)

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 ...

`√(8)((√(8))^4-1)/(√(8)-1)=√(8)((64-1)(√(8)+1))/((\sqrt(8))^2-(1)^2)\\\\=√(8)(63)/(7)(√(8)+1)=9(8+√(8))\\\\=18(4+√(2))`

Or, you could add up the 4 terms:

`2√(2)+8+16√(2)+64=72+18√(2)=18(4+√(2))`