Question

Find the tenth term of the sequence root 2 ,root 8, root18, . . . . . . .

Answer 1

√2, √8,√18
√2, 2√2,3√2
a = √2
d = 2√2 - √2
=√2

A10 = a + (n-1)d
       = √2 + (10 - 1)(√2)
      = √2 + 9√2
=10√2

Answer 2

$a = \sqrt{2} \\ r = \sqrt{8} \div \sqrt{2} = 2 \sqrt{2 } \div \sqrt{2 } = 2 \\ 10th \: term = \sqrt{2} \times {2}^{9} = 512 \sqrt{2}$