Question

Find the 18 term of the AP √2,√18 ,√50,√98...​

Answer 1

Answer:

$\sqrt{2} = 1 \sqrt{2} \\ \sqrt{18} = 3 \sqrt{2} \\ \sqrt{50} = 5 \sqrt{2} \\ \sqrt{98} = 7 \sqrt{2} \\ . \\ . \\ . \\ . \\ . \\ . \\ . \\ a = \sqrt{2} \: \: \: \: \: d = 3 \sqrt{2} - \sqrt{2} = 2 \sqrt{2} \\ \\ \pink{ \huge{tn = a + (n - 1)d}} \\ \\ t18 = \sqrt{2} + (18 - 1)2 \sqrt{2} \\ \\ = \sqrt{2} + 17 \times 2 \sqrt{2} \\ \\ = \sqrt{2} + 34 \sqrt{2} \: \: \: \: \: \: \: \: \\ \\ = 35 \sqrt{2} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

Answer 2

Answer:

please make me brainlist

Step-by-step explanation:

AP = √2,√18,√50,√98

d= a² –a¹

d = √18–√2

d=√16

a18= a+( n–1)d

= √2+(18–1) √16

= √2 + (17)√16

= 17√18