Question

find the 16th term of AP √2,3√2,5√2

Answer 1

for given AP
$d = 3 \sqrt{2} - \sqrt{2} = 2 \sqrt{2} \\ a = \sqrt{2} and \: d = 2 \sqrt{2} \\ {16}^{th} term \: = a + 15d = \sqrt{2} + 15(2 \sqrt{2} ) = \sqrt{2} + 30 \sqrt{2} = 31 \sqrt{2}$
16thterm is 31√2