Question

under root 2, 3under root 3, 5 under root 2 find the 18th term of AP​

Answer 1

Answer:

$t18 = 51 \sqrt{3} - 16 \sqrt{2}$

Step-by-step explanation:

Given:

$AP: \: \sqrt{2}, 3 \sqrt{3}, 5 \sqrt{2}.....$

Solution:

$We have: tn = a + (n-1)d \\ t2 = \sqrt{2} + d = 3 \sqrt{3} \\ d = 3 \sqrt{3 - \sqrt{2} } \\ Now \\ t18 = a + 17d \\ t18 = ( \sqrt{2}) + (17)(3 \sqrt{3} - \sqrt{2} ) \\ t18 = 51 \sqrt{3} - 16 \sqrt{2}$

Answer 2

Answer:

refers to the attachment....