Answer :

Given A.P. √2, 3√2, 5√2 …


Here, first term, a = √2


We know that common difference, d = a1 – a.


d = 3√2 - √2 = 2√2


We know that nth term of A.P. , tn = a + (n – 1) d.


t12 = √2 + (12 – 1) (2√2)


= √2 + 11 (2√2)


= √2 + 22 √2


= 23√2


12th term of A.P. i.e. t12 = 23√2


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