Find the 12th term of the A.P.
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
∴ 12th term of A.P. i.e. t12 = 23√2
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