Q. 4

# 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

= 23√2

12th term of A.P. i.e. t12 = 23√2

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