Answer :

If 'a' is the first term and 'd' is the common difference in an AP,

We know that,

nth term of an AP, a_{n} = a + (n - 1)d

Sum of first 'n' terms of an AP,

Now, given

Sum of first six terms, S_{6} = 42

⇒ 3[2a + 5d] = 42

⇒ 2a + 5d = 14 …[1]

Also, given ratio of 10^{th} and 30^{th} term is 1 : 3

⇒ 3a + 27d = a + 29d

⇒ 2a = 2d

⇒ a = d

Putting d = a in [1]

⇒ 2a + 5a = 14

⇒ 7a = 14

⇒ a = 2

i.e. First term of AP is 2.

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