The n^th term of an A.P. is 2n - 5, then the sum of its first 6 term is :

1. 12

2. 15

3. -48

4. -12

Given,

⇒ a_n = 2n - 5

⇒ a₁ = 2(1) - 5 = 2 - 5 = -3,

⇒ a₂ = 2(2) - 5 = 4 - 5 = -1.

So, first term (a) = -3 and common difference (d) = -1 - (-3) = -1 + 3 = 2.

By formula,

Sum of n terms (S) = $\dfrac{n}{2}[2a + (n - 1)d]$

Substituting values we get :

$S_6 = \dfrac{6}{2}[2 \times -3 + (6 - 1) \times 2] \\[1em] = 3[-6 + 5 \times 2] \\[1em] = 3 \times [-6 + 10] \\[1em] = 3 \times 4 \\[1em] = 12.$

Hence, Option 1 is the correct option.