4^th term of an A.P. is equal to 3 times its first term and 7^th term exceeds twice the 3^rd term by 1. Find the first term and the common difference.

Let first term be a and common difference be d.

According to question,

⇒ a₄ = 3a

⇒ a + (4 - 1)d = 3(a)

⇒ a + 3d = 3a

⇒ 2a = 3d

⇒ a = $\dfrac{3d}{2}$ ……..(i)

Also,

⇒ a₇ - 2a₃ = 1

⇒ a + (7 - 1)d - 2[a + (3 - 1)d] = 1

⇒ a + 6d - 2(a + 2d) = 1

⇒ a + 6d - 2a - 4d = 1

⇒ a - 2a + 2d = 1

⇒ -a + 2d = 1

Substituting value of a from (i) in above equation,

$\Rightarrow -\dfrac{3d}{2} + 2d = 1 \\[1em] \Rightarrow \dfrac{-3d + 4d}{2} = 1 \\[1em] \Rightarrow \dfrac{d}{2} = 1 \\[1em] \Rightarrow d = 2.$

Substituting value of d in (i) we get,

a = $\dfrac{3 \times 2}{2}$ = 3.

Hence, first term = 3 and common difference = 2.