Given,

4^th term of an A.P. is three times the first and the 7^th term exceeds twice the third term by 1

⇒ a₄ = 3a and a₇ = 2a₃ + 1

We know, a_n = a + (n – 1)d where a is first term or a₁ and d is common difference and n is any natural number

When n = 4:

∴ a₄ = a + (4 – 1)d

⇒ a₄ = a + 3d

When n = 7:

∴ a₇ = a + (7 – 1)d

⇒ a₇ = a + 6d

When n = 3:

∴ a₃ = a + (3 – 1)d

⇒ a₃ = a + 2d

According to question:

a₇ = 2a₃ + 1

⇒ a + 6d = 2(a + 2d) + 1

⇒ a + 6d = 2a + 4d + 1

⇒ a – 2a + 6d – 4d = 1

⇒ -a + 2d = 1

⇒ a – 2d = -1……(i)

a₄ = 3a

⇒ a + 3d = 3a

⇒ 3d = 3a – a

⇒ 3d = 2a

Put this value of d in equation (i):

⇒ a = 3

Now, put this value of a in equation (ii):

⇒ d = 2

Hence, the value of a and d are 3 and 2 respectively