If the 6th term of an A.P is equal to four times its first term, and the sum of first six terms is 75, find the first term and the common difference.

Let a be the first term and d be the common difference.

We know that,

∴ a_n = a + (n - 1)d

Given,

6th term is equal to four times the first term.

⇒ a₆ = 4a₁

⇒ a + (6 - 1)d = 4a

⇒ 5d = 4a - a

⇒ 5d = 3a ………(1)

We know that,

Sum of n terms of an A.P. is given by,

∴ S_n = $\dfrac{n}{2}$ [2a + (n - 1)d]

Given,

Sum of first six terms is 75.

⇒ S₆ = 75

⇒ $\dfrac{6}{2}$ [2a + (6 - 1)d] = 75

⇒ 3[2a + 5d] = 75

⇒ 2a + 5d = $\dfrac{75}{3}$

⇒ 2a + 5d = 25 ….(2)

Substituting value of 5d from equation (1) in (2), we get:

⇒ 2a + 3a = 25

⇒ 5a = 25

⇒ a = $\dfrac{25}{5}$

⇒ a = 5.

Substitute a = 5 into Equation 1, we get :

⇒ 5d = 3a

⇒ 5d = 3(5)

⇒ 5d = 15

⇒ d = $\dfrac{15}{5}$

⇒ d = 3.

Hence, a = 5 and d = 3.