In an Arithmetic Progression (A.P.), the fourth and sixth terms are 8 and 14 respectively. Find the :

(i) first term

(ii) common difference

(iii) sum of the first 20 terms.

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

We know that,

∴ a_n = a + (n - 1)d

Given,

The 4th term of an A.P. is 8.

⇒ a₄ = a + (4 - 1)d

⇒ 8 = a + 3d

⇒ a + 3d = 8 ….(1)

Given,

The 6th term of an A.P. is 14.

⇒ a₆ = a + (6 - 1)d

⇒ 14 = a + 5d

⇒ a + 5d = 14 ….(2)

Subtracting Equation (1) from Equation (2), we get:

⇒ a + 5d - (a + 3d) = 14 - 8

⇒ a + 5d - a - 3d = 6

⇒ 2d = 6

⇒ d = $\dfrac{6}{2}$

⇒ d = 3.

Substituting value of d in equation (1):

⇒ a + 3(3) = 8

⇒ a + 9 = 8

⇒ a = 8 - 9

⇒ a = -1.

We know that,

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

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

Now we have,

a = -1

d = 3

n = 20

⇒ S₂₀ = $\dfrac{20}{2}$ [2(-1) + (20 - 1)3]

= 10 [-2 + (19)3]

= 10 [-2 + 57]

= 10 × (55)

= 550.

Hence, a = -1, d = 3 and S₂₀ = 550.