The 5th term and the 9th term of an Arithmetic Progression are 4 and −12 respectively. Find:

(i) the first term

(ii) common difference

(iii) sum of first 16 terms of the AP.

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

We know that,

∴ a_n = a + (n - 1)d

Given,

The 5th term of an A.P. is 4.

⇒ a + (5 - 1)d = 4

⇒ a + 4d = 4 ….(1)

Given,

The 9th term of an A.P. is -12.

⇒ a + (9 - 1)d = -12

⇒ a + 8d = -12 ….(2)

Subtracting Equation (1) from Equation (2):

⇒ a + 8d - (a + 4d) = -12 - 4

⇒ a + 8d - a - 4d = -16

⇒ 4d = -16

⇒ d = $\dfrac{-16}{4}$

⇒ d = -4.

Substituting d = -4 in Equation (1) :

⇒ a + 4(-4) = 4

⇒ a - 16 = 4

⇒ a = 4 + 16

⇒ a = 20.

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 = 20

d = -4

n = 16

⇒ S₁₆ = $\dfrac{16}{2}$ [2(20) + (16 - 1)-4]

= 8[40 + (15)-4]

= 8[40 - 60]

= 8.(-20)

= -160.

Hence, a = 20, d = -4 and S₁₆ = -160.