Determine the arithmetic progression whose 3^rd term is 5 and 7^th term is 9.

We know that,

n^th term of an A.P. is given by,

a_n = a + (n - 1)d

Given, 3^rd term is 5

∴ a₃ = a + (3 - 1)d

⇒ 5 = a + 2d

⇒ a + 2d = 5 ……..(i)

Given, 7^th term is 9

∴ a₇ = a + (7 - 1)d

⇒ 9 = a + 6d

⇒ a + 6d = 9 ……..(ii)

Subtracting (i) from (ii) we get,

⇒ a + 6d - (a + 2d) = 9 - 5

⇒ 4d = 4

⇒ d = 1.

Substituting value of d in (i) we get,

⇒ a + 2(1) = 5

⇒ a + 2 = 5

⇒ a = 3.

A.P. = a, (a + d), (a + 2d), (a + 3d) ……….

= 3, 4, 5, 6, 7, ………..

Hence, A.P. = 3, 4, 5, 6, 7, ………..