We have, a3 = 5
⇒ a + (3 – 1) d = 5
⇒ a + 2d = 5 ...(i)
and a7 = 9
⇒ a + (7– n) d = 9
⇒ a + 6d = 9 ...(ii)
Subtracting (i) from (ii), we get
(a + 6d) – (a + 2d) = 4
⇒ a + 6d – a – 2d = 4
⇒ 4d = 4
⇒ d = 1
Putting the value of ‘d’ in (i), we get
a + 2d = 5 ⇒ a + 2(1) = 5
⇒ a + 2 = 5 ⇒ a = 3
Hence, the required A .P. be 3, 4, 5, 6, 7,..........