Here, first term = a =3

Common difference = d = 7 - 3 = 4

Let first n terms of the AP sums to 406.

∴ S_n = 406

To find: n

Now, S_n = (n/2) × [2a + (n - 1)d]

Since, S_n = 406

∴ (n/2) × [2a + (n - 1)d] = 406

⇒ (n/2) × [2(3) + (n - 1)4] = 406

⇒ (n/2) × [6 + 4n - 4] = 406

⇒ (n/2) × [(2 + 4n] = 406

⇒ n[1 + 2n] = 406

⇒ n + 2n² = 406

⇒ 2n² + n - 406 = 0

⇒ 2n² - 28n + 29n - 406 = 0

⇒ 2(n - 14) + 29(n - 14)= 0

⇒ (2n + 29)(n - 14) = 0

⇒ n = 14 or n = - 29/2

∴ n= 14 (∵ n can’t be a fraction or negative number)