**Concept: The question tests on applications of Sequences**

**Solution:**

Given a1 = b and an=a _{n−1}+1

a2 = a1 + 1 = b + 1

a3 = a2 + 1 = (b + 1) + 1 = b + 2

a4 = (b + 2) + 1 = b + 3

So a _{n}=b+(n−1)

a10 = b + 9 and a7 = b + 6

(a10−a7)^2=(a10+a7)(a10−a7)

= (b + 9 + b + 6)(b + 9 - b - 6) = (2b + 15)(3)

= 6b + 45

**Option B**