Answer:
Let A1=a1+a2+a3, A2=a2+a3+a4, and so on, A10=a10+a1+a2. Then A1+A2+⋯+A10=3(a1+a2+⋯+a10)=(3)(55)=165, so some Ai≥165/10=16.5, so some Ai≥17.
Step-by-step explanation:
we know that f(1)=2 so f^-1(2)=1
so B is correct!
<D is correct its confusing on the solving but