Puzzle ranking numbers

Question

There are 10 distinct numbers in a list. Except the 1st and 10th numbers, the value of each number lies between the value of its immediate neighbors. If the 4th number is less than 7th, what is the smallest number in the list ?

(A)1st (B)4th (C)10th (D)Can't be determined

My thinking:-

let the numbers be $a_{1},a_{2},a_{3},...,a_{10}$, Now ATQ $a_2<a_3 ; a_2>a_3, a_3<a_4 $, now we have $a_2>a_3$ and $a_2<a_3$, how is it possible if they are distinct. Is this question correctly framed or am I interpreting it wrongly

Answer given in the book is (A)1st

Answer 1

Let $x_1,x_2,x_3,....,x_{10}$ be all the $10$ numbers. Then according to question except $x_1,x_{10}$ every number lies between it's immediate neighbours. Then, $$x_1<x_2<x_3, x_2<x_3<x_4, x_3<x_4<x_5... ,x_8<x_9<x_{10}$$ Combining all those equations we have $$x_1<x_2<x_3<...<x_{10}$$ Hence the least element is $x_1$ and greatest is $x_{10}$.

Answer 2

Consider any three consecutive numbers in the list: $p, q, r$. For $q$ to be between $p$ and $r$, either $p<q<r$ or $r<q<p$. Extending this further in both directions eventually gives you either $x_1<x_2<...<x_9<x_{10}$ or $x_{10}<x_9<...<x_2<x_1$.

Because $x_4<x_7$ the smallest must be $x_1$.