4.075 The sequence of numbers $1,8,22,43, \ldots$ has the property that the differences between two consecutive terms (the subsequent and the previous) form an arithmetic progression: $7,14,21, \ldots$. Find the index of the term in the sequence that equals 35351.
Solution. Let the numbers of the given sequence $1, 8, 22, \ldots$ be denoted as $a_{1}, a_{2}, a_{3}, \ldots$.

By the condition, $a_{2}-a_{1}=7 \cdot 1, a_{3}-a_{2}=7 \cdot 2, \ldots, a_{n}-a_{n-1}=7(n-1)$.

Adding these equations term by term, we have: $\left(a_{2}-a_{1}\right)+\left(a_{3}-a_{2}\right)+\ldots+$

$+\left(a_{n-1}-a_{n-2}\right)+\left(a_{n}-a_{n-1}\right)=7(1+2+\ldots+(n-1)) \Leftrightarrow a_{n}-a_{1}=7 \cdot \frac{n(n-1)}{2} \Rightarrow$

$\Rightarrow a_{n}=a_{1}+\frac{7(n-1) n}{2}$. Since $a_{n}=35351$, then $n(n-1)=10100 \Rightarrow n=101$.

Answer: 101.