This question provides the values of terms in an arithmetic sequence and asks for the ninth term. Here’s a breakdown of the problem:

1. Given values:
   • $a$: First term
   • $a+1$: Second term
   • $5$: Third term
   • $a - 29$: Fourth term
   • $a - 49$: Fifth term

Since this sequence is arithmetic, the difference between consecutive terms, called the common difference $d$, should be the same.

2. Finding the common difference $d$:
   • From the first and second terms: $d = (a + 1) - a = 1$
   • From the second and third terms: $d = 5 - (a + 1) = 4 - a$

Since $d$ should be consistent across all terms, equate the two expressions for $d$: $1 = 4 - a$ Solving for $a$, we get: $a = 3$

3. Verifying $a = 3$ with other terms:
   • First term: $a = 3$
   • Second term: $a + 1 = 4$
   • Third term: $5$
   • Fourth term: $a - 29 = 3 - 29 = -26$
   • Fifth term: $a - 49 = 3 - 49 = -46$

The common difference $d = 1$ is consistent.

4. Finding the ninth term: The formula for the $n$-th term in an arithmetic sequence is: $a_n = a + (n - 1) \cdot d$ Substituting for $a = 3$, $d = 1$, and $n = 9$: [ a_9 = 3 + (9 - 1) \cdot 1 = 3 + 8 = -57