Verify if 36 Belongs to the Arithmetic Sequence: 51, 47, 43, 39,...

To solve this problem, we need to establish whether $36$ is a term in the given sequence $51, 47, 43, 39, \ldots$.

First, let's determine the pattern of the sequence. Observe the differences between consecutive terms:

• From $51$ to $47$: $51 - 47 = 4$
• From $47$ to $43$: $47 - 43 = 4$
• From $43$ to $39$: $43 - 39 = 4$

The sequence decreases by a constant difference of $4$. Therefore, it is an arithmetic sequence with the first term $a_1 = 51$ and common difference $d = -4$.

The formula for the $n$-th term of an arithmetic sequence is given by:

$a_n = a_1 + (n-1) \cdot d$

Substitute the known values:

$a_n = 51 + (n-1) \cdot (-4)$

Simplify the expression:

$a_n = 51 - 4(n-1)$

$a_n = 51 - 4n + 4$

$a_n = 55 - 4n$

Now, let's find out if $36$ is in the sequence by setting $a_n = 36$:

$55 - 4n = 36$

Solve for $n$:

$55 - 36 = 4n$

$19 = 4n$

$n = \frac{19}{4}$

$n = 4.75$

Since $n = 4.75$ is not an integer, $36$ is not a term of the sequence.

Therefore, the answer to the question is No.