Check whether -150 is a term of the AP: 11, 8, 5, 2, ...
No
Step 1: Identify the first term and common difference.
The AP is: 11, 8, 5, 2, ...
First term: \(a = 11\)
Common difference: \(d = 8 - 11 = -3\)
Step 2: Use the general term formula of an AP.
The \(n\)-th term of an AP is given by:
\[a_n = a + (n - 1)d\]
We check whether for some positive integer \(n\), the term equals \(-150\):
\[a_n = -150\]
Substitute values of \(a\) and \(d\):
\[11 + (n - 1)(-3) = -150\]
Step 3: Solve for \(n\).
Expand:
\[11 - 3(n - 1) = -150\]
Distribute the -3:
\[11 - 3n + 3 = -150\]
Simplify left side:
\[14 - 3n = -150\]
Move 14 to the right:
\[-3n = -150 - 14 = -164\]
Divide both sides by -3:
\[n = \frac{164}{3}\]
Step 4: Interpret the result.
\(n = \frac{164}{3}\) is not a whole number.
Since the term number must be a positive integer, no such \(n\) exists.
Conclusion: \(-150\) is not a term of the given AP.