Check whether – 150 is a term of the AP : 11, 8, 5, 2 . . .
Step-by-Step Solution
Key Concept: Use the nth‑term formula of an arithmetic progression, $a_n = a + (n-1)d$, and verify whether the required number can be obtained for a positive integer value of $n$.
1. Identify the first term and common difference of the given AP.
\[ a = 11, \quad d = 8-11 = -3 \]
2. Write the general (nth) term of the AP.
\[ a_n = a + (n-1)d = 11 + (n-1)(-3) = 11 - 3(n-1) = 14 - 3n \]
3. Set $a_n$ equal to the required term $-150$ and solve for $n$.
\[ -150 = 14 - 3n \]
\[ -150 - 14 = -3n \]
\[ -164 = -3n \]
\[ n = \frac{164}{3} \]
4. Since $n = \frac{164}{3} \approx 54.67$ is not a positive integer, the required term cannot be obtained from the AP.
5. Conclude that $-150$ is not a term of the given AP.
Correct Answer: No, –150 is not a term of the AP 11, 8, 5, 2, …