Arithmetic Progressions
NCERT Exemplar Ch 05
CBSE_NCERT_EXEMPLAR_CH05
Grade 10
Question:
Check whether $-150$ is a term of the AP: $11, 8, 5, 2, \dots$
Step-by-Step Solution
Key Concept: Set $a_n = a + (n-1)d = -150$. If $n$ is a positive integer, it is a term; otherwise not.
Stepwise Solution:
Here $a = 11, d = 8 - 11 = -3$. Let $a_n = -150$.
$11 + (n - 1)(-3) = -150 \Rightarrow (n - 1)(-3) = -161 \Rightarrow n - 1 = \dfrac{161}{3} \Rightarrow n = \dfrac{164}{3} = 54.67$. [1.0 Mark]
Since $n$ is not a natural number (positive integer), $-150$ is NOT a term of the given AP. [1.0 Mark]
Marking Scheme:
• Setting $a_n = -150$ and finding $n = 164/3$: 1.0 Mark
• Concluding $-150$ is not a term since $n
otin \mathbb{N}$: 1.0 Mark
Correct Answer:
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