Paste your question to our Mathbee AI Mentor below to get an instant step-by-step solution.
Arithmetic Progressions
NCERT Exemplar
CBSE
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.
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]
--- 🎯 Official CBSE 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:
Mathbee AI Mentor (Free Demo)
Confused by the solution? Ask the AI to explain a specific step, tell you where you went wrong, or break down the key trap in this question.
Master Arithmetic Progressions with Mathbee
Practice this topic under real exam conditions with strict timers, or ask our AI Mentor to explain the concepts step-by-step.