- AA is true, R is true, and R explains A.
- BA is false, but R is true.
- CA is true, but R is false.
- DBoth A and R are false.
Step-by-Step Solution
1, 4, 9, 16 has differences 3, 5, 7 — not constant — so it is NOT an AP. Assertion is false. The Reason correctly states the definition of an AP, so R is true.
Answer: B) A is false, but R is true.
Key idea: Recognising an AP
A sequence is an AP iff the difference of consecutive terms is constant.
**Test**: Compute a₂ − a₁ and a₃ − a₂. If they're equal, it's likely an AP — verify with a₄ − a₃.
**Examples**:
- 4, 7, 10, 13: differences 3, 3, 3 → AP, d = 3.
- 1, 4, 9, 16: differences 3, 5, 7 → NOT an AP (these are squares).
- 5, 5, 5, 5: differences 0, 0, 0 → AP with d = 0 (a constant sequence is an AP).
- 2, 6, 12, 20: differences 4, 6, 8 → NOT an AP.
The common difference d can be positive (increasing AP), negative (decreasing AP), or zero (constant AP).