Q1. [1] § 5.5 Summary
If $a_n$ represents $n$th term of the A.P. $-\frac{15}{4}, -\frac{10}{4}, -\frac{5}{4}, \ldots$ then value of $a_{16} - a_{12}$ is
- (A) $4$
- (B) $\frac{5}{4}$
- (C) $5$
- (D) $\frac{25}{4}$
Previously asked in CBSE board exam
2026 30/5/1 Q8
Generated by claude-sonnet-4-6 · 2026-06-15 10:33 · grounding rag
Model Answer
Here, $d = -\frac{10}{4} - (-\frac{15}{4}) = \frac{5}{4}$. So $a_{16} - a_{12} = (16-12) \times d = 4 \times \frac{5}{4} = \mathbf{5}$.
(C) 5
Explanation
Since $a_n - a_m = (n-m)d$ for any AP, you only need the common difference and the difference in term numbers. No need to calculate individual terms separately. Common difference here is $\frac{5}{4}$, and $16 - 12 = 4$, giving $4 \times \frac{5}{4} = 5$.
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