Step 1: Understand the trapezium structure.
We are given $PQRS$ with $PS \parallel QR$. Lines $PQ$ and $SR$ are extended to meet at point $A$ above the trapezium. We want to calculate $\angle PAS$. Step 2: Using Statement I.
- From the figure, $\triangle AQR \sim \triangle APS$ (because $PS \parallel QR$).
- This similarity gives ratio:
\[ \frac{AQ}{AP} = \frac{QR}{PS} = \frac{AR}{AS} = k \tag{1} \] Now, we are given $PQ=3$, $RS=4$, and $\angle QPS=60^\circ$. Let us drop a perpendicular from $Q$ onto $PS$, meeting at $M$. In right $\triangle PQM$: \[ \sin 60^\circ = \frac{QM}{PQ} = \frac{QM}{3} \] \[ QM = 3\cdot \frac{\sqrt{3}}{2} = \frac{3\sqrt{3}}{2} \] Similarly, drop a perpendicular from $R$ onto $PS$, meeting at $N$. Then $RN = QM = \tfrac{3\sqrt{3}}{2}$. Now, in right $\triangle RSN$: \[ \sin \angle RSN = \frac{RN}{RS} = \frac{\tfrac{3\sqrt{3}}{2}}{4} = \frac{3\sqrt{3}}{8} \] Hence, \[ \angle RSN = \sin^{-1}\!\left(\frac{3\sqrt{3}}{8}\right) \] Finally, in $\triangle APS$, \[ \angle PAS = 180^\circ - (\angle APS + \angle PSA) \] But $\angle APS = 120^\circ$ (straight line property at $P$ since $\angle QPS=60^\circ$) and $\angle PSA = \angle RSN$. Thus, \[ \angle PAS = 120^\circ - \sin^{-1}\!\left(\frac{3\sqrt{3}}{8}\right) \] Therefore, Statement I alone gives us a clear value of $\angle PAS$.
Step 3: Using Statement II.
If $PS=10$ and $QR=5$, then the ratio of parallel sides is $2:1$. But without knowing any angle (like $\angle QPS$), the trapezium can scale differently, so $\angle PAS$ cannot be uniquely determined.
Final Decision:
- Statement I alone $\Rightarrow$ sufficient.
- Statement II alone $\Rightarrow$ insufficient.
- Combined use is not required since I alone works.
\[ \boxed{\text{Answer: (A)}} \]
Step 1: Observe the recurrence.
The rule is: \[ A_{n+1} = A_n^2+1 \] This means once the initial term is fixed, the entire sequence is uniquely determined.
Step 2: Using Statement I ($A_0=1$).
Start with $A_0=1$: \[ A_1 = 1^2+1=2, \quad A_2=2^2+1=5, \quad A_3=5^2+1=26, \ldots \] Thus, the sequence is completely determined. With a computer or theoretical methods, $\gcd(A_{900}, A_{1000})$ can be found.
Hence, Statement I alone is sufficient.
Step 3: Using Statement II ($A_1=2$).
From the recurrence: \[ A_1 = A_0^2+1 \] So $2=A_0^2+1 \Rightarrow A_0=1$. Thus Statement II actually implies Statement I, and again the sequence is uniquely fixed. Therefore, Statement II alone is also sufficient.
Step 4: Conclude sufficiency.
Since either statement alone fixes the entire sequence and hence allows computation of $\gcd(A_{900}, A_{1000})$, each statement alone is sufficient. \[ \boxed{\text{Answer: (D)}} \]
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