Reason (R): The ends of string are nodes, so the number of nodes is one more than the number of antinodes.
Solution:
Assertion (A) is true. For a stretched string fixed at both ends, the \(n^{\text{th}}\) normal mode indeed has \(n\) antinodes and \(n+1\) nodes.
Reason (R) is true and correctly explains this: the fixed ends are always nodes, leading to one more node than antinode.
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