## Solution :

Let a be the first term and d (d \(\ne\) 0) be the common difference of the given AP, then

\(T_{100}\) = a + (100 – 1)d = a + 99d

\(T_{50}\) = a + (50 – 1)d = a + 49d

\(T_{150}\) = a + (150 – 1)d = a + 149d

Since 100 times the 100th term = 50 times its 50th term

\(\implies\) 100(a + 99d) = 50(a + 49d)

\(\implies\) 2a + 198d = a + 49d

\(\implies\) a + 149d = 0

\(\therefore\) \(T_{150}\) = 0

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