# If 100 times the 100th term of an AP with non-zero common difference equal to the 50 times its 50th term, then the 150th term of AP is

## 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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