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Question
The 4th term of an AP is zero. Prove that its 25th term is triple its 11th term.
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Solution
In the given AP, let the first be a and the common difference be d.
Then, tn = a + (n-1)d
Now , T4 = a + (4-1) d
⇒ a +3d = 0 ....................(1)
⇒ a = -3d
Again T11 = a + (11-1) d = a + 10 d
= -3d + 10d = 7d [ Using (1)]
Also , T25 = a + ( 25-1) d = a + 24d = - 3d + 24 d = 21d [Using (1)]
i.e . , T25 = 3 × 7d = ( 3 × T 11)
Hence, 25th term is triple its 11th term.
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