if 9th term of an ap is zero prove that its 29 term is double the 19th term
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In A.P the first term = a and common difference = d.
Given that 9th term of an A.P. is 0.
Therefore t9 = 0
⇒ a + 8d = 0 ⇒ a = -8d --------------(1)
We have to prove that t29 = 2 t19.
t19 = a + 18d = -8d + 18d = 10d [ from (1) ]
t29 = a + 28d = -8d + 28d = 20d [ from (1) ]
t29 = 2 x 10d = 2 x t19
∴ t29 = 2 x t19 .
raghavsharma62:
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