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Question
If the 9th term of an A.P. is zero then show that the 29th term is twice the 19th term?
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Solution
It is given that,
t9 = 0
Now,
tn = a + (n - 1)d
t9 = a + (9 - 1)d
⇒ 0 = a + 8d
⇒ a = -8d ...(1)
t29 = a + (29 - 1)d
= -8d + 28d
= 20d ...(2)
t19 = a + (19 - 1)d
= -8d + 18d
= 10d ...(3)
From (2) and (3), we get
t29 = 2t19
Hence, the 29th term is twice the 19th term.
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