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Question
If the 9th term of an AP is zero, prove that its 29th term is twice its 19th term.
Solution
Let the first term, common difference and number of terms of an AP are a, d and n respectively.
Given that,
9th term of an AP,
T9 = 0 ...[∵ nth term of an AP, Tn = a + (n – 1)d]
⇒ a + (9 – 1)d = 0
⇒ a + 8d = 0
⇒ a = – 8d ...(i)
Now, its 19th term,
T19 = a + (19 – 1)d
= – 8d + 18d ...[From equation (i)]
= 10d ...(ii)
And its 29th term,
T29 = a + (29 – 1)d
= – 8d + 28d ...[From equation (i)]
= 2 × (10d)
= 20d
⇒ T29 = 2 × T19
Hence, its 29th term is twice its 19th term.
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