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If 9th term of an A.P is zero, prove that its 29th term is double the 19th term?

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Let 'a' be the first term and 'd' be the common difference of the given A P. If Tn stands for nth term of this A P, then we get T9 = a + 8d= 0 ( Given). T19 = a + 18 d = ( a + 8 d) + 10 d = T9 + 10 d = 0 +10 d = 10d T29 = a + 28 d = (a + 8 d) +20 d = T9 + 20 d = 0 + 20 d = 20d =2 *10 d= 2 * T19.
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