Answer :

Here, t9 = 0


We have to prove that t29 = 2t19


We know that nth term of A.P. , tn = a + (n – 1) d.


t9 = a + (9 – 1) d = a + 8d


But t9 = 0


a + 8d = 0


a = -8d … (i)


Now, t29 = a + (29 – 1) d


= a + 28d


= -8d + 28d [From (i)]


= 20d


t29 = 20d … (1)


Now, t19 = a + (19 – 1) d


= a + 18d


= -8d + 18d [From (i)]


= 10d


t19 = 10d … (2)


Equating (1) and (2),


20d = 10d


20d = 10d (2)


t29 = 2(t19)


Hence proved.


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