# If 9th term of an A.P. is zero, prove that its 29th term is double (twice) the 19th term.

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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