Q. 74.7( 8 Votes )

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

Answer :

Here, t_{9} = 0

We have to prove that t_{29} = 2t_{19}

__We know that nth term of A.P. , t _{n} = a + (n – 1) d.__

⇒ t_{9} = a + (9 – 1) d = a + 8d

But t_{9} = 0

⇒ a + 8d = 0

∴ a = -8d … (i)

Now, t_{29} = a + (29 – 1) d

= a + 28d

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

= 20d

∴ t_{29} = 20d … (1)

Now, t_{19} = a + (19 – 1) d

= a + 18d

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

= 10d

∴ t_{19} = 10d … (2)

Equating (1) and (2),

⇒ 20d = 10d

⇒ 20d = 10d (2)

∴ t_{29} = 2(t_{19})

Hence proved.

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