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

Given:

9th term of an A.P. is zero.

To do:

We have to prove that 29th term of the given A.P. is double the 19th term.
Solution:

Let the required A.P. be $a, a+d, a+2d, ......$

Here,

$a_1=a, a_2=a+d$ and Common difference $=a_2-a_1=a+d-a=d$

We know that,

$a_n=a+(n-1)d$

Therefore,

$a_9=a+(9-1)d$

$0=a+8d$

$a=-8d$.....(i)

$a_{19}=a+(19-1)d$

$=-8d+18d$    (From(i))

$=10d$....(ii)

$a_{29}=a+(29-1)d$

$=-8d+28d$    (From(i))

$=20d$....(iii)

$=2(10d)$

$=2(a_{19})$    (From (ii))

This implies,

$a_{29}=2\times a_{19}$

29th term is double the 19th term.

Hence proved.

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Updated on: 10-Oct-2022

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