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


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


In the given problem, the 9th term of an A.P. is zero.

Here, let us take the first term of the A.P as a and the common difference as d

So, as we know,

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

We get,

$a_{9}=a+(9-1) d$

$0=a+8 d$

$a=-8 d$ .......(1)

Now, we need to prove that $29^{\text {th }}$ term is double of $19^{\text {th }}$ term. So, let us first find the two terms.


For $19^{\text {th }}$ term $(n=19)$,

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

$=-8 d+18 d \quad($ Using 1 $)$

$=10 \mathrm{~d}$

For 29th term (n = 29),

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

$=-8 d+28 d$

$=20 d$

$=2 \times 10 d$


$=2 \times a_{19}$

Therefore, for the given A.P. the 29th term is double of the 19th term.

Hence proved.

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