Question

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

Answer 1

Let the First term of $A.P=a$

Ans the difference $=d$

Given,

$T_{9^{th}}=0$

$a+(9-4)d=0$

$a+8d=0.....\left(1\right)$

We have to prove that

$T_{{29}^{th}}=2T_{{19}^{th}}$

Now, $T_{{19}^{th}}=a+(19-1)d$

$=a+18d$

$=-8d+18d$ (From eq $1$)

$=10d......\left(2\right)$

$T_{{29}^{th}}=a(29-1)d$

$=a+28d$

$=-8d+28d$

$=20d.......\left(3\right)$

From eq $2$ and eq $3$

$T_{{29}^{th}}=2T_{{19}^{th}}$