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

It is given that the $9^{th}$ term of an A.P is $T_9=0$

We know that the general term of an arithmetic progression with first term $a$ and common difference $d$ is $T_n=a+(n-1)d$, therefore, the $9^{th}$, ${19}^{th}$ and ${29}^{th}$ terms are as follows:

$T_9=a+(9-1)d=a+8d.......\left(1\right)$

$T_{19}=a+(19-1)d=a+18d......\left(2\right)$

$T_{29}=a+(29-1)d=a+28d.......\left(3\right)$

Now since $T_9=0$, therefore, equation 1 becomes

$0=a+8d$

$\Rightarrow a=-8d........\left(4\right)$

Substitute the value of equation (4) in equation (3):

$T_{29}=-8d+28d=20d=2\left(10d\right)=2(-8d+18d)=2(a+18d)=2\left[T_{19}\right]$ (Using equations 1 and 2)

Hence, the $29$th term of A.P is twice the $19$ term.