Question

If the 9^th term of an A.P. is zero then show that the 29^th term is twice the 19^th term.

Answer 1

It is given that,
a₉ = 0

Now,
$$\begin{gathered} a_n=a+\left(n-1\right)d \\ {a}_9=a+\left(9-1\right)d \\ \Rightarrow 0=a+8d \\ \Rightarrow a=-8d...\left(1\right) \\ {a}_{29}=a+\left(29-1\right)d \\ =-8d+28d \\ =20d...\left(2\right) \\ {a}_{19}=a+\left(19-1\right)d \\ =-8d+18d \\ =10d...\left(3\right) \\ \mathrm{From}\left(2\right)\mathrm{and}\left(3\right),\mathrm{we}\mathrm{get} \\ {a}_{29}=2a_{19} \end{gathered}$$

Hence, the 29^th term is twice the 19^th term.