Question

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

Answer 1

In the given problem, the 9^th 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,

We get,

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

For 19^th term (n = 19),

(Using 1)

For 29^th term (n = 29),

$$\begin{gathered} a_{29}=a+\left(29-1\right)d \\ =-8d+28d \\ =20d \\ =2\times 10d \\ =2\times a_{19} \end{gathered}$$ (Using 1)

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

Hence proved.