Question

If the ${9^{th}}$ term of an A.P. is zero, then prove that ${29^{th}}$ term is double of ${19^{th}}$ term.

Answer 1

We know that the first statement is true. So our ultimate goal must be to bring ${9^{th}}$ term as zero by solving ${29^{th}}$ term which is double of ${19^{th}}$ term.

Complete step by step answer:
We know that the general term of an AP is given by ${A_n} = a + (n - 1)d$ where n is for the ${n^{th}}$ term d is for the common difference and a is for the first term.
So now we know that

{\therefore {A_9} = a + (9 - 1)d}\\\ { \Rightarrow {A_9} = a + 8d} \end{array}$$ Let us assume that $${A_{29}} = 2{A_{19}}$$ So let us use this and see what possible relation we can get in a and d $$\begin{array}{l} {A_{29}} = 2{A_{19}}\\\ \Rightarrow a + (29 - 1)d = 2[a + (19 - 1)d]\\\ \Rightarrow a + 28d = 2(a + 18d)\\\ \Rightarrow a + 28d = 2a + 36d\\\ \Rightarrow 2a - a + 36d - 28d = 0\\\ \Rightarrow a + 8d = 0 \end{array}$$ Clearly $${{A_9} = a + 8d = 0}$$ which is true $$\therefore {A_{29}} = 2{A_{19}}$$ is proved. **Note:** We can also do this question by substituting either d or a in $$a + 28d = 2(a + 18d)$$ this equation, but it will make the solution too much lengthy, still we will get the same result, i.e., $${A_9} = 0$$