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

Hint: 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
\[\begin{array}{*{20}{l}}
{\therefore {A_9} = a + (9 - 1)d}\\
{ \Rightarrow {A_9} = a + 8d}
\end{array}\]
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., \[{A_9} = 0\]