Question

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

Answer 1

Hint: We know that a sequence of numbers in which difference between any two successive numbers is a constant is called the arithmetic progression (A.P.). For example, the series $ 2,4,6,8,10,..... $ is an arithmetic progression. We know that the general term of an arithmetic progression is expressed as a relation between the first term and common difference. We will use this relation to solve this question.
Formula used:
 $ {T_n} = a + \left( {n - 1} \right)d $ , where. $ {T_n} $ is the $ {n^{th}} $ term of an arithmetic progression, $ a $ is the first term of an arithmetic progression and $ d $ is the common difference

Complete step-by-step answer:
Here, we are given that the ninth term of an A.P. is zero.
By using the formula $ {T_n} = a + \left( {n - 1} \right)d $ , the ninth term can be given as:
 $
  {T_9} = a + \left( {9 - 1} \right)d \\
   \Rightarrow {T_9} = a + 8d \;
  $
But, the ninth term of an A.P. is zero. From this, we will get the relation between the first term $ a $ and the common difference $ d $ .
 $
   \Rightarrow 0 = a + 8d \\
   \Rightarrow a = - 8d \;
  $
Now, the $ {19^{th}} $ term of the arithmetic progression is given by:
  $
  {T_{19}} = a + \left( {19 - 1} \right)d \\
   \Rightarrow {T_{19}} = a + 18d \;
  $
But we have determined from the given information that $ a = - 8d $ . Putting this value in the equation of the $ {19^{th}} $ term, we get
 $
   \Rightarrow {T_{19}} = - 8d + 18d \\
   \Rightarrow {T_{19}} = 10d \;
  $
The $ {29^{th}} $ term of the arithmetic progression is given by:
 $
  {T_{29}} = a + \left( {29 - 1} \right)d \\
   \Rightarrow {T_{29}} = a + 28d \;
  $
Putting the value $ a = - 8d $ in the equation of the $ {29^{th}} $ term, we get
 $
   \Rightarrow {T_{29}} = - 8d + 28d \\
   \Rightarrow {T_{29}} = 20d \\
   \Rightarrow {T_{29}} = 2\left( {10d} \right) \;
  $
But, we know that $ 10d $ is the $ {19^{th}} $ term of the arithmetic progression.
 $ {T_{29}} = 2{T_{19}} $
Hence, it is proved that if the ninth term of an A.P. is zero, prove that its $ {29^{th}} $ term is double the $ {19^{th}} $ term.
So, the correct answer is “ $ {T_{29}} = 2{T_{19}} $ ”.

Note: In this type of question, the important thing is that first we have to determine the relation between the first term and the common difference off the arithmetic progression. This is done by the given information about a certain term of the A.P. which is about the ninth term in this equation. After that we simply put this value in the formula of the two terms between which we are asked to find a relationship.