Question

If the ${{19}^{th}}$ term of a non-zero A.P is zero, then its \[\left( {{49}^{th}}\text{ }term\text{ }:\text{ }{{29}^{th}}\text{ }term \right)\] is
(A) 3:1
(B) 4:1
(C) 2:1
(D) 1:3

Answer 1

Hint: We solve this problem by first equating the ${{19}^{th}}$ term to zero and using the formula for the ${{n}^{th}}$ term of an A.P, ${{a}_{n}}=a+\left( n-1 \right)d$. Then we get a relation between $a$ and $d$. Then we use the same formula and find the values of ${{49}^{th}}$ term and ${{29}^{th}}$ term and then use the relation obtained between $a$ and $d$ to find their values in a single variable. Then we take the ratio of ${{49}^{th}}$ term and ${{29}^{th}}$ term and substitute their values to find the required value.

Complete step-by-step solution:
First let us consider the formula for the ${{n}^{th}}$ term of an A.P with the first term $a$ and with common difference $d$.
${{a}_{n}}=a+\left( n-1 \right)d$
We are given that the ${{19}^{th}}$ term of an A.P is zero, which is ${{a}_{19}}=0$.
Let us assume that the first term of given A.P is $a$ and the common difference of the A.P be $d$.
Now, let us apply the formula for the ${{n}^{th}}$ term of an A.P for the ${{19}^{th}}$ term.
$\begin{align}
  & \Rightarrow {{a}_{19}}=a+\left( 19-1 \right)d \\
 & \Rightarrow {{a}_{19}}=a+18d \\
\end{align}$
Now, as the ${{19}^{th}}$ term of an A.P is zero, let us equate the above value to zero. Then we get,
$\begin{align}
  & \Rightarrow {{a}_{19}}=a+18d=0 \\
 & \Rightarrow a=-18d...........\left( 1 \right) \\
\end{align}$
We need to find the ratio of ${{49}^{th}}$ term and ${{29}^{th}}$ term.
First let us find the value of ${{49}^{th}}$ term.
Using the formula for the ${{n}^{th}}$ term of an A.P for the ${{49}^{th}}$ term, we can write it as,
\[\begin{align}
  & \Rightarrow {{a}_{49}}=a+\left( 49-1 \right)d \\
 & \Rightarrow {{a}_{49}}=a+48d \\
\end{align}\]
Now, let us substitute the value of $a$ obtained above in the equation (1) here. Then we get,
\[\begin{align}
  & \Rightarrow {{a}_{49}}=\left( -18d \right)+48d \\
 & \Rightarrow {{a}_{49}}=30d..............\left( 2 \right) \\
\end{align}\]
Now let us consider the ${{29}^{th}}$ term.
Using the formula for the ${{n}^{th}}$ term of an A.P for the ${{29}^{th}}$ term, we can write it as,
\[\begin{align}
  & \Rightarrow {{a}_{29}}=a+\left( 29-1 \right)d \\
 & \Rightarrow {{a}_{29}}=a+28d \\
\end{align}\]
Now, let us substitute the value of $a$ obtained above in the equation (1) here. Then we get,
\[\begin{align}
  & \Rightarrow {{a}_{29}}=\left( -18d \right)+28d \\
 & \Rightarrow {{a}_{29}}=10d.............\left( 3 \right) \\
\end{align}\]
As we need to find the ratio of ${{49}^{th}}$ term and ${{29}^{th}}$ term, let us divide them
$\Rightarrow \dfrac{{{a}_{49}}}{{{a}_{29}}}$
Now let us substitute the values of those terms from equation (2) and equation (3). Then we get,
$\begin{align}
  & \Rightarrow \dfrac{{{a}_{49}}}{{{a}_{29}}}=\dfrac{30d}{10d} \\
 & \Rightarrow \dfrac{{{a}_{49}}}{{{a}_{29}}}=\dfrac{3}{1} \\
\end{align}$
So, ratio of ${{49}^{th}}$ term and ${{29}^{th}}$ term is 3:1. Hence answer is Option A.

Note: There is a possibility of one making a mistake while solving this problem by taking the formula for the ${{n}^{th}}$ term as ${{a}_{n}}=\dfrac{n}{2}\left( 2a+\left( n-1 \right)d \right)$. But it is the formula for the sum of first $n$ terms of an A.P not for the ${{n}^{th}}$ term of the A.P.