Question

Which term of the A.P. $100,90,80,...$ is Zero?
A. ${{5}^{th}}$
B. ${{6}^{th}}$
C. ${{10}^{th}}$
D. ${{11}^{th}}$

Answer 1

Hint: First we will write the given series in the form of $a,a+d,a+2d,...$ to know the values of first term$\left( a \right)$ and the value of common difference$\left( d \right)$ of the series. Now we will assume that ${{m}^{th}}$ term of A.P is zero. So, we have the value of ${{n}^{th}}$ term in A.P. as ${{a}_{n}}=a+\left( n-1 \right)d$. From this formula we will calculate the value of ${{m}^{th}}$ term and equate it to given value zero to get the value of $m$.

Complete step-by-step solution
Given that, $100,90,80,...$ are in A.P.
Now we are going to write a relation between the first and second term of the given series as
$\begin{align}
  & 100=90+10 \\
 & \Rightarrow 90=100-10 \\
\end{align}$
And also, we can write the relation between the second and third term of the given series as
$\begin{align}
  & 90=80+10 \\
 & \Rightarrow 80=90-10 \\
\end{align}$
From these two relations we can write a relation between the first and third term of the given series as
$\begin{align}
  & 80=90-10 \\
 & \Rightarrow 80=\left( 100-10 \right)-10 \\
 & \Rightarrow 80=100-10-10 \\
 & \Rightarrow 80=100-20 \\
\end{align}$
Now the given series is written as
$\begin{align}
  & 100,90,80,... \\
 & \Rightarrow 100,100-10,100-20,... \\
\end{align}$
To find the value of first term and common difference we are going to write the above series as $a,a+d,a+2d,...$, then
$\begin{align}
  & 100,100-10,100-20,... \\
 & \Rightarrow 100,100+\left( -10 \right),100+2\left( -10 \right),... \\
\end{align}$
Comparing the above series with $a,a+d,a+2d,...$, the values of $a,d$ are given by
$a=100$, $d=-10$.
Let ${{m}^{th}}$ term of the given series be Zero. We know the value of ${{n}^{th}}$ in A.P. as ${{a}_{n}}=a+\left( n-1 \right)d$, from this the value of ${{m}^{th}}$ term is given by
${{a}_{m}}=a+\left( m-1 \right)d$
Here we have values ${{a}_{m}}=0$, $a=100$, $d=-10$ and substituting these values in the above equation then we will get
$\begin{align}
  & 0=100+\left( m-1 \right)\left( -10 \right) \\
 & \Rightarrow -100=-10\left( m-1 \right) \\
 & \Rightarrow 10=m-1 \\
 & \Rightarrow m=10+1 \\
 & \Rightarrow m=11 \\
\end{align}$
$\therefore $ ${{11}^{th}}$ term of A.P. $100,90,80,...$ is Zero.

Note: Students may be confused at the value of $d$ for this kind of problem. We have a general form of A.P as $ a, a+d,a+2d,...$ but we got the series as $100,100-10,100-20,...$, so students may stop their process by thinking that something is wrong with the problem. But you have to remember that the value $d$ may be negative, so modify the given series according to our convenience.