Question

If the $3^{rd}$ and the $9^{th}$ terms of an A.P are $4$ and $-8$ respectively. Which term of the A.P is zero?

Answer 1

Step 1. Find the equations

For given A.P.

Let the first term $=a$

Common difference $=d$

Number of terms $=n$

$\therefore n^{th}$ term of an A.P. $\left(a_n\right)$ $=a+(n-1)d$

The $3^{rd}$ term $=a+(3-1)d$

$=a+2d$

The $9^{th}$ term $=a+(9-1)d$

$=a+8d$

The given condition is, the $3^{rd}$ term is $4$

$\therefore a+2d=4...............................\left(1\right)$

Another given condition, is the $9^{th}$ term is $-8$.

$\therefore a+8d=-8.....\left(2\right)$

Step 2. Find the first term and common difference

Subtracting equation $\left(2\right)$ from equation $\left(1\right)$ we get

$$\begin{gathered} a+2d=4 \\ \underline{-a-8d=8} \\ -6d=12 \\ \Rightarrow d=-2 \end{gathered}$$

Therefore the common difference is $-2$

Putting the value of $d$ in equation $\left(1\right)$ we get

$$\begin{gathered} a+2d=4 \\ \Rightarrow a+2·\left(-2\right)=4 \\ \Rightarrow a-4=4 \\ \Rightarrow a=8 \end{gathered}$$

Therefore the first term is $8$

Step 3. Find the term that is zero

Let the $n^{th}$ term be $0$

$$\begin{gathered} a_n=a+\left(n-1\right)d \\ \Rightarrow 0=a+\left(n-1\right)d\left[\because a_n=0\right] \\ \Rightarrow 0=8+\left(n-1\right)\left(-2\right) \\ \Rightarrow 2\left(n-1\right)=8 \\ \Rightarrow n-1=4 \\ \Rightarrow n=5 \end{gathered}$$

Hence, the $5^{th}$ term of the A.P is $0$