Question

In an AP, the $p^{th}$ term is$q$and ${(p+q)}^{th}$ term is $0$. Show that its $q^{th}$ term is$p$.

Answer 1

Find the $q^{th}$ term of the AP

Given, $p^{th}$ term is $q$and ${(p+q)}^{th}$ term is $0$.

Let, First term$=a$ and Common difference of the given A.P.$=d$

$\therefore a+(p-1)d=q......\left(1\right)$

and $a+(p+q-1)d=0$

$$\begin{gathered} \Rightarrow a+(p-1)d+qd=0 \\ \Rightarrow q+qd=0\left[\because a+(p-1)d=q\right] \\ \Rightarrow qd=-q \\ \Rightarrow d=-1 \end{gathered}$$

Now, from equation. $\left(1\right)$

$$\begin{gathered} \Rightarrow a+(p-1)d=q \\ \Rightarrow a+(p-1)\times \left(-1\right)=q\left[\because d=-1\right] \\ \Rightarrow a-p+1=q \\ \Rightarrow a=p+q-1 \end{gathered}$$

$\therefore q^{th}$ term $=a+(q-1)d$

$$\begin{gathered} =p+q-1+(q-1)\times (-1)\left[a=p+q-1\mathrm{and}d=-1\right] \\ =p+q-1-q+1 \\ =p \end{gathered}$$

Hence, the $q^{th}$ term of the given AP is$p$.