Question

If the pth term of an AP is q and the qth term is p, prove that its nth term is (p+q-n).

Answer 1

Hint: Use the general mth term for an AP ${{\text{t}}_{\text{m}}}\text{ = a + (m }-\text{ 1)d}$, where ‘a’ is the first term of the AP and ‘d’ is its common difference, to formulate the two conditions given in the question to form two equations. Now use this to prove ${{\text{t}}_{\text{n}}}\text{ = (p + q }-\text{ n)}$.

Complete step-by-step answer:
We know the general mth term for an AP is ${{\text{t}}_{\text{m}}}\text{ = a + (m }-\text{ 1)d}$. Here, a is the first term of the AP and d is its common difference. Now, it is given in the question that the pth term of an AP is q. So, putting m = p in the general term, we get,
$\begin{align}
  & {{\text{t}}_{\text{p}}}\text{ = q} \\
 & \therefore \text{ a + (p }-\text{ 1)d = q }....\text{(i)} \\
\end{align}$
Again, it is given that the qth term of the AP is p. So putting m = q in the general term, we get,
 $\begin{align}
  & {{\text{t}}_{\text{q}}}\text{ = p} \\
 & \therefore \text{ a + (q }-\text{ 1)d = p }....\text{(ii)} \\
\end{align}$
Now, subtracting equation (i) from (ii), we get,
\[\begin{align}
  & \text{a + (p }-\text{ 1)d }-\text{ a }-\text{ (q }-\text{ 1)d = q }-\text{ p} \\
 & \Rightarrow \text{ }\left( \text{p }-\text{ q} \right)\text{d = q }-\text{ p} \\
 & \therefore \text{ d = }-1 \\
\end{align}\]
Thus, we obtain the value of the common difference to be –1.
Now, putting the value of common difference d = –1 in equation (i), we get,
$\begin{align}
  & \text{a + }\left( \text{p }-\text{ 1} \right)\cdot \left( -1 \right)\text{ = q} \\
 & \Rightarrow \text{ a }-\text{ p + 1 = q} \\
 & \therefore \text{ a = p + q }-\text{ 1} \\
\end{align}$
Hence, the first term of this arithmetic progression is (p + q – 1).
Thus, in order to get the nth term of the AP, we simply need to put the values of the first term and common difference of the AP in the general term. Thus, in the formula for the general mth term ${{\text{t}}_{\text{m}}}\text{ = a + (m }-\text{ 1)d}$, putting m = n, a = (p + q – 1) and d = –1, we get,
$\begin{align}
  & {{\text{t}}_{\text{n}}}\text{ = a + }\left( \text{n }-\text{ 1} \right)\text{d} \\
 & \text{ = }\left( \text{p + q }-\text{ 1} \right)\text{ + }\left( \text{n }-\text{ 1} \right)\cdot \left( -1 \right) \\
 & \text{ = p + q }-\text{ 1 }-\text{ n + 1} \\
 & \text{ = p + q }-\text{ n } \\
\end{align}$
Thus, it is proved that the nth term of the arithmetic progression is (p + q – n).

Note: In this solution, two simultaneous equations are formed where the variables are ‘a’ and ‘d’, which are solved explicitly. Without finding the values of ‘a’ and ‘d’, it is impossible to prove that the nth term of the AP is (p + q – n).