Question

If the ${p}^{\text{th}}$ term of an AP is q and the ${q}^{\text{th}}$ term is p, prove that its ${n}^{\text{th}}$ term is (p+q-n).

Answer 1

Hint:
Use the general ${m}^{\text{th}}$ term for an AP
${t_m} = a + (m - 1)d$
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,
${t_n} = (p + q - n)$

Complete step by step solution:
We know the general ${m}^{\text{th}}$ term for AP
 ${t_m} = a + (m - 1)d$

Here, a is the first term of the AP and d is its common difference. Now, it is given in the question that the ${p}^{\text{th}}$ term of an AP is q. So, putting m=p in the general term, we get,
 $\
  {t_p} = q \\
  \therefore a + (p - 1)d = q......(1) \\
\ $

Again, the ${q}^{\text{th}}$ term of the AP is p.
So putting m=q in the general term, we get,
 $\
  {t_q} = p \\
  \therefore a + (q - 1)d = p.......(2) \\
\ $

Now, subtracting equation (1) and (2), we get,
 $\
  a + (p - 1)d - a - (q - 1)d = q - p \\
   \Rightarrow (p - q)d = q - p \\
  \therefore d = - 1 \\
\ $

Thus, we obtain the value of the common difference to be -1.
Now,putting the value of common difference d=-1 in equation(1), we get,
 $\
  a + (p - 1)( - 1) = q \\
   \Rightarrow a - p + 1 = q \\
  \therefore a = p + q - 1 \\
\ $

Hence, the first term of the arithmetic progression is $\left( {p + q - 1} \right)$.
Thus, in order to get the ${n}^{\text{th}}$ term of the AP, we simply need to put the value of the first term and common difference of the AP in the general term.
Thus, in the formula for the general ${m}^{\text{th}}$ term

Putting $m = n,a = (p + q - 1)$
And $d = - 1$
 $\
  {t_n} = a + (n - 1)d \\
   = (p + q - 1) + (n - 1)( - 1) \\
   = p + q - 1 - n + 1 \\
   = p + q - n \\
\ $

Thus, it is proved that the ${n}^{\text{th}}$ term of the AP is $(p + q - 1)$.

Note:
Two simultaneous equations are formed where the variables are ‘a’ and ‘d’ which are solved explicitly. Without finding the value of ‘a’ and ‘d’, it is impossible to prove that the ${n}^{\text{th}}$ term of the AP is $(p + q - 1)$.