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Question: If the \({p}^{\text{th}}\) term of an AP is q and the \({q}^{\text{th}}\) term is p, prove that its ...

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

Explanation

Solution

Use the general mth{m}^{\text{th}} term for an AP
tm=a+(m−1)d{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,
tn=(p+q−n){t_n} = (p + q - n)

Complete step by step solution:
We know the general mth{m}^{\text{th}} term for AP
tm=a+(m−1)d{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 pth{p}^{\text{th}} term of an AP is q. So, putting m=p in the general term, we get,
 tp=q ∴a+(p−1)d=q......(1)  \ {t_p} = q \\\ \therefore a + (p - 1)d = q......(1) \\\ \

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

Hence, the first term of the arithmetic progression is (p+q−1)\left( {p + q - 1} \right).
Thus, in order to get the nth{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 mth{m}^{\text{th}} term

Putting m=n,a=(p+q−1)m = n,a = (p + q - 1)
And d=−1d = - 1
 tn=a+(n−1)d =(p+q−1)+(n−1)(−1) =p+q−1−n+1 =p+q−n  \ {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 nth{n}^{\text{th}} term of the AP is (p+q−1)(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 nth{n}^{\text{th}} term of the AP is (p+q−1)(p + q - 1).