Mathematics — Standard (041) — CBSE Class 10 board question

Given: $p$th term $= q$ and $q$th term $= p$

Using $a_n = a + (n-1)d$:

$$a_p = a + (p-1)d = q \quad \text{...(1)}$$

$$a_q = a + (q-1)d = p \quad \text{...(2)}$$

Subtracting (2) from (1):

$$(p-1)d - (q-1)d = q - p$$

$$d(p - q) = -(p - q)$$

$$d = -1$$

Substituting $d = -1$ in (1):

$$a + (p-1)(-1) = q$$

$$a = q + p - 1 = p + q - 1$$

nth term:

$$a_n = a + (n-1)d = (p + q - 1) + (n-1)(-1)$$

$$a_n = p + q - 1 - n + 1 = \boxed{p + q - n}$$

Hence proved.

Source: Chapter 5, Section 5.3 (nth term formula $a_n = a + (n-1)d$)

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• Examiners award marks for: setting up two equations (1 mark), finding $d = -1$ by subtraction (1 mark), and correctly arriving at $a_n = p+q-n$ (1 mark).
• The key trick is subtracting the two equations to eliminate $a$ and find $d$ first, then back-substitute to get $a$.
• Write each step clearly; do not skip the substitution step as it carries a mark.