If the sum of the first $p$ terms of an AP is equal to the sum of the first $q$ terms, then show that the sum of its first $(p + q)$ terms is $0$.
Step-by-Step Solution
Key Concept: Set $S_p = S_q \Rightarrow \dfrac{p}{2}[2a + (p-1)d] = \dfrac{q}{2}[2a + (q-1)d]$, rearrange to get $2a + (p+q-1)d = 0$.
Given $S_p = S_q \Rightarrow p[2a + (p-1)d] = q[2a + (q-1)d]$. [0.5 Mark]
$2ap + p(p-1)d = 2aq + q(q-1)d \Rightarrow 2a(p - q) + [p^2 - p - (q^2 - q)]d = 0$. [1.0 Mark]
$2a(p - q) + [(p^2 - q^2) - (p - q)]d = 0 \Rightarrow (p - q)[2a + (p + q - 1)d] = 0$. [1.0 Mark]
Since $p
eq q$, $2a + (p + q - 1)d = 0$.
Now $S_{p+q} = \dfrac{p+q}{2}[2a + (p+q-1)d] = \dfrac{p+q}{2}(0) = 0$. Proved! [0.5 Mark]
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🎯 Official CBSE Marking Scheme:
Expanding $S_p = S_q$: 0.5 Mark
Grouping and factoring out $(p - q)$: 1.0 Mark
Deducing $2a + (p+q-1)d = 0$: 1.0 Mark
Evaluating $S_{p+q} = 0$: 0.5 Mark
Correct Answer: