If the $m^{\text{th}}$ term of an A.P. is $\dfrac{1}{n}$ and the $n^{\text{th}}$ term is $\dfrac{1}{m}$, show that the sum of first $mn$ terms is $\dfrac{1}{2}(mn + 1)$.
Step-by-Step Solution
Key Concept: $a + (m-1)d = 1/n$ and $a + (n-1)d = 1/m$. Subtract: $(m-n)d = (m-n)/(mn) \Rightarrow d = 1/(mn) \Rightarrow a = 1/(mn)$. $S_{mn} = \dfrac{mn}{2}\left[2\left(\dfrac{1}{mn}\right) + (mn-1)\dfrac{1}{mn}\right] = \dfrac{1}{2}[2 + mn - 1] = \dfrac{1}{2}(mn + 1)$.
Subtract equations: $(m-n)d = \dfrac{m-n}{mn} \Rightarrow d = \dfrac{1}{mn}$. [1.0 Mark]
Substitute $d$: $a + \dfrac{m-1}{mn} = \dfrac{1}{n} \Rightarrow a = \dfrac{1}{mn}$. [1.0 Mark]
$S_{mn} = \dfrac{mn}{2}\left[\dfrac{2}{mn} + \dfrac{mn-1}{mn}\right] = \dfrac{1}{2}(mn + 1)$. Proved! [1.0 Mark]
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🎯 Official CBSE Marking Scheme:
Solving $d = 1/(mn)$: 1.0 Mark
Solving $a = 1/(mn)$: 1.0 Mark
Evaluating $S_{mn} = (mn + 1)/2$: 1.0 Mark
Correct Answer: