If the $m^{\text{th}}$ term of an AP is $\dfrac{1}{n}$ and the $n^{\text{th}}$ term is $\dfrac{1}{m}$, show that its $(mn)^{\text{th}}$ term is $1$.
Step-by-Step Solution
Key Concept: Subtract $a_m$ and $a_n$ to find $d = \dfrac{1}{mn}$, then show $a = \dfrac{1}{mn}$ and compute $a_{mn} = a + (mn - 1)d$.
$a_m = a + (m-1)d = \dfrac{1}{n}$ -- (eq 1)
$a_n = a + (n-1)d = \dfrac{1}{m}$ -- (eq 2). [0.5 Mark]
Subtracting eq 2 from eq 1: $(m - n)d = \dfrac{1}{n} - \dfrac{1}{m} = \dfrac{m - n}{mn} \Rightarrow d = \dfrac{1}{mn}$. [1.0 Mark]
Substitute $d = \dfrac{1}{mn}$ into eq 1: $a + (m-1)\left(\dfrac{1}{mn}\right) = \dfrac{1}{n} \Rightarrow a + \dfrac{1}{n} - \dfrac{1}{mn} = \dfrac{1}{n} \Rightarrow a = \dfrac{1}{mn}$. [1.0 Mark]
$a_{mn} = a + (mn - 1)d = \dfrac{1}{mn} + (mn - 1)\left(\dfrac{1}{mn}\right) = \dfrac{1 + mn - 1}{mn} = \dfrac{mn}{mn} = 1$. Proved! [0.5 Mark]
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🎯 Official CBSE Marking Scheme:
Writing equations for $a_m$ and $a_n$: 0.5 Mark
Finding common difference $d = 1/(mn)$: 1.0 Mark
Finding first term $a = 1/(mn)$: 1.0 Mark
Evaluating $a_{mn} = 1$: 0.5 Mark
Correct Answer: