If $m$ times the $m^{\text{th}}$ term of an A.P. is equal to $n$ times its $n^{\text{th}}$ term, show that the $(m + n)^{\text{th}}$ term of the A.P. is zero.
Step-by-Step Solution
Key Concept: $m [a + (m-1)d] = n [a + (n-1)d] \Rightarrow (m-n)a + [m(m-1) - n(n-1)]d = 0 \Rightarrow (m-n)a + (m^2 - n^2 - m + n)d = 0 \Rightarrow (m-n)[a + (m+n-1)d] = 0$. Since $m <br>eq n$, $a + (m+n-1)d = 0 \Rightarrow a_{m+n} = 0$.
$m[a + (m-1)d] = n[a + (n-1)d] \Rightarrow (m-n)a + [m^2 - m - n^2 + n]d = 0$. [1.0 Mark]
$(m-n)a + [(m-n)(m+n) - (m-n)]d = 0 \Rightarrow (m-n)[a + (m+n-1)d] = 0$. [1.0 Mark]
Since $m
eq n$, $a + (m+n-1)d = 0 \Rightarrow a_{m+n} = 0$. Proved! [1.0 Mark]
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🎯 Official CBSE Marking Scheme:
Expanding equation: 1.0 Mark
Factoring out $(m - n)$: 1.0 Mark
Concluding $a_{m+n} = 0$: 1.0 Mark
Correct Answer: