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Question
If m times the mth term of an A.P. is eqaul to n times nth term then show that the (m + n)th term of the A.P. is zero.
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Solution
We know,
\[a_n = a + \left( n - 1 \right)d\]
According to the question,
\[m\left( a_m \right) = n\left( a_n \right)\]
\[ \Rightarrow m\left( a + \left( m - 1 \right)d \right) = n\left( a + \left( n - 1 \right)d \right)\]
\[ \Rightarrow am + \left( m - 1 \right)md = an + \left( n - 1 \right)nd\]
\[ \Rightarrow am + m^2 d - md = an + n^2 d - nd\]
\[ \Rightarrow am - an = n^2 d - nd - m^2 d + md\]
\[ \Rightarrow a\left( m - n \right) = d\left( n^2 - m^2 \right) + d\left( m - n \right)\]
\[ \Rightarrow a\left( m - n \right) = d\left( m + n \right)\left( n - m \right) + d\left( m - n \right)\]
\[ \Rightarrow a\left( m - n \right) = d\left[ \left( m + n \right)\left( n - m \right) + \left( m - n \right) \right]\]
\[ \Rightarrow a\left( m - n \right) = d\left[ - \left( m + n \right)\left( m - n \right) + \left( m - n \right) \right]\]
\[ \Rightarrow a\left( m - n \right) = d\left( m - n \right)\left[ 1 - m - n \right]\]
\[ \Rightarrow a = d\left( 1 - m - n \right) \left( \because m \neq n \right)\]
\[ \Rightarrow a = d\left( 1 - m - n \right) . . . \left( 1 \right)\]
Now,
\[a_{m + n} = \left( a + \left( m + n - 1 \right)d \right)\]
\[ = \left( \left( 1 - m - n \right)d + \left( m + n - 1 \right)d \right) \left( \text{from } \left( 1 \right) \right)\]
\[ = d\left( 1 - m - n + m + n - 1 \right)\]
\[ = 0\]
Hence, the (m + n)th term of the A.P. is zero.
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