JEE MAIN - Mathematics (2004 - No. 28)
Let $${{T_r}}$$ be the rth term of an A.P. whose first term is a and common difference is d. If for some positive integers m, n, $$m \ne n,\,\,{T_m} = {1 \over n}\,\,and\,{T_n} = {1 \over m},\,$$ then a - d equals
$${1 \over m} + {1 \over n}$$
1
$${1 \over {m\,n}}$$
0
Explanation
$${T_m} = a + \left( {m - 1} \right)d = {1 \over n}...........\left( 1 \right)$$
$${T_n} = a + \left( {n - 1} \right)d = {1 \over m}..........\left( 2 \right)$$
$$\left( 1 \right) - \left( 2 \right) \Rightarrow \left( {m - n} \right)d$$
$$ = {1 \over n} - {1 \over m} \Rightarrow d = {1 \over {mn}}$$
From $$\left( 1 \right)$$ $$a = {1 \over {mn}} \Rightarrow a - d = 0$$
$${T_n} = a + \left( {n - 1} \right)d = {1 \over m}..........\left( 2 \right)$$
$$\left( 1 \right) - \left( 2 \right) \Rightarrow \left( {m - n} \right)d$$
$$ = {1 \over n} - {1 \over m} \Rightarrow d = {1 \over {mn}}$$
From $$\left( 1 \right)$$ $$a = {1 \over {mn}} \Rightarrow a - d = 0$$
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