JEE MAIN - Mathematics (2019 - 12th January Evening Slot - No. 7)
If nC4, nC5 and nC6 are in A.P., then n can be :
11
12
9
14
Explanation
2.nC5 = nC4 + nC6
2.$${n \over {\left| 5 \right|n - 5}} = {n \over {\left| 4 \right|n - 4}} + {n \over {\left| 6 \right|n - 6}}$$
$${2 \over 5}.{1 \over {n - 5}} = {1 \over {\left( {n - 4} \right)\left( {n - 5} \right)}} + {1 \over {30}}$$
$$n = 14$$ satisfying equation.
2.$${n \over {\left| 5 \right|n - 5}} = {n \over {\left| 4 \right|n - 4}} + {n \over {\left| 6 \right|n - 6}}$$
$${2 \over 5}.{1 \over {n - 5}} = {1 \over {\left( {n - 4} \right)\left( {n - 5} \right)}} + {1 \over {30}}$$
$$n = 14$$ satisfying equation.
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