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Question

# If the seventh terms from the beginning and the end in the expansion of (3√2+13√3)n are equal then, n=?

A
9
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B
14
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C
22
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D
12
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Solution

## The correct option is B 12Solution:Seventh term from beginning i.e. T7 in the expansion of (3√2+13√3)n=nC6×263×3−(n−6)3and Seventh term from end i.e. Tn−7+2=Tn−5 in the expansion of (3√2+13√3)n=nCn−6×2n−63×3−63A/qT7=Tn−5or, nC6×263×3−(n−6)3 =nCn−6×2n−63×3−63or, 2−n+123×3−n+123=1or, 6−n+123=60From above, −n+123=0or, n=12Hence, D is the correct option.

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