NIMCET 2009 Question Paper with solutions

Question : 85 of 120

Marks: +1, -0

A_{1}=\{3\}, A_{2}=\{5,7,9\}, A_{3}=\{11,13,15,17,19\}

A_{4}=\{21,23,25,27,29,31,33\}

and so on. What is the average of the numbers of the set

A_{20}

Solution:

A_{20}

, there will be

2 × 20-1=39

numbers in AP with common difference 2
Also, the first term of

A_{20}

will be

3+2 × 3+2 × 5+2 × 7+. . .+2 × 37+2

=3+2(1+3+5+7+. . .+37)

=3+2 × \;{19}/{2}[2+18 × 2]=3=2 × 19 × 19=725

Average of 39 numbers of

A_{20}

=\;{39/{2}[2 × 725+(39-1) × 2]}/{39}=725+38=763

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