1, 9, 25, 49, 81, ?

100

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112

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121

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144

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Solution

The correct option is C 121
The series consists of squares of consecutive odd numbers i.e. $1^{2}, 3^{2}, 5^{2}, 7^{2}, 9^{2}, . . . . .$
So, missing term = $11^{2} = 121$ .

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Similar questions

1, 9, 25, 49, 81, ?

4 \times 27 \times 3125, 8 \times 9 \times 25 \times 7, 16 \times 81 \times 5 \times 11 \times 49 ?

The H.C.F. of 4 $\times$ 27 $\times$ 3125, 8 $\times$ 9 $\times$ 25 $\times$ 7 & 16 $\times$ 81 $\times$ 5 $\times$ 11 $\times$ 49 is ____.

4 \times 27 \times 3125, 8 \times 9 \times 25 \times 7

16 \times 81 \times 5 \times 11 \times 49

{(\frac{9}{25})}^{- 3 / 2} \times {(\frac{7}{5})}^{3} \times {(\frac{81}{49})}^{1 / 2}

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