Given set : (49, 25, 9)

(36, 16, 4)

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(36 , 25, 16)

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(39, 26, 13)

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(64, 27, 8)

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Solution

The correct option is A (36, 16, 4)
each set contains squares of three consecutive alternative numbers in reverse order

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Similar questions

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

13, 16, 19, 25, 27, 30, 32, 35, 38, 44, 49

{\sqrt{8 + 2 \sqrt{16}}} + {\sqrt{26 - 2 \sqrt{25}}}

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