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Logic: Subtract 2 from the squares of the consecutive numbers to get the numbers in the given sets. (34, 47, 62) → (6)2−2=34;(7)2−2=47;(8)2−2=62→ (6)2−2=34;(7)2−2=47;(8)2−2=62 (6, 7, and 8 are the consecutive numbers.) (119, 142, 167) → (11)2−2=119;(12)2−2=142;(13)2−2=167→ (11)2−2=119;(12)2−2=142;(13)2−2=167 (11, 12, and 13 are the consecutive numbers.) Explanation: (a) (27, 38, 51) → (5)2−2=23≠27;(6)2−2=34≠39;(7)2−2=47≠51→ (5)2−2=23=27;(6)2−2=34=39;(7)2−2=47=51 (b) (48, 65, 82) → (7)2−2=47≠48;(8)2−2=62≠65;(9)2−2=79≠82→ (7)2−2=47=48;(8)2−2=62=65;(9)2−2=79=82 (c) (64, 78, 99) → (8)2−2=62≠64;(9)2−2=79≠78;(10)2−2=98≠99→ (8)2−2=62=64;(9)2−2=79=78;(10)2−2=98=99 (d) (79, 98, 119) → (9)2−2=79;(10)2−2=98;(11)2−2=119→ (9)2−2=79;(10)2−2=98;(11)2−2=119 Thus, only option (d) follows the same logic. Thus, the correct option is (d).
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