(1) 22, 4, 5 (2) 34, 4, 8
(3) 37, 4, 9 (4) 54, 4, 13

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Solution

The correct option is C 3
In the given sets, add 2 to the product of the second and the third numbers to get the first number except 37, 4, 9.

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Similar questions

\frac{4}{1!} + \frac{11}{2!} + \frac{22}{3!} + \frac{37}{4!} + \frac{56}{5!} + . . . \infty = b e - 1

\frac{4}{1!} + \frac{11}{2!} + \frac{22}{3!} + \frac{37}{4!} + \frac{56}{5!} + . . . \infty = 6 e - 1

\frac{4}{1!} + \frac{11}{2!} + \frac{22}{3!} + \frac{37}{4!} + \frac{56}{5!} + . . . .

\frac{4}{1!} + \frac{11}{2!} + \frac{22}{3!} + \frac{37}{4!} + \frac{56}{5!} + . . . .

4, 9, 19, 34, 54, ?, 109

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