2, 3, 4, 4, 6, 8, 9, 12, 16

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Solution

The correct option is C 9
The given sequence is a combination of three series:
I. 1st, 4th, 7th terms i.e. 2, 4, 9,.....
II. 2nd, 5th, 8th terms i.e. 3, 6, 12,.....
III. 3rd, 6th, 9th terms i.e. 4, 8, 16,.....
In each one of I, II and III, each term is twice the preceding term. So, 9 is wrong and must be replaced by (4 x 2) i.e. 8.

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Similar questions

Q. If

3

4

8 = 6

and

9

4

3 = 12

, then

12

6

24 = ?

Q. Solve:

\frac{12^{3} \times 9^{2} \times 4}{6^{4} \times 8^{2} \times 8} .

Solve: $\frac{12^{4} \times 9^{3} \times 4}{6^{3} \times 8^{2} \times 27}$

Verify the following :

(i) $\frac{3}{7} \times (\frac{5}{6} + \frac{12}{13}) = (\frac{3}{7} \times \frac{5}{6}) + (\frac{3}{7} \times \frac{12}{13})$
(ii) $\frac{- 15}{4} \times (\frac{3}{7} + \frac{- 12}{5}) = (\frac{- 15}{4} \times \frac{3}{7}) + (\frac{- 15}{4} \times \frac{- 12}{5})$
(iii) $(\frac{- 8}{3} + \frac{- 13}{12}) \times \frac{5}{6} = (\frac{- 8}{3} \times \frac{5}{6}) + (\frac{- 13}{12} \times \frac{5}{6})$
(iv) $\frac{- 16}{7} \times (\frac{- 8}{9} + \frac{- 7}{6}) = (\frac{- 16}{7} \times \frac{- 8}{9}) + (\frac{- 16}{7} \times \frac{- 7}{6})$

Q. simplify:

\frac{18 \sqrt{3}}{12 - 6 \sqrt{2}} + \frac{3 \sqrt{2} - 2 \sqrt{3}}{8 - \sqrt{6}} - \frac{12 \sqrt{2} + 4 \sqrt{6}}{8 - \sqrt{12}}

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