D - 4, G - 7, I - 9, M - 13, .....

P - 14

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O - 15

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Q - 16

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M - 17

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Solution

The correct option is B O - 15
In the given series, all the alphabets with their positions.

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Similar questions

Question 9.
Sum of $\frac{4}{17}$ and $\frac{15}{17}$ is
(a) $\frac{19}{17}$
(b) $\frac{11}{17}$
(c) $\frac{19}{34}$
(d) $\frac{2}{17}$

| z_{i} | = i, i = 1, 2, 3, 4

| 16 z_{1} z_{2} z_{3} + 9 z_{1} z_{2} z_{4} + 4 z_{1} z_{3} z_{4} + z_{2} z_{3} z_{4} | = 48,

∣ ∣ ∣ \frac{1}{{¯ ¯ ¯ z}_{1}} + \frac{4}{{¯ ¯ ¯ z}_{2}} + \frac{9}{{¯ ¯ ¯ z}_{3}} + \frac{16}{{¯ ¯ ¯ z}_{4}} ∣ ∣ ∣ .

1) \frac{18}{23} \times (\frac{4}{9} \times \frac{6}{7}) = (\frac{18}{23} \times \frac{4}{9}) \times \frac{6}{7}

2) \frac{16}{17} \times (\frac{3}{4} + \frac{20}{31}) = (\frac{16}{17} \times \frac{3}{4}) + (\frac{16}{17} \times \frac{20}{31})

If $15 %$ of $A = 20 %$ of $B$ , then $A : B = ?$
(a) $3 : 4$
(b) $4 : 3$
(c) $17 : 16$
(d) $16 : 17$

z_{1}, z_{2}, z_{3}

3

\frac{3}{| z_{2} - z_{3} |} = \frac{4}{| z_{3} - z_{1} |} = \frac{5}{| z_{1} - z_{2} |}

\frac{9}{z_{2} - z_{3}} + \frac{16}{z_{3} - z_{1}} + \frac{25}{z_{1} - z_{2}}

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