A, Z, X, B, V, T, C, R, ?, ?

Question

Let \[ S=\frac{1}{2!5!}+\frac{1}{3!2!3!}+\frac{1}{5!2!1!}+\cdots \text{ up to 13 terms}. \] If $13S=\dfrac{2^k}{n!}$, $k\in\mathbb{N}$, then $n+k$ is equal to

Answer 1

An irregular interleaving of two series is presented: Series 1 (positions 1, 4, 7, 10...): A, B, C... D is at position 10. Series 2 (positions 2, 3, 5, 6, 8, 9...): Z(26), X(24), V(22), T(20), R(18)... the pattern is -2. P(16) is at position 9. The missing elements are P (at 9) and D (at 10). This corresponds to option (3).

A, Z, X, B, V, T, C, R, ?, ?

Show Hint

The Correct Option is C

Solution and Explanation

Top Questions on Series

Questions Asked in JEECUP exam