There are four numbers of which the first three are in GP and the last three are in AP, whose common difference is 6. If the first and the last numbers are equal, then the two other numbers are:

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

Let three numbers in an arithmetic progression (AP) be $a, (a + 6), (a + 12)$.

Furthermore, the first and last of four numbers are equal. Therefore, the four numbers are $(a + 12), a, (a + 6), (a + 12)$.

Given that the first three numbers form a geometric progression (GP), the following equation holds:

\[ \Rightarrow a^2 = (a + 12)(a + 6) \]

\[ \Rightarrow a^2 = a^2 + 18a + 72 \]

\[ \Rightarrow a = \frac{-72}{18} = -4 \]

There are four numbers of which the first three are in GP and the last three are in AP, whose common difference is 6. If the first and the last numbers are equal, then the two other numbers are:

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The Correct Option is B

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