BITSAT 2024 | Sequences and Series Question 3 | Mathematics

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 two other numbers are

$ -2,4 $

$ -4,2 $

2,6

None of the above

BITSAT 2024

MCQ (Single Correct Answer)

-1

The coefficient of $ x^{n} $ in the expansion of $ \frac{e^{7 x}+e^{x}}{e^{3 x}} $ is

$ \frac{4^{n-1}+(-2)^{n}}{n!} $

$ \frac{4^{n-1}+2^{n}}{n!} $

$ \frac{4^{n}+(-2)^{n}}{n!} $

$ \frac{4^{n-1}+(-2)^{n-1}}{n!} $

BITSAT 2023

MCQ (Single Correct Answer)

-1

Let $$\frac{1}{16}, a$$ and $$b$$ be in GP and $$\frac{1}{a}, \frac{1}{b}, 6$$ be in AP, where $$a, b>0$$. Then, $$72(a+b)$$ is equal to

BITSAT 2023

MCQ (Single Correct Answer)

-1

If $$a_1, a_2, \ldots, a_n$$ are in HP, then the expression $$a_1 a_2+a_2 a_3+\ldots+a_{n-1} a_n$$ is equal to

$$n\left(a_1-a_n\right)$$

$$(n-1)\left(a_1-a_n\right)$$

$$n a_1 a_n$$

$$(n-1) a_1 a_n$$

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