1
BITSAT 2024
MCQ (Single Correct Answer)
+3
-1
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
A
$ -2,4 $
B
$ -4,2 $
C
2,6
D
None of the above
2
BITSAT 2024
MCQ (Single Correct Answer)
+3
-1
The coefficient of $ x^{n} $ in the expansion of $ \frac{e^{7 x}+e^{x}}{e^{3 x}} $ is
A
$ \frac{4^{n-1}+(-2)^{n}}{n!} $
B
$ \frac{4^{n-1}+2^{n}}{n!} $
C
$ \frac{4^{n}+(-2)^{n}}{n!} $
D
$ \frac{4^{n-1}+(-2)^{n-1}}{n!} $
3
BITSAT 2023
MCQ (Single Correct Answer)
+3
-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

A
12
B
14
C
16
D
18
4
BITSAT 2023
MCQ (Single Correct Answer)
+3
-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

A
$$n\left(a_1-a_n\right)$$
B
$$(n-1)\left(a_1-a_n\right)$$
C
$$n a_1 a_n$$
D
$$(n-1) a_1 a_n$$
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