1
JEE Main 2026 (Online) 28th January Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

The value of $\sum\limits_{k=1}^{\infty}(-1)^{k+1}\left(\frac{k(k+1)}{k!}\right)$ is

A

e/2

B

$\sqrt{e}$

C

2/e

D

1/e

2
JEE Main 2026 (Online) 28th January Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

The common difference of the A.P.: $a_1, a_2, \ldots, a_{\mathrm{m}}$ is 13 more than the common difference of the A.P.: $b_1, b_2, \ldots, b_n$. If $b_{31}=-277, b_{43}=-385$ and $a_{78}=327$, then $a_1$ is equal to

A

21

B

19

C

24

D

16

3
JEE Main 2026 (Online) 24th January Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

Let $a_1, a_2, a_3, a_4$ be an A.P. of four terms such that each term of the A.P. and its common difference $l$ are integers. If $a_1+a_2+a_3+a_4=48$ and $a_1 a_2 a_3 a_4+l^4=361$, then the largest term of the A.P. is equal to

A

27

B

24

C

23

D

21

4
JEE Main 2026 (Online) 24th January Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

$\left(\frac{1}{3}+\frac{4}{7}\right)+\left(\frac{1}{3^2}+\frac{1}{3} \times \frac{4}{7}+\frac{4^2}{7^2}\right)+\left(\frac{1}{3^3}+\frac{1}{3^2} \times \frac{4}{7}+\frac{1}{3} \times \frac{4^2}{7^2}+\frac{4^3}{7^3}\right)+\ldots$ upto infinite terms, is equal to

A

$\frac{7}{4}$

B

$\frac{4}{3}$

C

$\frac{6}{5}$

D

$\frac{5}{2}$

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