List of top Mathematics Questions on Sum of First n Terms of an AP asked in JEE Main

For positive integers \( n \), if \( 4 a_n = \frac{n^2 + 5n + 6}{4} \) and $$ S_n = \sum_{k=1}^{n} \left( \frac{1}{a_k} \right), \text{ then the value of } 507 S_{2025} \text{ is:} $$ 

Let \[ f(x) = \lim_{n \to \infty} \sum_{r=0}^{n} \left( \frac{\tan \left( \frac{x}{2^{r+1}} \right) + \tan^3 \left( \frac{x}{2^{r+1}} \right)}{1 - \tan^2 \left( \frac{x}{2^{r+1}} \right)} \right) \] Then, \( \lim_{x \to 0} \frac{e^x - e^{f(x)}}{x - f(x)} \) is equal to:

$$ \lim_{n \to \infty} \frac{(1^2 - 1)(n-1) + (2^2 - 2)(n-2) + \ldots + ((n-1)^2 - (n-1))}{(1^3 + 2^3 + \ldots + n^3) - (1^2 + 2^2 + \ldots + n^2)} $$ is equal to: 

If \(S_n=3+7+11....\) upto \(n\) terms and \(40<\frac {6}{n(n+1)}\displaystyle\sum_{k=1}^n S_k<45\). Then \(n\) is

Let the first term of a series be \( T_1 = 6 \) and its \( r^\text{th} \) term \( T_r = 3T_{r-1} + 6^r \), \( r = 2, 3, \dots, n \). If the sum of the first \( n \) terms of this series is \[ \frac{1}{5} \left(n^2 - 12n + 39\right) \left(4 \cdot 6^n - 5 \cdot 3^n + 1\right), \] then \( n \) is equal to ______.

Let \(a_1,a_2,a_3\), ..., an, be in A. P. and \(S_n\) denotes the sum of first \(n\) terms of this A. P. is \(S_{10}\) =\(390, \frac{a_{10}}{a_{50}} =\frac{15}{7}\), then \(S_{15} -S_5 =\) _________.

Let \( S_n \) denote the sum of the first \( n \) terms of an arithmetic progression. If \( S_{20} = 790 \) and \( S_{10} = 145 \), then \( S_{15} - S_5 \) is:

The number of $3$-digit numbers, that are divisible by either $2$ or $3$ but not divisible by $7$ , is_________

The sum $1+ \frac{1^{3} +2^{3}}{1+2} + \frac{1^{3}+2^{3}+3^{3}}{1+2+3} +.... + \frac{1^{3} +2^{3}+3^{3} +....+15^{3}}{1+2+3+...+15} - \frac{1}{2} \left(1+2+3+...+15\right)$

The sum $\displaystyle\sum^{10}_{r=1}(r^2 + 1) \times (r!)$ is equal to :