Sequence and Series

The sum of the first $20$ natural numbers is

The number of terms of an A.P. is even; the sum of all the odd terms is 24, the sum of all the even terms is 30 and the last term exceeds the first by $\frac{21}{2}$. Then the number of terms which are integers in the A.P. is:

Let $a_1,a_2,a_3,\dots$ be in an A.P. such that $\displaystyle\sum_{k=1}^{12}a_{2k-1}=-\tfrac{72}{5}a_1$, $a_1\neq0$. If $\displaystyle\sum_{k=1}^n a_k=0$, then $n$ is:

The sum $1 + \tfrac{1+3}{2!} + \tfrac{1+3+5}{3!} + \tfrac{1+3+5+7}{4!} + \dots$ up to infinitely many terms is equal to:

The sum $1 + 3 + 11 + 25 + 45 + 71 + \dots$ up to 20 terms is equal to:

Let $a_1, a_2, a_3, \dots$ be a G.P. of increasing positive numbers. If $a_3 a_5 = 729$ and $a_2 + a_4 = \tfrac{111}{4}$, then $24(a_1 + a_2 + a_3)$ is equal to:

If the sum of the first 20 terms of the series $\displaystyle\frac{4\cdot1}{4+3\cdot1^2+1^4}+\frac{4\cdot2}{4+3\cdot2^2+2^4}+\frac{4\cdot3}{4+3\cdot3^2+3^4}+\dots$ is $\tfrac{m}{n}$, where $m$ and $n$ are coprime, then $m+n$ is equal to:

Consider two sets A and B, each containing three numbers in A.P. Let the sum and the product of the elements of A be 36 and p respectively, and the sum and the product of the elements of B be 36 and q respectively. Let d and D be the common differences of the A.P. in A and B respectively such that D = d + 3, d > 0. If $\frac{p+q}{p-q} = \frac{19}{5}$, then p - q is equal to:

Let A = \{1, 6, 11, 16, \ldots\} and B = \{9, 16, 23, 30, \ldots\} be the sets consisting of the first 2025 terms of two arithmetic progressions. Then n(A \cup B) is:

1 + 3 + 5² + 7 + 9² + \dots up to 40 terms is equal to: