Let $a_{1}=1$ and for $n \ge 1$, $a_{n+1} = \frac{1}{2}a_{n} + \frac{n^{2}-2n-1}{n^{2}(n+1)^{2}}$. Then $|\sum_{n=1}^{\infty}(a_{n}-\frac{2}{n^{2}})|$ is equal to ........... .

Let $\langle a_n \rangle$ be a sequence such that $a_0 = 0, a_1 = \frac{1}{2}$ and $2a_{n+2} = 5a_{n+1} - 3a_n$ for $n = 0, 1, 2, 3, \ldots$. Then $\sum_{k=1}^{100} a_k$ is equal to:

If $A_1, A_2$ are two arithmetic means,$G_1, G_2$ are two geometric means,and $H_1, H_2$ are two harmonic means between two numbers,then $\frac{A_1 + A_2}{H_1 + H_2} \cdot \frac{H_1 H_2}{G_1 G_2} = \dots$

The number of common terms in the progressions $4, 9, 14, 19, \ldots$ up to $25^{\text{th}}$ term and $3, 6, 9, 12, \ldots$ up to $37^{\text{th}}$ term is:

If $|x| < 1, |y| < 1$ and $x \neq y,$ then the sum to infinity of the following series $(x+y)+(x^{2}+xy+y^{2})+(x^{3}+x^{2}y+xy^{2}+y^{3})+\ldots$ is:

Three numbers form a $G.P.$ If the $3^{rd}$ term is decreased by $64$,the three numbers thus obtained constitute an $A.P.$ If the second term of this $A.P.$ is decreased by $8$,a $G.P.$ is formed again. Find the numbers.