Arun Sharma Solutions for Exercise 3: Level of Difficulty

The sum to $17$ terms of the series $\frac{3}{1^2·2^2}+\frac{5}{2^2·3^2}+\frac{7}{3^2·4^2}+......$ is:

Find the value of $S$ if $S=\frac{4}{11}+\frac{44}{{11}^2}+\frac{444}{{11}^3}+\frac{4444}{{11}^4}+......\infty$

$$\begin{gathered} A=a+A_1+A_2+.......+A_N+b \\ B=a+G_1+G_2+.......+G_N+b \end{gathered}$$

$A$ is the sum of $n+2$ terms of an A.P. with first term $a$ and last term $b$.
$B$ is the sum of $n+2$ terms of a G.P. with first term $a$ and last term $b$.

Then, what can be said about the relative values of $A$ and $B$?

$\frac{9}{2}+\frac{25}{6}+\frac{49}{12}+.........+\frac{9801}{2450}=$?

The product of $1^{st}$ five terms of an increasing A.P. is $3840$. If the $1^{st}, 2^{nd}$ and $4^{th}$ terms of the A.P. are in G.P. Find ${10}^{th}$ term of the series.

If $S=\left[1+(-\frac{1}{3})\right]\left[1+{(-\frac{1}{3})}^2\right]\left[1+{(-\frac{1}{3})}^4\right]\left[1+{(-\frac{1}{3})}^8\right].....$ then $S$=?

Let the positive numbers $a, b, c, d$ be in A.P. Then the type of progression for the numbers $abc, abd, acd, bcd$ is:

Sum of the series $\frac{1^3}{1}+\frac{1^3+2^3}{1+3}+\frac{1^3+2^3+3^3}{1+3+5}+......$ to $10$ terms: