Let $\left\{x\right\}$ and $\left[x\right]$ denote the fractional part of $x$ and the greatest integer $\le x$ respectively of $a$ real number $x$ . if ${\int }_0^n \left\{x\right\}dx,{\int }_0^n \left[x\right]dx$ and $10\left(n^2-n\right),(n\in N,n>1)$ are three consecutive terms of a G.P. then $\mathrm{n}$ is equal to(Integration)

Three consecutive integers belong to 98 sets and four consecutive integers belongs to 97 sets.

Þ Number of permutations of b₁ b₂ b₃ b₄ = number of permutations when b₁ b₂ b₃ are consecutive + number of permutations when b₂, b₃, b₄ are consecutive – b₁ b₂ b₃ b₄ are consecutive = 98 * 97 * 98 * 97 – 97

Last two digit must be in form

$\begin{array}{l}23,4,516\\ 324\\ 32\\ 52\end{array}\}3\times 4=12$

Total number of required number = 12 + 18 = 30

Kindly consider the following figure

Sum of digits

1 + 2 + 3 + 5 + 6 + 7 = 24

So, either 3 or 6 rejected at a time

Case 1 Last digit is 2

……….2

no. of cases = ²C₁ × 4! = 48

Case 2 Last digit is 6

……….6

= 4! = 24

Total cases = 72

 $A=\left[\begin{array}{ccc}\hfill a_1\hfill & \hfill a_2\hfill & \hfill a_3\hfill \\ \hfill b_1\hfill & \hfill b_2\hfill & \hfill b_3\hfill \\ \hfill c_1\hfill & \hfill c_2\hfill & \hfill c_3\hfill \end{array}\right],a_2,b_2,c_2\in \left\{0,1\right\}$

S = $a_1+a_2+a_3+b_1+b_2+b_3+c_1+c_2+c_3$ is prime

$0\le s\le 9$

Prime value = 2,3, 5, 7

$ForS=2,1+1+0+0+0+0+0+0+0\to \frac{9!}{2!7!}=36$

Total number of matrices

= 36 + 84 + 126 + 36 = 282