Sets

T = {9, 10, 11, 12, …., 1000}

S = {4, 6, 9}

A = $\left\{a_1+a_2+....+a_k:k\in N,a_i\in S\right\}$

Let 4 appear x no. of times

6 appear y no. of times

9 appear z no. of times

10 then set A has n element 4x + 6y + 9z

Concept : Let a and b are co-prime numbers, then members from (a - ) (b – 1) and more can be expressed in the form ax + by where x

$\Delta =p!\left(p+1\right)!\left(p+2\right)!{\Delta }_1=2p!\left(p+1\right);\left(p+2\right)!=2{\left(p!\right)}^3.{\left(p+1\right)}^2.{\left(p+2\right)}^1\to \alpha +\beta =3+1=4$

${\Delta }_1=\left|\begin{array}{ccc}\hfill 1\hfill & \hfill p+1\hfill & \hfill \left(p+2\right)\left(p+1\right)\hfill \\ \hfill 1\hfill & \hfill p+2\hfill & \hfill \left(p+3\right)\left(p+2\right)\hfill \\ \hfill 1\hfill & \hfill p+3\hfill & \hfill \left(p+4\right)\left(p+3\right)\hfill \end{array}\right|=\left|\begin{array}{ccc}\hfill 0\hfill & \hfill -1\hfill & \hfill \left(p+2\right)\left(-2\right)\hfill \\ \hfill 0\hfill & \hfill -1\hfill & \hfill \left(p+3\right)\left(-2\right)\hfill \\ \hfill 1\hfill & \hfill p+3\hfill & \hfill \left(p+4\right)\left(p+3\right)\hfill \end{array}\right|$

= 2(p + 3) – 2 (p + 2) = 2

$S=2+\frac{6}{7}+\frac{12}{7^2}+\frac{20}{7^3}+\frac{30}{7^4}+........(i)$  

$\frac{1}{7}S=\frac{2}{7}+\frac{6}{7^2}+\frac{12}{7^3}+\frac{20}{7^4}+........(ii)$               

(i) - (ii)

$\frac{6}{7}S=2+\frac{4}{7}=\frac{6}{7^2}+\frac{8}{7^3}+...........(iii)$               

$\frac{6}{7^2}S=\frac{2}{7}+\frac{4}{7^2}+\frac{6}{7^3}+........(iv)$              

(iii) – (iv)

$=2+\left[\frac{1}{1-\frac{1}{7}}\right]=2\left(\frac{7}{6}\right)$

$\therefore 4S=8{\left(\frac{7}{6}\right)}^3={\left(\frac{7}{3}\right)}^3$      

Yes, Sandip University employs a normalization process for SU JEE (Sandip University Joint Entrance Examination) scores to ensure fairness across different question paper sets and candidate categories. This process adjusts scores to account for variations in difficulty levels among different exam ve

66. Given series is 1× 2× 3 + 2× 3 ×4 + 3× 4 ×5 + … to n term

a_n = (n^th term of A. P. 1, 2, 3, …) ´× (n^th terms of A. P. 2, 3, 4) ×

i e, a = 1, d = 2- 1 = 1i e, a = 2, d = 3- 2 = 1

(nth term of A. P. 3, 4, 5)

i e, a = 3, d = 3 -4 = 1.

= [1 + (n -1) 1] ×[2 + (n -1):1]× [3 + (n- 1) 1]

= (1 + n -1)×(2 + n -1

43. ${\displaystyle \sum _{}^{11}}\left(2+3^x\right)=\left(2+3^1\right)+\left(2+3^2\right)+.........+\left(2+3^{11}\right)$

$x=1$  [2+2+ ….+ (11lines)] + (31+ 32+……+311)

= $11\times 2+\frac{3\left(3^{11}-1\right)}{3-1}\left[?s_n=\frac{a\left(r^n-1\right)}{r-1},r\ge 1\right]$

= $22+\frac{3\left(3^{11}-1\right)}{2}$

63. Let H, T and I be of people who reads newspaper H, T and I respectively.

Then,

number of people who reads newspaper H, n (H) = 25.

number of people who T, n (T) = 26.

number of people who I, n (I) = 26

number of people who both H and T, n (H$\cap$I) = 9

number of people who both H and T, n (H $\cap$T) = 11

n

62. Let H and E be set of students who known Hindi and English respectively.

Then, number of students who know Hindi = n (H) = 100

Number of students who know English = n (E) = 50

Number of students who know both English & Hindi = 25 = n (H$\cap$E)

As each of students knows either Hindi or English,

Total

61. Let T and C be sets of students taking tea and coffee.

Then, n (T) = 150, number of students taking tea

n (C) = 225, number of students taking coffee

n (T$\cap$C) = 100, number of students taking both tea and coffee.

So, Number of students taking either tea or coffee is.

n (T$\cup$C) = n (T) + n (C) n (T$\cap$C)

=

60. Let A = {x, y}

B = {y, z}

C = {x, z}

So, A$\cap$B = {x, y}$\cap$ {y, z} = {y}≠$?$

B$\cap$C = {y, z} $\cap$ {x, z} = {z}≠$?$

A$\cap$C = {x, y}$\cap$ {x, z} = {x}≠$?$

But A$\cap$B$\cap$C = (A$\cap$B)$\cap$ C

= {y} $\cap$ (x, z}

=$?$