SOLUTION IN 101 TO 110

101.           Given  exp. = (a + b)² - 4ab, where a = 476 and b=425

= (a –b)² = (475 – 425)² =(50)² = 2500.

102.           20z = (64)² (36)² 20z = (64 + 36) (64 – 36)

                                         20z = 100 x 28 z = 100 x 28/20 = 140.

103.           Let (46)² - X² = 4398 – 3066.

Then, (46)² - X² = 1332 X² = (46)² - 1332 = (2116 – 1332)

                                         X² = 784 x= 784 = 28.

104.           Given exp.   = ( a + b )²  ( a – b)²     = 2 ( a² + b²)        =2.

                                                                                     (a ²+ b² )                      ( a² + b² )

105.           Given exp. =   = ( a + b )²  ( a – b)²     =  4ab                  =2.

                                                                                         ab                            ab

106.  We know that  ( 1 + 2 + 3 + ….+ n)=  n (n + 1)/2

(1 + 2 + 3+…=(45x 46)/2 = 1035.

107.  Required numbers are 2, 3 4, 6 …, 30.

This is an A.P. containing 15 terms.

 Required sum = n/2 (first term + last term) = (2 + 30) = 240.

108.  (1 + 52 +53 +... +  100)

= (1+2 = 3+…. +100)  - (1 + 2 + 3 + …+ 50).

= (100 x 101\2 – 50 x 51 \2) = (5050 – 1275) = 3775.

109.  Every such number must be divisible by L. C. M. of 4, 5, 6, i.e. 60.

Such numbers are 240, 300, 360, 420, 480, and 540.

Clearly, there are 6 such numbers.

110.  Required numbers are 102, 108, and 114,996.

This is an A.P. with a = 102 and d = 6,

Let the number of it s terms be n. then,

a + (n – 1) d= 996 102 +( n -1 )x 6 = 996   n =150.