1. the value of 6+6+6+...................is

2.(xb-c)1bybc (xc-a)1byca(xa-b)1byab=

3.a sector is cut from a circle exceeds the diameter by 16.8 cm,then the length of arc is?

4.in an A.P 3rd term is 5 and the 7th term is 9 then 11th term is?
5.in a cyclic quadrilateral ABCD,cosA+cosB+cosC+cosD=

6.the value of log [(tan1)tan2)(tan3)...................(tan89)]is
7.median of xby 5,xby4,xby2,xby3is 8.if x>0,then thevalueof x is

8. the roots of the equation a(b-c)x2+b(c-a)x+c(a-b)=0 are

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2015-04-20T17:03:21+05:30

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1.)     6 + 6 + 6 + ....  n terms =  6 * n
2)    (xb - c) /bc  + (xc -a )/ca + (xa - b) / ab
     = [ ax b - ac + x bc - ab + xac - bc ] / abc
     =   (x - 1) ( ab + bc + ca) / abc
3)
  case 1)   perimeter of the sector exceeds the diameter by 16.8 cm
            (2 R + arc length) - 2 R = 16.8 cm      => arc length = 16.8 cm
   case 2)  arc length exceeds
           R Ф - 2 R = 16.8 cm.      Here there are two parameters R & Ф. we need more information to solve this.

4)    T3 = 5 = a + 2 d
       T7 = a + 8 d = 9
           subtract equations,  6 d = 4      =>  d = 2/3
               =>  a = 11/3

5)    In a cyclic quadrilateral,  sum of opposite angles is 180 deg.
   Cos A + Cos  C = 2 Cos (A+C)/2 Cos (A-C)/2    = 0
                  As (A+C)/2  = 90 deg
   similarly  Cos B + Cos D = 0  as  (B+D)/2 = 90 deg.
   hence the answer is 0.

6)  tan 1 tan2 tan 3  ...... tan 88 tan 89
       = tan1  tan2 tan 3 .... tan 43  tan 44  tan 45  cot 44 cot 43 .... cot 3 cot 2  cot 1
       = tan 1 cot 1  tan 2 cot2  tan3  cot3    .... tan43 cot43  tan44 cot44  * 1
       = 1
  Log  1 = 0

7)
      x / 5  , x/4  x/3  ,  x/2
         median is the average of  x/4 and  x/3
                 so x/4 + x/3 = 2 * 8 = 16
                 7 x / 12 = 16         
         x = 12 * 16 / 7

8)
       D =  discriminant = b² (c-a)² - 4 ac (b-c) (a-b)
                   = b² (c² + a² - 2 a c) - 4 a c ( a b - b² - a c + b c)
                   = b² c² + a² b² - 2 a c b²  - 4 a² b c + 4 a c b² + 4 a² c² - 4 a b c²
                   = b² c² + a² b² + 2 a c b²  - 4 a² b c + 4 a² c² - 4 a b c²
                   = b² (a +c)² - 4 a c (a b - a c + b c)
                   = (a+c)² [ b² - 4 abc ] + 4 a² c²

       x = [ - b (c - a) + - √ D ]  /  [ 2 a (b-c)  ] 

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