Maths-
General
Easy

Question

If fraction numerator 1 over denominator b minus a end fraction plus fraction numerator 1 over denominator b minus c end fraction equals fraction numerator 1 over denominator a end fraction plus fraction numerator 1 over denominator c end fraction a n d b not equal to a plus c comma blank t h e n a comma b comma c are in

  1. AP    
  2. GP    
  3. HP    
  4. None of these    

The correct answer is: HP


    Since, fraction numerator 1 over denominator b minus a end fraction plus fraction numerator 1 over denominator b minus c end fraction equals fraction numerator 1 over denominator a end fraction plus fraction numerator 1 over denominator c end fraction
    rightwards double arrow fraction numerator b minus a plus b minus c over denominator b to the power of 2 end exponent minus open parentheses a plus c close parentheses b plus a c end fraction equals fraction numerator a plus c over denominator a c end fraction
    rightwards double arrow 2 a b c minus open parentheses a plus c close parentheses a c equals b to the power of 2 end exponent open parentheses a plus c close parentheses minus b open parentheses a plus c close parentheses to the power of 2 end exponent plus a c left parenthesis a plus c right parenthesis
    rightwards double arrow 2 a c open parentheses b minus a minus c close parentheses equals b open parentheses a plus c close parentheses left parenthesis b minus a minus c right parenthesis
    rightwards double arrow fraction numerator 2 a c over denominator a plus c end fraction equals b
    rightwards double arrow a comma b comma c are in HP.

    Related Questions to study

    General
    maths-

    An infinite GP has first term x and sum 5, then

    An infinite GP has first term x and sum 5, then

    maths-General
    General
    maths-

    Area of a triangle whose vertices are left parenthesis a c o s invisible function application theta comma b s i n invisible function application theta right parenthesis comma left parenthesis negative a s i n invisible function application theta comma b c o s invisible function application theta right parenthesis text  and  end text left parenthesis negative a c o s invisible function application theta comma negative b s i n invisible function application theta right parenthesis is

    Area of a triangle whose vertices are left parenthesis a c o s invisible function application theta comma b s i n invisible function application theta right parenthesis comma left parenthesis negative a s i n invisible function application theta comma b c o s invisible function application theta right parenthesis text  and  end text left parenthesis negative a c o s invisible function application theta comma negative b s i n invisible function application theta right parenthesis is

    maths-General
    General
    maths-

    In a geometric progression (GP) the ratio of the sum of the first three terms and first six terms is 125:152 the common ratio is

    In a geometric progression (GP) the ratio of the sum of the first three terms and first six terms is 125:152 the common ratio is

    maths-General
    parallel
    General
    maths-

    The sum of the first and third term of an arithmetic series is 12 and the product of first and second term is 24, then first term is

    The sum of the first and third term of an arithmetic series is 12 and the product of first and second term is 24, then first term is

    maths-General
    General
    maths-

    If fraction numerator a over denominator b plus c end fraction comma fraction numerator b over denominator c plus a end fraction comma fraction numerator c over denominator a plus b end fraction are in AP, then

    If fraction numerator a over denominator b plus c end fraction comma fraction numerator b over denominator c plus a end fraction comma fraction numerator c over denominator a plus b end fraction are in AP, then

    maths-General
    General
    maths-

    If a subscript 1 end subscript comma a subscript 2 end subscript comma a subscript 3 end subscript comma horizontal ellipsis comma a subscript n end subscript are the n arithmetic means between a and b, then 2 not stretchy sum from i equals 1 to n of a subscript i end subscript equals

    If a subscript 1 end subscript comma a subscript 2 end subscript comma a subscript 3 end subscript comma horizontal ellipsis comma a subscript n end subscript are the n arithmetic means between a and b, then 2 not stretchy sum from i equals 1 to n of a subscript i end subscript equals

    maths-General
    parallel
    General
    Maths-

    The entries in a two-by-two determinant open vertical bar table row cell a blank b end cell row cell c blank d end cell end table close vertical bar are integers that are chosen randomly and independently, and, for each entry, the probability that the entry is odd is p. If the probability that the value of the determinant is even is 1/ 2, then the value of p, is

    The entries in a two-by-two determinant open vertical bar table row cell a blank b end cell row cell c blank d end cell end table close vertical bar are integers that are chosen randomly and independently, and, for each entry, the probability that the entry is odd is p. If the probability that the value of the determinant is even is 1/ 2, then the value of p, is

    Maths-General
    General
    Maths-

    Statement-I : Cot to the power of negative 1 end exponent invisible function application open parentheses x close parentheses minus tan to the power of negative 1 end exponent invisible function application open parentheses x close parentheses greater than 0 text  for all  end text x less than 1
    Statement-II : Graph of c o t to the power of negative 1 end exponent invisible function application left parenthesis x right parenthesis is always above the graph of t a n to the power of negative 1 end exponent invisible function application left parenthesis x right parenthesis text  for all  end text x less than 1 text  . end text

    Statement-I : Cot to the power of negative 1 end exponent invisible function application open parentheses x close parentheses minus tan to the power of negative 1 end exponent invisible function application open parentheses x close parentheses greater than 0 text  for all  end text x less than 1
    Statement-II : Graph of c o t to the power of negative 1 end exponent invisible function application left parenthesis x right parenthesis is always above the graph of t a n to the power of negative 1 end exponent invisible function application left parenthesis x right parenthesis text  for all  end text x less than 1 text  . end text

    Maths-General
    General
    Maths-

    If c o s to the power of negative 1 end exponent invisible function application x minus c o s to the power of negative 1 end exponent invisible function application fraction numerator y over denominator 2 end fraction equals alpha comma then 4 x to the power of 2 end exponent minus 4 x y c o s invisible function application alpha plus y to the power of 2 end exponent text end textis equal to-

    If c o s to the power of negative 1 end exponent invisible function application x minus c o s to the power of negative 1 end exponent invisible function application fraction numerator y over denominator 2 end fraction equals alpha comma then 4 x to the power of 2 end exponent minus 4 x y c o s invisible function application alpha plus y to the power of 2 end exponent text end textis equal to-

    Maths-General
    parallel
    General
    Maths-

    The solution of the equation 2 c o s to the power of negative 1 end exponent invisible function application x equals s i n to the power of negative 1 end exponent invisible function application open parentheses 2 x square root of 1 minus x to the power of 2 end exponent end root close parentheses

    The solution of the equation 2 c o s to the power of negative 1 end exponent invisible function application x equals s i n to the power of negative 1 end exponent invisible function application open parentheses 2 x square root of 1 minus x to the power of 2 end exponent end root close parentheses

    Maths-General
    General
    Maths-

    The solution of the inequality open parentheses tan to the power of negative 1 end exponent invisible function application x close parentheses to the power of 2 end exponent minus 3 t a n to the power of negative 1 end exponent invisible function application x plus 2 greater or equal than 0 text  is - end text

    The solution of the inequality open parentheses tan to the power of negative 1 end exponent invisible function application x close parentheses to the power of 2 end exponent minus 3 t a n to the power of negative 1 end exponent invisible function application x plus 2 greater or equal than 0 text  is - end text

    Maths-General
    General
    Maths-

    The value of taninvisible function application open square brackets s i n to the power of negative 1 end exponent invisible function application open parentheses fraction numerator 3 over denominator 5 end fraction close parentheses plus t a n to the power of negative 1 end exponent invisible function application open parentheses fraction numerator 2 over denominator 3 end fraction close parentheses close square brackets is

    The value of taninvisible function application open square brackets s i n to the power of negative 1 end exponent invisible function application open parentheses fraction numerator 3 over denominator 5 end fraction close parentheses plus t a n to the power of negative 1 end exponent invisible function application open parentheses fraction numerator 2 over denominator 3 end fraction close parentheses close square brackets is

    Maths-General
    parallel
    General
    Maths-

    The image of the interval [- 1, 3] under the mapping specified by the function f left parenthesis x right parenthesis equals 4 x to the power of 3 end exponent minus 12 x text end textis :

    Hence the correct option in  [-8,72]

    The image of the interval [- 1, 3] under the mapping specified by the function f left parenthesis x right parenthesis equals 4 x to the power of 3 end exponent minus 12 x text end textis :

    Maths-General

    Hence the correct option in  [-8,72]

    General
    maths-

    Let f(x)equals fraction numerator X over denominator 1 plus x end fraction defined from left parenthesis 0 comma infinity right parenthesis rightwards arrow left square bracket 0 comma infinity right parenthesis, then by f(x) is -

    Let f(x)equals fraction numerator X over denominator 1 plus x end fraction defined from left parenthesis 0 comma infinity right parenthesis rightwards arrow left square bracket 0 comma infinity right parenthesis, then by f(x) is -

    maths-General
    General
    Maths-

    If f : R rightwards arrow R is a function defined by f(x) = [x] c o s invisible function application pi open parentheses fraction numerator 2 x minus 1 over denominator 2 end fraction close parentheses, where [x] denotes the greatest integer function, then f is :

    The correct answer is choice 2

    If f : R rightwards arrow R is a function defined by f(x) = [x] c o s invisible function application pi open parentheses fraction numerator 2 x minus 1 over denominator 2 end fraction close parentheses, where [x] denotes the greatest integer function, then f is :

    Maths-General

    The correct answer is choice 2

    parallel

    card img

    With Turito Academy.

    card img

    With Turito Foundation.

    card img

    Get an Expert Advice From Turito.

    Turito Academy

    card img

    With Turito Academy.

    Test Prep

    card img

    With Turito Foundation.