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    Mathematics
    Algebra
    Easy

    Question

    text  Let  end text a comma b text  and  end text c text  be in G.P. with common ratio  end text r comma text  where  end text a not equal to 0 text  and  end text 0 less than r less or equal than 1 half. text  If  end text 3 a comma 7 b text  and  end text 15 c text  are the first three  end text t e r m s space o f space a n space A. P. comma space t h e n space t h e space 4 t h space t e r m space o f space t h i s space A. P. space i s space colon

    1. a
    2. 7 over 3 a
    3. 2 over 3 a
    4. 5a

    The correct answer is: a

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    text  If  end text a subscript 1 comma a subscript 2 comma a subscript 3 comma horizontal ellipsis horizontal ellipsis horizontal ellipsis horizontal ellipsis horizontal ellipsis a subscript n text  are in A.P. and  end text a subscript 1 plus a subscript 4 plus a subscript 7 plus horizontal ellipsis horizontal ellipsis horizontal ellipsis.. plus a subscript 16 equals 114 comma text  then  end text a subscript 1 plus a subscript 6 plus a subscript 11 plus a subscript 16 text  is equal to :  end text

    text  If  end text a subscript 1 comma a subscript 2 comma a subscript 3 comma horizontal ellipsis horizontal ellipsis horizontal ellipsis horizontal ellipsis horizontal ellipsis a subscript n text  are in A.P. and  end text a subscript 1 plus a subscript 4 plus a subscript 7 plus horizontal ellipsis horizontal ellipsis horizontal ellipsis.. plus a subscript 16 equals 114 comma text  then  end text a subscript 1 plus a subscript 6 plus a subscript 11 plus a subscript 16 text  is equal to :  end text

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    T h e space s u m stack space space sum with k equals 1 below and 20 on top k 1 over 2 to the power of k space i s space e q u a l space t o

    T h e space s u m stack space space sum with k equals 1 below and 20 on top k 1 over 2 to the power of k space i s space e q u a l space t o

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    I f space t h e space s u m space a n d space p r o d u c t space o f space t h e space f i r s t space t h r e e space t e r m s space i n space a n space A. P. space a r e space 33 space a n d space 1155 comma space r e s p e c t i v e l y comma space t h e n space a space v a l u e space o f space i t s space 11 t h space t e r m space i s space colon

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    text  Some identical balls are arranged in rows to form an equilateral triangle. The first row consists of one ball,  end text text  the second row consists of two balls and so on. If 99  more identical balls are added to the total number of  end text text  balls used in forming the equilateral triangle, then all these balls can be arranged in a square whose each side  end text text  contains exactly  end text 2 text  balls less than the number of balls each side of the triangle contains. Then the number of  end text text  balls used to form the equilateral triangle is:  end text

    text  Some identical balls are arranged in rows to form an equilateral triangle. The first row consists of one ball,  end text text  the second row consists of two balls and so on. If 99  more identical balls are added to the total number of  end text text  balls used in forming the equilateral triangle, then all these balls can be arranged in a square whose each side  end text text  contains exactly  end text 2 text  balls less than the number of balls each side of the triangle contains. Then the number of  end text text  balls used to form the equilateral triangle is:  end text

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    I f space capital delta subscript 1 equals open vertical bar table row straight x cell sin theta end cell cell cos theta end cell row cell negative sin theta end cell cell negative straight x end cell 1 row cell cos theta end cell 1 straight x end table close vertical bar space a n d space capital delta subscript 2 equals open vertical bar table row straight x cell sin 2 theta end cell cell cos 2 theta end cell row cell negative sin 2 theta end cell cell negative straight x end cell 1 row cell cos 2 theta end cell 1 straight x end table close vertical bar comma space straight x not equal to 0 space t h e n space f o r space a l l space theta element of open parentheses 0 comma pi over 2 close parentheses colon

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    text  Let  end text A equals open parentheses table attributes columnalign center center columnspacing 1em rowspacing 4 pt end attributes row cell cos invisible function application alpha minus sin invisible function application alpha end cell blank row cell sin invisible function application alpha end cell cell cos invisible function application alpha end cell end table close parentheses comma left parenthesis a element of R right parenthesis text  such that  end text A to the power of 32 equals open parentheses table attributes columnalign center center columnspacing 1em rowspacing 4 pt end attributes row 0 cell negative 1 end cell row 1 0 end table close parentheses. text  Then a value of  end text alpha text  is -  end text

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    T h e space g r e a t e s t space v a l u e space o f space c space element of space R space f o r space w h i c h space t h e space s y s t e m space o f space l i n e a r space e q u a t i o n s x space – space c y space – space c z space equals space 0 space c x space – space y space plus space c z space equals space 0 space c x space plus space c y space – space z space equals space 0 space h a s space a space n o n minus t r i v i a l space s o l u t i o n comma space i s space minus

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    L e t space M equals open square brackets table attributes columnalign left end attributes row 0 1 a row 1 2 3 row 3 b 1 end table close square brackets space a n d space adj M equals open square brackets table row cell negative 1 end cell 1 cell negative 1 end cell row 8 cell negative 6 end cell 2 row cell negative 5 end cell 3 cell negative 1 end cell end table close square brackets space text  where a and b are real numbers. Which of the  end text text  following options is/are correct?  end text

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    L e t space M equals open square brackets table attributes columnalign center center columnspacing 1em rowspacing 4 pt end attributes row cell sin to the power of 4 theta end cell cell negative 1 minus sin squared theta end cell row cell 1 plus cos squared theta end cell cell cos to the power of 4 theta end cell end table close square brackets equals alpha I plus beta M to the power of negative 1 end exponent comma text where  end text alpha equals alpha left parenthesis theta right parenthesis text  and  end text beta equals beta left parenthesis theta right parenthesis space text  are real numbers, and I is the   end text 2 cross times 2 text  identity matrix. If  end text alpha to the power of ∗ text  is the minimum of the set   end text left curly bracket alpha left parenthesis theta right parenthesis colon theta element of left square bracket 0 comma 2 pi right parenthesis right curly bracket text  and  end text beta to the power of ∗ space text  is the minimum of the  end text set left curly bracket beta left parenthesis theta right parenthesis colon theta element of left square bracket 0 comma 2 pi right parenthesis right curly bracket space comma space text  then value of  end text alpha to the power of ∗ plus beta to the power of ∗ text  is -  end text

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    text  The sum of the series   end text 1 plus 2 cross times 3 plus 3 cross times 5 plus 4 cross times 7 plus horizontal ellipsis text  upto  end text 11 to the power of th text  term is :  end text

    text  The sum of the series   end text 1 plus 2 cross times 3 plus 3 cross times 5 plus 4 cross times 7 plus horizontal ellipsis text  upto  end text 11 to the power of th text  term is :  end text

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