CoCubes Series Question and Answers Quiz-1

Exp.  a + 4d  = 10 …………..  (1)

         a + 20d = 42 …………..  (2)

         Eqn (2) – Eqn (1)  gives  16d  =  32 ,  d  = 2

         Substituting  d =2 in either (1) or  (2),  a =  2.

         31 st  term  =  a + 30d  = 2 + 30\times 2  =  62

 

Explanation:  Let  the  terms  be  \frac{a}{r} , a ,  ar

\frac{a}{r} \times a\times  ar  =  a^{3} = 216

a = 6

(\frac{a}{r}\times a)+ (a\times ar) + (ar\times\frac{a}{r}) =114

\frac{a^{2}}{r} + a^{2}r + a^{2} =114

a^{2}(\frac{1}{r} + r  +1 )  =  114

36[\frac{( 1+ r^{2} +r )}{r}]  = 114

6 [\frac{( 1 +r^{2}+ r  )}{r} ]=19

6 r^{2} -13 r +  6  =  0,

On  solving,   r  =  \frac{2}{3}  or  \frac{3}{2}

Let   the  angles  be  a, ar, ar^{2}, ar^{3}


       Sum  of  the angles = a \frac{( r^{4}- 1 )} {r -1} = a ( r ² + 1 ) ( r + 1 ) = 360

       a< 90 , and  ar< 90,  Therefore,  a ( 1 + r ) <  180,  or   ( r ² + 1 ) > 2

       Therefore, r  is  not  equal  to  1.  Trying  for  r  =  2  we  get  a  = 24

       Therefore, The  angles  are  24, 48, 96  and  192.

let the numbers be a-d, a and a+d

then, their sum i.e. 3a=33

a=11

and sum of their squares,

(a-d)²+a²+(a+d)²=491

(11-d)²+121+(11+d)²=491

(11-d)²+(11+d)²=370

121+d²-22d + 121+d²+22d =370

2d²= I28

d²=64

d=8

hence, the numbers are 3,11 and 19

 Exp.  As  the sum of  the  first  three terms will be thrice the  second term, 3(a + d)  =21,

Also  the  sum of  13 th, 14th and  15th = 3(a+13d)= 93

On  solving  the  two  eqns,  d = 2,   

 3(a+2)=21,   a=5

Exp. Arithmetic mean =\frac{552}{n}=\frac{1}{2}(3 +45) = 2


  n=\frac{552}{24} =23

a \frac{(r 6- 1)}{(r -  1 )} =  1001 a \frac{(r 3 –1)}{(r – 1)}
r^{3}+1  = 1001,
r  = 10

5th Term = a\times r^{n-1}

= 3\times 2^{5-1}

= 48

Exp.  It  is  a  G.P  with  infinite  terms.  The  common  ratio  is (\frac{4}{5}) and  this  is  less  than  1. 
The  sum  to  infinity  of  a  series  with  first  term  ‘a ‘  and    common  ratio  ‘r’ =\frac{ a}{( 1-r)} ,    
The  sum  to  infinity  =   \frac{ 1}{( 1-\frac{4}{5})} = 5

 Exp.  4th  term  =  sum  of  4  terms  -  sum  of 3 terms

                         = ( 6\times 4^{2} + 6\times 4 )-( 6\times 3^{2} +6\times 3),

                         = ( 96+24 ) – ( 54+18 )

                         = 120 - 72  =  48