Jee 2014 Booklet4 Hwt Solutions Sequence & Series

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Solutions to Home Work Test/Mathematics

1. log32, log62, log122 can be written as

2 2 2

3 3 2 3 2 2

log log log

, ,

log log log log log log

 H.P. with common difference 1.

2. Use A.M. ≥ G.M.

3. Let first five terms be 2

2 a a , , a, ar, ar r r 4 a (given)

Product of first five terms = a5

8.





2 5 2 3 1 1 1 3 _{3 3} 0 16. log . . . . .           



_{0 16.}



log2 5. 2 ₂log2 5. 0 16.

 

₂2 ₄    

9. Coefficient of x99= – (sum of roots)

1. 2 a b A  , G ab  b G2 a   2A a G2 a    2 2 2 0 a  Aa G   2 2 a A A G 2.



2

  

1





2 n n S  a  n d

















1 1 1 1 2 2 2 2 1 2 1 3 18 ₂ 1 2 1 7 15 2 n n n a d a n d S n n S a n d n a d      _{ }  _       Since 12th term = a11d Put 1 11

2 n _{ } n = 23 

 

 

12 1 1 12 2 2 23 3 8 23 77 7 23 7 23 15 176 16 T a d T a d   _  _{  }   

3. Product reduces to ‘x’. 4. Derivation in module. 5. In Illustrations

6. In Module 7. In Module 8. In Illustrations

9. In Module

Sequence & Series

HWT - 1

1. In Illustration 2. In INE

3. abc, abd, acd, bcd can be written as

abcd abcd abcd abcd

, , , d c b a  in H.P. 4. In INE 7. 2 2 2 1 1 4 2 4

log x log x log x. . . . . 

 2 1 1 1 4 2 4 log x_   . . . . . _    2log x2 4  log x2 2  x224

9. Coefficient of

 

x0  

 

1No. of roots (product of roots)

1. Lengths are in A.P.  sides are ad , a, ad







2 2





2

ad a  ad  2ada22ad

 4d = a

 Sides are 3d, 4d, 5d  sin of acute angles are 3

5 and 4 5

2. log a



 c



log a



 c 2b



2log a c











 







2 2 log ac a c b log ac 





2



 



2 2 ac  b ac  ac  2 2 2 2 2 2 2 2 a c  ac ab bcc  aca  4ac2



ab bc



 2ac b ac  in H.P. 5.



 



1 1 1 2 12 n n n r S t n n n  



  



 

 

 





 

1 1 2 1 1 1 3 12 12 n n n n n n n n n n n T S S _        





1 4 1 n T n n 

_

_

1 1 1 1 1 1 1 4 4 1 1 n n n n r r r r S T n n n n          







1 4 4 1 1 1 n n n    _  _    

Sequence & Series

HWT - 3

8. False as n! = 1 . 2 . 3 . 4 . 5 . . . n and nn= n . n . n . n . n . . . n Clearly n! < nn

9. Yes if common ratio of a G.P. is –ve we can have this kind of G.P.

10. b 2ac a c    2 2 2 2

2

2

3

3

2

2

ac

ac

a

c

a

ac

c

ac

a

c

a

c

ac

ac

_ac

_a

_ac

_c

a

c

a

c

a

c



 









 



_

_







 











_

 

_









_

 

_





 



 a 3c c 3a 2c 2a 2

_



c a

_



2 c a a c c a c a   _  _  _{ }      False. 1. We have





2 2 4 2 1 a r r s and a



1 r r2



s  2 2

 

2 1 1 r r f r r R r r         

Check range of f (r) which is 1 1

 

1 3

3, ,  _{ }     2 1   as for 2 0 1

r ,  (Similar question as in quadratic equation) If  then2 2 2 2 1 2 1 r r r r       2 3 1 0 r  r   3 5 2 r  Take 3 5 2 r   [r] = 2 Now if r = 2 2 1 2 4 7 1 2 4 3         2 2   _{ }   6. In Module 7. In Module 8. Since ax2 dividesc ax3bx2cxd  x i c a

 is the zero of equation. Now 3 2 3 2 1 2 c c bc c f i i i d a a a a                    0 bc d a   

 bcad  a, b, c, d are in G.P.  statement is true 9. 1 x x2x . . . x3  230  24 1 0 1 x x  _  2 3 19 1 x x x . . . x 0  20 1 0 1 x x  _   24 1 x  and x201  4 1 x   roots are 1,1, i, i  statement is false.

1. Use concept of insertion of A.M’s and G.M’s between two numbers.

4. Add terms until terms come out to be negative to get maximum sum.

5. In Module 6. In Illustration 7. 11 11 11 . . . (91 times) = 99



91 times



9 . . . .



91



10 1 9  

 







 



13 7 7 7 10 1 10 1 9 10 1  _    



7 21 12 7







1 10 10 10 10 9 9 9 9 9 9 9 9 . . . .      



7 4 21 12 7







1 10 10 10 . . . . .10  1111111    

 Not a prime number.

9. In INE

10. Maximum value of x y z is whena b c axbycz

 maximum of ab c is when 2b2  a c

 4b 4 1

 a2b c 2  statement is false.

1. General term is 8

2n1 2. Answer is

2 2

2 0

x  Ax G 

3. In Illustrations 4. In Illustrations 5. In INE

6. In Illustrations 7. In INE 8. Use AM ≥ GM

9. In INE

10. This is G.P. with common ratio ex.

 Sum = 1 1 1 x x x e e e   

   positive always and less than 1.

1. 1 1 2

1

1 x 1 x  x  they are in A.P.

2. In objective worksheet 3. In INE 4. In INE 5. In INE

6. Series is



1 05.

 

 1 05.



2 



1 05.



3 



1 05.



4. . . .



1 05.



49



 

1 05



49 1



1 05



50 1 05 1 05 1 05 1 0 05 . . . . . .  _  _          11 658 1 05 0 05 . . .   10 608 0 05 . .  = 212.16 7. Sum = 5 1 x r   Where   1 r 1  1 5 x r   1 1 1 5 x      0x and x10  x



0 10,



8. (AM) (HM) = (GM)2 9. Sum of progression



2



2 n

 (sum of first and last term)





2 2 38 2 n   = 20(n + 2) and 20(n + 2) = 200  n + 2 = 10  n = 8

Sequence & Series

HWT - 7

1. P117 21 25, , , . . . . .417

2 16 21 26 466

P  , , , . . . . .

General term for P1 =174



n 1



4n13

General term for P2 =16 5



k 1



5k11

Now 4n135k11

 5 2 2

4 4

k k

n   k 

As k and n both are integers 2 4 k I   4 2 k I and 0 x 111 0,  k 91  k



2 6 10 14, , , . . . .86



We can’t include k

 

90 as for this n exceeds 111.  k has 22 values  n has 22 values.

2. This an AGP. 4. In Module (Objective Worksheet)

5. Objective Worksheet Q. No. 35 6. Objective Worksheet Q. No. 34

8. Objective Worksheet Q. No. 14 9. Objective Worksheet Q. No. 18

10. Objective Worksheet Q. No. 67

3. This is a geometric progression with common difference 3 .



 





 









 

10 5 10 2 3 1 2 3 1 _{2 242} 3 1 3 1 3 1 S  _            

 





2 242 3 1 2  





121 6 2  

4. Harmonic mean = 2 2 product of roots





sum of roots ab a b 



0



2 coeff. of coeff. of x x _           5. S1 a arar2ar . . . . .3  20 2 2 2 2 4 2 6 2 100 S a a r a r a r . . . . . 

Sequence & Series

HWT - 9

1 20 1 a S r    2 2 ₂ 100 1 a S r    





2 2 2 2 1 ₄₀₀ 4 100 1 a r a r      _   _{  }      _    





2 2 1 4 1 r r  _   1 4 1 r r  _   3 5 r 7.



1



2 n n n T  











2 1 1 1 1 2 1 1 1 2 2 6 2 n n n r r n r r r r n n n n n S T             _  _  

 





1

 

2



6 n n n  8. n ,! 3

  

n , n! 1 !



are in G.P. 



3

 

n!



2



n1 !

  

n!  9

     

n! n!  n! n1!



 9 = n + 1  n = 8 Now 2 5



 

n!



n!



n1 !



 these are in A.P.

9. This is an AGP.

10. b2ac  2log blog alog c . . . .(i)

and a1xb1yc1z

 1log a 1log b 1log c k

x  y  z  . . . .(ii)

Using (i) and (ii) we have

 

2 y kxkzk

 2 y x z