Jee 2014 Booklet4 Hwt Solutions Sequence & Series

Vidyamandir Classes

Aggarwal Corporate Heights, 3rd Floor, Plot No. A - 7, Netaji Subhash Place, Pitam Pura, Delhi - 110034 Phone: 011-45221190-93

Solutions to Home Work Test/Mathematics Sequence & Series

1.

log 2 log 2 log 2 , , log 3 log 3  log 2 log 3  log 2  log 2

log3 2 , log6 2 , log12 2 can be written as



HWT - 1

H.P. with common difference 1.

2.

Use A.M. ≥ G.M.

3.

Let first five terms be

a r

2

,

a , a, ar, ar 2 r

a  4 (given) Product of first five terms = a5 1

8.

1

 0.16 log2.5  3  32   0.16 

9.



1 33

 . . . .   

log2.5 2

 2log2.5 0.16   2   4 2

Coefficient of x99 = – (sum of roots)

Sequence & Series 1.

2.

A

ab , 2

G  ab G2 a



b



a 2  2 Aa  G 2  0



a  A  A2  G 2

Sn 

HWT - 2





n 2  a    n  1 d 2

2A  a 

G2 a



 n  1 2a1   n  1 d1 3n  18 a1  2 d1 Sn    Sn 2a2   n  1 d 2 7n  15  n  1 d a2  2 2 Since 12th term = a  11d 

Put

n 1  11 2



n = 23

  23 3  8  77  7 T12 a  23d1  1   a2  23d 2 7  23  15 176 16 T12

3.

Product reduces to ‘x’.

4.

Derivation in module.

5.

In Illustrations

6.

In Module

7.

In Module

8.

In Illustrations

9.

In Module

VMC/Sequence & Series

1

HWT-Solutions/Mathematics

Vidyamandir Classes Sequence & Series 1.

In Illustration

3.

abc, abd, acd, bcd can be written as abcd abcd abcd abcd , , , d c b a

4.

In INE

7.

log 2 x 

9.

HWT - 3

2.

In INE



in H.P.

1 1 log 2 x  log 2 x. . . . .   4 2 4



 1 1  log 2 x 1   . . . . .   4  2 4 



2 log 2 x  4



log 2 x  2



 

 No. of

Coefficient of x 0   1

roots 

x  22  4

 (product of roots)

Sequence & Series 1.

2.

sides are a  d , a, a  d



 a  d 2  a 2   a  d 2



4d = a



Sides are 3d, 4d, 5d



sin of acute angles are



2ad  a 2  2ad

3 4 and 5 5

log  a  c   log  a  c  2b   2 log  a  c 



log  a  c   a  c  2b   log  a  c 



 a  c  2  2b  a  c    a  c  2



a 2  c 2  2ac  2ab  2bc  c 2  2ac  a 2



4ac  2  ab  bc 



2ac b ac n

5.



Lengths are in A.P.

HWT - 4

Sn 

t r 1

n





2

in H.P.

1 n  n  1  n  2  12

Tn  Sn  Sn  1 

n  n  1  n  2    n  1 n  n  1 12



n  n  1 3 12

1 4  Tn n  n  1 n



Sn 

T r 1

1

n

r

n

 n  n  1  4 n  n  1

4

1

r 1

1

1

r 1

1  4n   4 1    n 1 n 1

VMC/Sequence & Series

2

HWT-Solutions/Mathematics

Vidyamandir Classes 8.

False as n! = 1 . 2 . 3 . 4 . 5 . . . . . . n nn = n . n . n . n . n . . . . . . n

and

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



     

  a  ac  2ac   a   ac  



a  3c c  3a 2c  2a 2  c  a     2 ca ac ca c  a 

2ac

  c  a 2  3ac c 2  3ac ac   2 2ac  ac  c 2  c  ac  a ac  2ac



False.

Sequence & Series 1.

HWT - 5

We have





a2 1  r 2  r 4  s2





and

a 1 r  r2   s



2 

1  r  r2 1  r  r2

 f r  r  R

1  Check range of f (r) which is  , 1  1, 3 3   2  1 as for r  0,  2  1 If  2  2 then 2 

1 r  r2 1 r  r2



r 2  3r  1  0



r

3 5 2

Take

r

3 5 2

Now if

r=2 2 



(Similar question as in quadratic equation)



[r] = 2

7.

In Module

1 2  4 7  1 2  4 3

 2   2  

6.

In Module

8.

Since ax 2  c divides ax3  bx 2  cx  d 

xi

c is the zero of equation. a

 c3 2  c Now f  i  i  1 2   a a   

d

 bc c3 2 d  i  a a 

bc 0 a

VMC/Sequence & Series

3

HWT-Solutions/Mathematics

Vidyamandir Classes 

9.

bc  ad



a, b, c, d are in G.P.



statement is true

1  x  x 2  x3 . . . . .  x 23  0



x 24  1 0 x 1

1  x  x 2  x3 . . . . .  x19  0



x 20  1 0 x 1



x 24  1

x 20  1



x4  1



roots are 1,  1, i,  i



statement is false.

and

Sequence & Series

HWT - 6

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 

10

91





9

1

9

 10 1 9  10  1

10 

13  7

7

7

1  10 



1

7

 10  1021 . . . .  1012  7 9

 1  10  10  10 

7

 9 9 9 9 9 9 9

4

21

. . . . . 1012  7

 1111111

Not a prime number.

9.

In INE

10.

Maximum value of x a y b z c is when ax  by  cz



maximum of ab 2 c is when 2b  a  c



4b  4  1



a  2b  c  2

VMC/Sequence & Series



statement is false.

4

HWT-Solutions/Mathematics

Vidyamandir Classes Sequence & Series 8 2n  1

HWT - 7

2.

Answer is x 2  2 Ax  G 2  0

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 . 

positive always and less than 1.

1.

General term is

3.



Sum =

e x 1 e

x



1 e 1 x

Sequence & Series

HWT - 8

1.

1 1 2   1 x 1 x 1 x



they are in A.P.

2.

In objective worksheet

3.

In INE

6.

Series is 1.05  1.05  1.05  1.05  . . . . 1.05  2

3

4.

4

In INE

5.

In INE

49

 1.05 49  1  1.05 50  1.05   1.05    1.05  1  0.05  

7.



11.658  1.05 0.05



10.608 = 212.16 0.05

x 5 1 r Where 1  r  1 Sum =

x 5



r  1



0 x



x   0 , 10 

8.

(AM) (HM) = (GM)2

9.

Sum of progression 



 n  2 2



1  1 

and

x  10

x 1 5

(sum of first and last term)

n2  2  38 2

= 20(n + 2) and

20(n + 2) = 200



n + 2 = 10

VMC/Sequence & Series



n=8

5

HWT-Solutions/Mathematics

Vidyamandir Classes Sequence & Series 1.

HWT - 9

P1  17 , 21, 25, . . . . . 417 P2  16 , 21, 26 , . . . . . 466

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 

n

5k  2 k 2 k 4 4

As k and n both are integers k 2 I 4

k  4I  2 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

Sequence & Series 3.

This is a geometric progression with common difference 3 .



S10 

 3   1 2 35  1    3  1  3  1

 2 

10



2  242  3 1

2  242 



 121

4.

Harmonic mean =



2

6 2



3 1



2ab 2  product of roots   ab sum of roots



 2 coeff. of x 0   coeff. of x  5.

HWT - 10

   

S1  a  ar  ar 2  ar 3 . . . . .   20 S2  a 2  a 2 r 2  a 2 r 4  a 2 r 6 . . . . .   100

VMC/Sequence & Series

6

HWT-Solutions/Mathematics

Vidyamandir Classes S1  S2 

a  20 1 r a2

 a2     1  r 2     400  4 100  a2     1 r2   



1 r2



7.

Tn 

 100

1 r2

1  r 

r

3 5

2



n

T

r



n

r2 

r 1

r r 1

2

r 1



 8.



n  n  1 n

Sn 

1 r 4 1 r



4

2

1  n  n  1 2n  1 n  n  1     2  6 2  n  n  1  n  2  6

n!, 3  n! ,  n  1! are in G.P.



2  3 n!    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.

b 2  ac





2log b  log a  log c

and

a1 x  b1 y  c1 z



1 1 1 log a  log b  log c  k x y z

. . . .(i)

. . . .(ii)

Using (i) and (ii) we have 2  y  k  xk  zk



2y  x  z



x, y, z are in A.P.

VMC/Sequence & Series

7

HWT-Solutions/Mathematics