CHAPTER PRACTICE PROBLEMS
PROGRESSION AND SERIES
Q.1.
If {a, b, c, d  R+  1}, then find the minimum value of logd a + logb d + loga c + logc b.
Q.2.
If A, G, H are arithmetic, geometric and harmonic means between three given numbers a, b
and c, then the equation whose roots are a, b and c.
Q.3.
If an be the nth term of an A.P. and if a7 = 15, then the value of the common difference that
would make a2a7a12 greatest.
Q.4.
Let a > 0, find the values of parameter a, for which the following three numbers
2x + 2-x, a, 22x + 2-2x are in A.P.
Q.5.
If a, b, c are in H.P., b, c, d are in G.P. and c, d, e are in A.P., then prove that e 
Q.6.
If n arithmetic means are inserted between x and 2y and then between 2x and y and rth
x
mean in each case is equal then find the ratio of .
y
Q.7.
If 2(a2x4+b2y4) =c4, then find the maximum value of x2y2, in terms of a, b and c.
Q.8.
5

55
555 5555
Find the sum of series  


 .........upoto   .
2
3
4
(13)
(13)
 13 (13)

Q.9.
Show that the value of x for which log3 (21x + 3), log9 4, log27 (2x  1)3 form an A.P. is 1.
ab 2
2a  b2
.
Q.10. The rth, sth and tth terms of a certain G.P. are R, S and T respectively. Prove that Rst.Str.Trx
= 1.
1
1
1
Q.11. If a, , c and , q, are in A.P. of same common difference, then show that if a, q, c are in
p
b
r
1 1 1
A.P., then , , is also in A.P.
p b r
1
Q.12. If the common ratio of an infinite G.P. be less than , show that each term will be greater
2
than the sum of all the terms that follow it.
Q.13. Find the sum of all the numbers of form n3, which lie between 100 and 10000.
Q.14. If cos ( – ), cos  and cos ( + ) are in harmonic progression,

show that cos  sec   2 .
2
Q.15. Find the sum of the series 1.1 + 2.3 + 3.6 + 4.10 + 5.15 + 6.21 + ……… n terms.
Q.16. If a, b > 0 (a > b) and n is a positive integer, prove that an – bn  n ab 
n 1
2
a  b  .
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Chapter Practice ProblemsP & S2
Q.17. If x, y, z  R+ are in H.P., prove that z.ex – y + x.ez – y 
Q.18. Prove that
2xz
.
y
ab
c b

 4 , if a, b, c are in H.P. (Given that a, b, c >0).
2a  b 2c  b
Q.19. If 1, logy x, logz y and –15logx z are in A.P then prove that x 
1
 z3 .
y
1
and an even number of AMs are inserted between the them
6
such that their sum exceeds their number by 1. Find the number of means inserted.
Q.20. The sum of two numbers is 2
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Chapter Practice ProblemsP & S3
SOLUTION
1.
2.
Applying A.M.  G.M.
logd a  logc b  loga c  logb d
loga logb logc logd
4



4
logd logc loga logb
Hence logd a + logb b + loga c + logc d  4.
1 1 1
 
abc
1
A=
, G = (abc)1/3 and  a b c
3
H
3
now, a + b + c = 3A, abc = G3 and ab + bc + ac =
3G3
H
now, equation is x3  (a) x2 + (ab) x  abc = 0
3G3
then equation will be x3  3Ax2 +
x  G3 = 0.
H
3.
Let d be the common difference
a2a7a12 = (15  5d), 15 (15 + 5d)
= 375 (9  d2)
R.H.S. is greatest when d = 0
 a2a7a12 will greatest when d = 0.
4.
Since the numbers are in A.P.

2x + 2-x + 22x + 2-2x = 2a
x
Let 2 = y
y+
1
1
+ y2 + 2 = 2a 
y
y

 y 

1 
  y 
y  
2
1
 = 2(a + 1)
y 
 y > 0  y + 1/y  2
 2 + 4  2a + 2  a  2.
5.
6.
2 1 1
  , c2 = bd, 2d = c + e
b a c
2
1
1
2a 2b
ab
ab 2
c2
e = 2d – c = 2
–c= 

=


.
2
b
b  2 1 2  2 1 
( 2a  b) 2a  b 2

2
a

b

  
  
b a
b a
rth arithmetic mean between x and 2y
 2y  x 
= x + r

 n1 
rth arithmetic mean between 2x and y
 y  2x 
= 2x + r 

 n1 
y  x 
From (1) and (2), x = r
n1
−
r)
x
=
ry
 (n + 1
…(1)
…(2)
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Chapter Practice ProblemsP & S4

7.
x
r

y n  1 r
Applying the inequality, A.M.  G.M.
We have
2a 2 x 4  2b 2 y 4
 4a 2 b 2 x 4 y 4
2

c4  4|ab|x2y2

x2y2 
c4
4 | ab |
Hence, maximum value of x2y2 is
8.
Sn = 5 
c4
4 | ab |

1 9
99
999 99999


 ........ 
 
2
3
4
9  13 (13)
(13)
(13)

=

5  (10  1) (100  1) (1000  1)


 ........ 

2
3
9  13
(13)
(13)

=
  1

5  10 100 1000
1
1


 ........   


 ........  

2
3
2
3
9  13 (13)
(13)
(13)
  13 (13)
 
 10   1  
5  13   13  
= 


10  
1 
9 
1

1

 


13  
13  
=
5  10 13 1 13 

 
 
9  13 3 13 12 
5  10 1 
 

9 3 2
5 
1
=
10  
9  3 
4
5 39 65


=
.
93 4
36
=
9.
2. log9 4 = log3 (21x + 3) + log27 (2x  1)3
making base 3,
2
3
log 4 = log (21x + 3) + log (2x  1)
2
3
2

4 =   3  (y  1), where y = 2x
y

4y = (3y + 2) (y  1)
 3y2  5y  2 = 0
1
y = 2,
3
1
x
2 = 2,
3
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Chapter Practice ProblemsP & S5
1
rejected then x = 1.
3
10.
11.
12.
13.
Let the common ratio be taken as k and a be the first term
R = Tr = akr1
Rst = ast. k(r1)(st)
Str = atr. k(s1)(ttr)
Trs = arx. k(t1)(rs)
Multiplying all three, we get
Rst. Str. Trx = 1.
1
1
1
1
Let d =
a=c
=q
= q
b
b
p
r
1 1
by (1) and (4), q  a = 
… (A)
r b
1 1
by (2) and (3), q  c = 
… (B)
p b
Adding equation (A) and (B)
1 1 2
2q  (a + c) =  
r p b
 if a, q, c are in A.P.
1 1 1
then, , , in A.P.
p b r
Tp > Tp+1 + Tp+2 + ……… 
ar p
arp1 >
1 r
where r < 1 or 1  r = positive
multiply the inequality (1) by quantity 1  r and cancel arp1,
1r>r
1
r< .
2
The first term will be 53 = 125
Now 203 = 400  20 = 8000
(21)3 = 441  21 = 9261
 then series will be 53 + 63 + 73 + ……… + (21)3
then, Sn = (13 + 23 + ……… + 213)  (13 + 23 + ……… + 43)
3
 n(n  1)   k(k  1) 
= 
 

 2   2 
= (21  11)2  (2  5)2
= 53261.
14.
2
 cos ( – ), cos  and cos ( + ) are in H.P
2 cos    cos    2 cos 2   sin 2 
 cos  =

cos     cos   
2 cos . cos 
or cos2  . cos  = cos2  -sin2 
or sin2  = cos2  (1 –cos )


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Chapter Practice ProblemsP & S6
sin2 
or cos2  =

1  cos 
or cos2  = 2 cos 2
or cos . sec
15.


cos2
2
2
2 
2 sin
2
4 sin2


, or cos2  . sec 2
=2
2
2

=  2.
2
Sn = 1 + 2.3 + 3.6 + 4.10 + 5.15 + 6.21 + ……… n terms
= 1.1 + 2(1 + 2) + 3(1 + 2 + 3) + 4(1 + 2 + 3 + 4) + ……… + n(1 + 2 + …..+ n)
 r (r  1) 
 Tr = r 

 2 
n
Sn =
n

Tr 
r 1
=
r 2 (r  1) 1 n 3 2

(r  r )
2
2 r 1
r 1


2
1  n(n  1) 
n(n  1)( 2n  1) 
1  n2 (n  1)2 n(n  1)(2n  1) 


 = 
 

2  2 
6
2
4
6


n(n  1)
n(n  1)
[3n(n + 1) + 2(2n + 1)] =
(n + 2)(3n + 1).
24
24
b b2
b n1
Let us consider numbers 1, , 2 ,............... n 1
a a
a
 AM GM
b b2
b n 1
1/ n
1   2  ....  n 1 
b b2
b n 1 
a
a
a

 1. . 2 .... n 1 
n
a 
 a a
n
b 
1/ n
n n 1
1/ n
1   n 


1 2  ... n 1
n
n


a


1 a b
a
b 2 
    b 
   
or
or 


 
n



  b    a 
n
a

b

 a
  a 
n 1    


  a 
=
16.


1
an  bn
b
or  n 1
 
n a a  b   a 
 an – bn  n ab 
17.
n1
2
n 1
n 1

2

or a  b  a  ba
n
n
n 1
b
2
. n 1
a 2
a  b  .
2xz
(given)
xz
We have to prove
 z ex – y + x ez – y  x + z
ze x  y  x ez  y

1
xz
Applying weighted A.M.  Weighted G.M. on ex – y and ez–y and having corresponding weights
as z and x
y=

ze x  y  x ez  y  x  y
 e

xz

 e  
z
zy x
1
xz
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Chapter Practice ProblemsP & S7

 e
 e 
 e xz  yz  xz  xy

1
x z
1
2 xz  y x  z  x  z

1
0 xz
18.
1
as 2xz = y(x + z) from (1)
Since a, b, c are in H.P.
1 2 1 2c  b
1 1 2
      
a c b
a b c
bc
1 2 1 2a  b
and
  
c b a
ab
From (1) and (2) , we have
c b
a b
ac  b  ca  b 



2c  b 2a  b
bc
ab
c c a a c a c a
=    =     
b a b c a c  b b
1  2ac 
c  a
>2+ 
 >2+ 

b b 
 b 
2 + 2.
19.
. . . . (1)
. . . . (2)
(since GM  HM)
logy x = 1 + d  x = y1+d
logz y = 1+ 2d  y = z1+2d
–15 logx z = 1 + 3d  z = x
from equation (1) and (2)
 x = z(1+d)(1+2d)
From equation (3)
……(1)
……(2)
 1 3 d 
 

 15 
where d = common difference
……(3)
 15 


 15
1  3d
 6d3 + 11d2 + 6d + 16 = 0  d = –2
From equation (1) and (2)
1
 x = = z3
y
 x = z 1 3 d   (1 + d)(1 + 2d) =
20.
Let n (being even) AMs inserted between the a and b,
 a, A1, A2,…., An, b are in AP and (n + 2) terms are there altogether.
Now since a + b = A1 + An = A2 + An-1 = …. constant .
Also A1 + A2 + …. + An = n +1 (given)
 13 
a b
 = n + 1  13n = 12n + 12  n = 12.
 n
 = n + 1  n 
 2 
 6. 2 
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