1.
If a, b, c, d are distinct integers in A. P. such that
d = a2
2.
r
–7
p
4 3
(b)
p
–7 < 4 3
r
B = b 2 c + b c 2 – a 2 b – a b2 and
C = a c 2 + a 2 c – c b 2 – c 2 b,
6.
where a > b > c > 0 and the equation A x 2 + Bx + C = 0 has
equal roots, then a, b, c are in
(a) A. P.
(b) G. P.
(c) H. P.
(d) None of these
If a, b, c are in G. P. and log a – log 2b, log 2b – log 3c and log
3c – log a are in A. P., then a, b, c are the sides of a triangle
which is
(a) acute angled
(b) obtuse angled
(c) right angled
(d) None of these
7.
If a1 , a2 , a3 , ..................., an are in H. P., then
(c) all p and r
(d) no p and r
In a geometric series, the first term = a, common ratio = r.
n
If Sn denotes the sum of n terms and Un =
Sn ,
n 1
4.
Suppose A, B, C are defined as
2
A = a b + a b2 – a 2 c – a c 2 ,
c 2 , then a + b + c + d is
(a) 0
(b) 1
(c) 2
(d) None of these
If p, q, r are three positive real numbers in A. P., then the
roots of the quadratic equation px2 + qx + r = 0 are all real for
(a)
3.
b2
5.
then rSn + (1 – r) U n equals
(a) 0
(b) n
(c) na
(d) nar
The line x + y = 1 meets x-axis at A and y-axis at B,
P is the mid point of AB,
P1 is the foot of the perpendicular from P to OA;
a2
a1
a2
,
, ............,
a3 ..... an a1 a3 ..... an
a1
a2
an
are in
..... an –1
(a) A. P.
(c) H. P.
(b) G. P.
(d) None of these
M1 is that of P1 to OP; P2 is that of M1 to OA;
If positive numbers a –1 , b –1 , c –1 are in A. P., then product
of roots of equation
M2 is that of P2 to OP; P3 is that of M2 to OA; and so on.
If Pn denotes the nth foot of the perpendicular on OA; then
OPn is
x 2 – kx + 2b101 – a101 – c101 = 0 (k R)
(a) has + ve sign
(b) has – ve sign
(c) equal to zero
(d) not definite
(a)
1
2
n –1
(b)
1
2
n
8.
9.
If a x = b y = c z , where x, y, z are unequal positive numbers
3
and a, b, c are in G. P., then x + z 3
(a)
(c)
1
2
n 1
MARK YOUR
RESPONSE
a3 + c3
(c) > 2 y3
(d) None of these
2 y3
(b)
(d) None of these
1.
2.
3.
4.
6.
7.
8.
9.
5.
10. Let a1 = 0 and a1 , a2 , a3 , ......... , an be real numbers such
that | ai | = | ai –1 1| for all i, then the AM of the numbers
x2
x4
+
+ .........
4 +
2
1– x
1 – x8
1– x
to infinite terms, if | x | < 1 is
a1 , a2 , a3 , ......... , an has the value A where
(a) A < –
(c) A
11.
1
2
(b) A < – 1
1
2
–
(d) A = –
1
2
Ar ; r = 1, 2, 3, ........... , n are n points on the parabola
y 2 = 4x in the first quadrant.
(a)
(c)
n 1
–2 2
( 2)n
1
(b)
2n
(d)
n
22
12. ABCD is a square of length a, a
1
( ALn 2
1 x
(d) 1
1– x
15. If a, b, c, d are positive real number such that
a + b + c + d = 2, then M = (a + b) (c + d) satisfies the relation:
(a) 0 < M
(b)
N, a > 1. Let L1 , L2 ,
(b)
(c)
(d)
(n 1) c1/ n
(c)
2nc1/ n
(d)
(n 1) (2c)1/ n
1
3
, then the value of a is
2
(b)
2 2
1 1
–
2
3
a2 n
1
a2 n
1
(a)
(d)
a1
a
+ 2n
a1
a2 n
1
2 3
1
1
–
2
2
19. If
6n 5)
a2 a3
a2
=
a1a 4
a1
a1 )
(b)
2
an
2
an
is equal to
n(n 1)
2
(d) None of these
a3
a – a3
=3 2
a4
a1 – a4
in
(a) A. P.
(c) H. P.
4
an
a2
+ ........ +
a2
an
n(n 1) . a2 a1
2
an 1
(c) (n + 1) (a2
n(n3 1)
8
MARK YOUR
RESPONSE
4
18. If a1 , a2 , a3 ,........., a2n 1 are in A. P. then
4n 2 10 n 8)
8
n2 (2n2
M
(b)
(c)
1) (2n 1)
3
n ( n3
(d) 3
n(2c)1/ n
If a < b < c and a + b + c =
3
13 + 3. 23 + 33 + 3. 43 + 5 + ................ is (n is even)
n( n
2
17. Suppose a, b, c are in A. P. and a 2 , b 2 , c 2 are in G. P..
1
a (a – 1) (4a – 1)
2
1
(a – 1) (2a – 1) (4a – 1) (d) None of these
2
13. Sum to n terms of the series
(a)
M
(a)
(a)
(c)
2
3
(b) 1
16. If a1 , a2 , ......, an are positive real numbers whose product
is a fixed number c, then the minimum value of
Ln M n2 ) is equal to
1
a(a –1)2
2
1
M
(c) 2
n 1
(a)
1
1– x
a1 a2 + .............. + an –1 2an is
= L2 L3 = ...... = 1 and M1 , M 2 , M 3 , ................ be points on
CD such that CM1 = M1M 2 = M 2 M 3 = ...... = 1.
Then
(b)
(c)
L3 , ............... be points on BC such that BL1 = L1L2
a –1
x
1– x
(a)
If Ar = ( xr , yr ) , where x1 , x2 , x3 , ..............., xn
are in G. P. and x1 = 1, x2 = 2, then yn is equal to
x
14. The sum of the series
then a1 , a2 , a3 , a4 are
(b) G. P.
(d) None of these
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
21.
th th
If p , q and r th terms of both an A. P. and a G. P. be
a
b
respectively a, b and c, then
b
c
(a) 2
(c) abc
(b) 1
(d) pqr
r 0
a
b
c
is
27.
quadratic equation whose roots are a and b is
mnx 2 + (m + n – 2mn )x + mn – m – n + 1 = 0
(b)
mnx 2 – (2mn – m – n)x + mn – m – n + 1 = 0
(c)
22.
23.
24.
mnx + (2mn + m + n)x + mn + m + n + 1 = 0
28.
2
(d) mnx – (2mn + m + n)x + mn + m + n + 1 = 0
One side of an equilateral triangle is 24 cm. The mid points of
its sides are joined to form another triangle whose mid points
are in turn joined to form still another triangle. This process
continues indefinitely. The sum of the perimeters of all the
triangles is
(a) 144 cm.
(b) 169 cm.
(c) 400 cm.
(d) 625 cm.
a be y
c de y
b ce y
=
=
, then a, b, c, d are
a – be y
c – de y
b – ce y
(a) in A. P.
(b) in G. P.
(c) in H.P.
(d) Equal
In a sequence of (4n + 1) terms, the first (2n + 1) terms are in
A. P., whose common difference is 2, and the last (2n + 1)
terms are in G. P. whose common ratio is 0.5. If the middle
terms of the A. P. and G. P. are equal then the middle term of
the sequence is
(a)
25.
2
n2
n 1
n.2
(b)
2n – 1
29.
(a) 2
(c) 2
MARK YOUR
RESPONSE
a
x
c
y
b
x
b
y
(d)
16
n arithmetic means are inserted between two sets of numbers
(a) n – m + 1 : m
(b) n – m + 1 : n
(c) m : n – m + 1
(d) n : n – m + 1
n
Let S k = lim
n
n
1
0 (k
i
1)i
kSk equals
. Then
k
1
(a)
n(n 1)
2
(b)
n(n 1)
2
(c)
n( n
(d)
n(n 3)
2
2)
2
The sum of
(a) 1 –
(c) 1
30.
1
2
3
3 1
1
1
4
5
+
+ ......... to n
. +
.
.
1 .2 2
2.3 2
3 .4 2
1
(b)
( n 1)2 n
1
n
2
(d) None of these
n.2n 1
In the given square, a diagonal is drawn, and equally spaced
parallel line segments joining points on the adjacent sides
are drawn on both sides of the diagonal. The length of the
segments be
is equal to
1
2
cm, then the sum of the lengths of all
possible line segments and the diagonal is
(b) – 4
(d) 4
(a) n (n + 1 ) 2 cm
(b)
n 2 cm
(c) n (n + 2 ) cm
(d)
n 2 2 cm
20.
21.
22.
23.
24.
25.
26.
27.
28.
29.
30.
1
n 1
diagonal is n 2 cm. If distance between consecutive line
(c) n.2 n
(d) None of these
If a, b, c are in G. P., x and y be the A. M.’s between a, b and
b, c respectively, then
(b) 4
(c) 8
terms is equal to
n 1
22 n
(a) 2
R. Suppose that mth mean
between these two sets of numbers is same, then the ratio
a : b equals
r 0
(a)
If log 2 (a + b) + log 2 (c + d) 4, where a, b, c, d are positive
numbers. Then the minimum value of the expression
a + b + c + d is
a, 2b and 2a, b, where a, b
b r , where 0 < a, b < 1, then the
ar , n =
If m =
c
a
26.
31. ABC is a right angled triangle in which
B = 90° and
BC = a . If n points L1 , L2 , ............ , Ln on AB is divided in
n
r
n + 1 equal parts and L1 M1 , L2 M 2 ,............., Ln M n are
AC, then the sum of the lengths of L1 M1 , L2 M 2 , ...........
(c)
(b)
a(n 1)
2
an
2
(d) None of these
r
(c)
33. If
(b) n (4n 2 1)
(c) n (4n 2
(d) None of these
n
2
m2
1
n(n 1)(n 2)(n 3)
MARK YOUR
RESPONSE
(b)
(d)
1
m
= an 4 + bn 3 + cn 2 + dn + e,
1
(a)
a
b
b
c
c
d
d
e
4. a
e
(b)
a
b
c
d
b
c
d
e
4.
a
b
c
d
e
+
+
+
+
b
c
d
e
a
(d) All three correct
(c)
(a)
(c)
k
39. If a1 , a2 , a3 , a4 are in H.P, then
1
=
– f (n). Then f (n) is equal to
18
(n 1)(n 2)(n 3)
when n is even
then a + b + c + d =
(a) 10
(b) 6
(c) 3
(d) 1
38. If a, b, c, d are four positive numbers, then
1
1
1
+
+ ....... to n terms
+
2.3.4.5 3.4.5.6
1 .2 .3 .4
(a)
N
(c) f (n) – 16 f
k
1
n
f (n) +
. Then f (n) is equal to
48
16(2n 1)
n
1
n 1
when n is odd
2
n
(a) n (n + 1) (2n + 1)
34. If
(2r 1)4 is equal to
r
(b) f (n) – 16 f
(d) None of these
6n 5)
n
(d) None of these
1
2
14
34
24
n4
+
+
+ ............ +
1 .3 3 . 5 5 .7
(2n 1)(2n 1)
=
a
(d) None of these
1
37. If
3
2
a nx
x
(b)
(a) f (2n) – 16 f (n) for all n
2
4
1
3
+
+ ........... is :
+
+
1.3 1.3.5 1.3.5.7 1.3.5.7.9
is
a (r 1) x
a nx
r 4 = f (n). Then
36. Let
32. Sum to infinite terms of the series
(b)
1
n( a nx – a)
x
n
a(n 1)
2
(a) 1
a
(c)
Ln M n is
a rx
n
(a)
line segments parallel to BC and M1 , M 2 , ....... M n are on
(a)
1
35. The value of
1
3(n 1)(n 2)( n 3)
n
3(n 1)(n 2)( n 3)
x2
2 x 15 = 0
2
a
e
5
3
1
a1a4
r 1
(b)
x2
ar ar
2
1
is a root of
2 x 15 = 0
(c) x
6x 8 = 0
(d) x 9 x 20 = 0
40. The sum of the integers lying between 1 and 100 (both
inclusive) and divisible by 3 or 5 or 7 is
(a) 2838
(b) 3468
(c) 2738
(d) 3368
31.
32.
33.
34.
35.
36.
37.
38.
39.
40.
41.
42.
If three successive terms of a G..P. with common ratio r (r > 1)
form the sides of a ABC and [r] denotes greatest integer
fuction, then [r] + [– r] =
(a) 0
(b) 1
(c) – 1
(d) None of these
Through the centroid of an equilateral triangle a line parallel
to the base is drawn. On this line, an arbitrary point P is
taken inside the triangle. Let h denote the distance of P from
46.
47.
the base of the triangle. Let h1 and h 2 be the distances of
P from the other two sides of the triangle,then
43.
(a) h is the H.M. of h1 , h 2
(b) h is the G..M. of h1 , h 2
(c) h is the A.M. of h1 , h 2
(d) None of these
If x15
then
x11 x9
x13
x7
(a)
x16 is equal to 15
(b)
x16 is less than 15
(c)
x16 is greater than 15
x5
x3
48.
n
ar
ak ,
49.
r 1
45.
50.
If x1 , x2 .....xn 1 , xn be the roots of
(b)
(c)
(d)
(b)
pq
(c)
p2
q2
p '2 q '2
(d)
p2
1 y1
2
1
1 y2
1
1 y1
1
1 y2
1
........
1 yn
1
1 yn
51.
1
1
1 yn
23.3
...........is
4.5
(c)
2n 1
1
n 2
(d) none of these
2n
1
1
ab
1
bc
1
is
ca
(b) 9
(d) none of these
The sum of the series
9
2
13
17
3
4
5 .2.1 5 .3.2
to infinite terms is equal to
(a) 1
(b)
9
5
1
5
(d)
2
5
The series
8
5
16
65
16
5
(b) 2
(c) 4
5 .4.3
......... up
24
.......... up to infinity has the sum
325
(d) 5
41.
42.
43.
44.
45.
46.
47.
48.
49.
50.
51.
n 1
n 2
n arithmetic means are inserted between the numbers 7 and
49. If the sum of these means be 364, then the sum of their
squares is
(a) 10380
(b) 11380
(c) 11830
(d) 18130
If a, b, c are positive numbers such that a + b + c = 1, then the
(a)
2(1 y1 )(1 y2 )........(1 yn )(1 yn 1 )
MARK YOUR
RESPONSE
22.2
3.4
equal to
1
q '2
(b)
(c)
1
1 yn
q2
2n 1
n 2
0 then (1 x1 )(1 x2 )......(1 xn )
........
p '2
(a)
(1 y1 )(1 y2 )........(1 yn )(1 yn 1 )
1
p 'q '
The sum to n terms of the series
0 and y1 , y2 ,....., yn , yn 1 be
1 x x 2 ...... x n 1
is equal to
(a)
p '2 q '2
(a) 3
(c) 27
(b) G.P.
(d) none of these
1 x x 2 ...... x n
those of equation
q2
minimum value of
a
a1
a
a
, 2 , 3 ........ n are in
f (n )
f (1) f (2) f (3),
(a) A.P.
(c) H.P.
p2
x = 7 , where x > 0,
If a1 , a2 , a3 ....... are in H.P. and f ( k )
then
(a)
2
2.3
(d) Nothing can be said regarding the value of x16
44.
If a, b, c, are in A.P. and p, p ' are respectively A.M. and G.M.
between a and b while q, q ' are respectively AM.and G.M.
between b and c, then
52. Let (al , a2 , a3 ....) be a sequence such that a1
2 and
58. The value of
20
an
an
1
2n for all n
2. Then
ai is
i 1
(a) 2960
(c) 3560
(b) 3080
(d) 4120
53. If w, x, y, z
R
3w 5 x 7 y 9 z
and 256 xywz
(w x
(a)
y
z )4 and
24. Then roots of the equation
( w x)t 2 zyt
x y
(a) real and integer
(c) imaginary
r
54. The sum of the series
r 0
n2
(b)
2n
2
n2
1
x3 y 3 z
( y2 z 2
3
x, y, z are in
(a) A.P.
(c) H.P.
57. Sum of n terms of series
25 x 2 y 2 )
5
y
3
,
z then
1
if ab = and(a b)
6
(b)
(c)
4c
(d) 6c
( 1)n .
(a)
( 1)n
( 1) n
(n 1)! (n 1)!
(c)
( 1)n 1
(n 1)!
(c)
3n
n2
( 1)n 1
(n 1)!
3c
n 1
is
n!
(b)
( 1)n
(n)!
( 1)n
(n 1)!
1
(d)
( 1)n
(n)!
( 1)n 1
(n 1)!
1
1 5 19 65 ...
1
2n
2
1
1
3n 1 2n 1 1
2
n
terms, then
(b)
(d)
3n
1
3n 2
2n
2
2
n
(1 n 2n 2 )
6
(b)
(c)
n
(1 2n 2n 2 )
6
(d) none of these
1
2n 3 1
2
n
Ck Sk is equal to
k 0
n
(1 2n)2
6
(a)
2n 2 [na1
(c)
2[na1 Sn ]
Sn ]
(b)
2n [a1 Sn ]
(d)
2n 1[a1 Sn ]
52.
53.
54.
55.
56.
57.
58.
59.
60.
61.
62.
0)
62. If a1 , a2 , a3 ,..., an are in A.P. with Sn as the sum of first ‘n’
1
, is
3
(a)
MARK YOUR
RESPONSE
(d) infinite
61. Sum Sn
(b) G.P.
(d) none of these
... (a (n 1))(b ( n 1))
1
2
2c
(a)
ab (a 1)(b 1) (a 2)(b 2)
(b)
60. Sum Sn where Tn
4n
15
x
3
2
is
r)
(a)
2
9 x2 z 2
r 2 1)(r 2
in H.P., then the value of a b is (given a, b, c
(c)
(d)
n 3n
4n
2n
55. The positive integral values of n such that
1.21 + 2.22 + 3.23 + 4.24 + 5.25 + ... + n.2n = 2(n+10) + 2 is
(a) 313
(b) 513
(c) 413
(d) 613
56. If
4
59. If log( a c ), log( a b), log(b c) are in A.P. and a, c, b are
is equal to
(a)
r 1 (r
(r 2 1)3
(c) 0
z w 0 are
(b) real and irrational
(d) none of these
( 1)r (n 2r )2 , (where n is even)
r3
PASSAGE-1
2
term I n is given by I n =
Suppose a series of n terms is given by
Sn
t1 t2
0
t3 ................. tn
Then Sn 1
4.
t1 t2 t3 ................. tn
Subtracting we get, Sn Sn
1
= tn , n
1, n > 1
2
5.
Further if we put n = 1 in the first sum then S1 = t1
Thus we can write tn
Sn
Sn
2 and t1
1, n
If the sum of n terms of a series is a.2n
2.
S1
(a) a
(b)
6.
(d)
k tn for all n
tn
2
then a1 a2
a3 .... an is
(a)
2n 1
(c)
n
(n 1
2
(2n 1)
(b)
(d)
)
n
A.P., as well as of a G.P., then the value of x y z , y z x , z x y
is
(a) 1
(b) – 1
(c) 0
(d) 2
N,
tn
k for all
0 , then again the sequence given above is
9..
Now consider a sequence I1 , I 2 , I 3 ,................I n where the nth
MARK YOUR
RESPONSE
sin(2n 1) x
dx
sin x
b(b 0), log b a1 , log b a2 ,..., logb an are in A.P. with common
difference logbr.
7.
If x, y, z are respectively the pth, qth and the rth terms of an
N , where k is a constant
1
0
are in G.P. with common ratio bd. If a1 , a2 ,..., an are positive and in
G.P. with common ratio r, then for any base
8.
Further, the sequence represents an A. P if tn 1
2tn
)
difference d, then for any b (> 0), the number b a1 , b a2 , b a3 ,...., ban
Consider a sequence of n terms : t1, t2, t3, .....................tn
If tn
A. P
n
(2n 2
4
equal to
PASSAGE-2
n
(d)
in A.P. and vice-versa. If a1 , a2 ,..., an are in A.P. with common
n(4n 2 1)
This sequence represents a G. P if tn 1
where k is a constant
2n 1
1 1
1
We know that, if a1 , a2 ,..., an are in H.P., then a , a ,..., a , are
1 2
n
1
n (n + 1) (n + 2) (n + 3) then the nth term of the series is
4
(a) n (n +1) (n + 2)
(b) (n + 1) (n + 2) (n + 3)
n(n 1) (2n 1)
6
(b)
PASSAGE-3
a
2
1
2
(d)
a
a
The sum of n terms of a series is given by
(c)
(2n 1)
If a sequence a1 , a2 , a3 ,...., an has general term given by
an
b then the sum
(c)
3.
4
(c) n (n + 1)
1
is
t
2 r
r
The sequence I1 , I 2 , I3 ,.............., I n represents
(a) A. P.
(b) G. P.
(c) A. P. and G. P. both (d) Neither A. P. nor G. P.
The sum of n terms of the sequence is
(a)
The sum of n terms of a series is a.2n – b, where a and b are
constants then the series is
(a) A. P
(b) G. P
(c) A. G. P
(d) G. P from second term onwards
1.
cos n x cos nx dx
If a, b, c, d are in G.P. and
are in
(a) A.P.
(c) H.P.
ax
by
cz
d v , then x, y, z , v
(b) G.P.
(d) none of these
1
1
1
If a, b, c are in H.P., then 4 a , 4 b , 4 c are in
(a) A.P.
(b) G.P
(c) H.P.
(d) none of these
1.
2.
3.
4.
6.
7.
8.
9.
5.
1.
Let ( , ) are the roots of x 2
the roots of x 2
of
1
2 x mx 2m
1
x
p
2 x mx 2m
Statement-1
: If
Statement-2
:
2.
0 and ( , ) are the roots
1
, (m, p, q, r
r
x q
0, ( , ) are
Let ai R {0}, i 1, 2,3,...n and all ai’s are distinct such
that
n 1
f ( x)
0)
n 1
ai2 x 2
i 1
then p, q, r are in A.P..
n
ai ai
2
1.
: If f ( x)
Statement-2
a1 , a2 ,...an are in G.P..
: For real
(a2 + b2 + c2 + 1) = 2(a + b + c + ab + bc + ca); then
(c)
2.
b
a
4.
1
. Then
64
(a) a = b =
1
1
,c=
4
2
(b) a = b =
1
1
,c=
4
2
5.
(d) a + b + c = 1
1
3
c2
ab bc ca
then
0
(b)
1
,1
3
(d)
1
,3
3
th
If the first and (2n –1) terms of an A. P., a G. P. and a H. P..
in G. P. is S 2 . If their sum is
interval
1.
S, then
2.
2
(a) a = b = c
(c) b2 = ac
For a positive integer n,
let a (n) = 1 +
The sum of squares of three distinct real numbers, which are
MARK YOUR
RESPONSE
b2
x R
th
of positive terms are equal and their n terms are a, b, c
respectively, then
1
(c) a = b = c =
3
3.
0,
(c) (1, 3)
(d) a, b, c are equal
a, b, c are three positive numbers and ab c 2 has the greatest
value
(a)
(b) a + b + c = 3
c
b
0 for some
2.
If a, b, c are non zero real numbers such that
(a) a = 1
ai2
i 2
Statement-1
2
1.
x
i 1
a, b, c; a 2
MARK YOUR
RESPONSE
1
1
2
(a) a (100) < 100
(c) a (200) > 100
can lie in the
3.
4.
1
3
(b) a b c
(d) a + c = 2b
1
1
......
. Then
n
4
(2 ) – 1
(b) a (200) < 200
(d) a (2010) > 1005
5.
6.
Let S1 , S2 ,.... be squares such that for each n
1, the
length of a side of Sn equals the length of a diagonal of
Sn
7.
8.
9.
1 . If the length of a side of
2
(a)
y
(c)
x y
y z
+
2y x
2y z
(b)
xz
4
xy
yz
2 xz
(d) None of these
13.
(c) r
14.
(b)
(d)
mr
m r
(c)
8
3
15.
8
3
(b)
(d) 8
16.
If Sn denotes the sum to n terms of series
cot
1
7
cot
4
1 19
4
cot
1 39
4
1
2
(d)
S
2
( 1)r 1 3r
(r 1)!
1
( 1)n 3n (n 2)
(n 1)!
If a, b, c, d are four unequal positive number which are in
A.P. then
(a)
1
a
1
d
1
b
1
c
(b)
1
a
1
d
1
b
1
c
(c)
1
a
1
d
1
b
1
c
(d)
1
b
1
c
4
a d
If a1 , a2 ,...., an are in A.P. with common difference d, then
1
1 a1a2
d
1
cot
1 a2 a3
d
(a)
tan
1
an
tan
1
(c)
cot
1
a1 cot
1
1
cot
1
1 an an
d
1
1 a3 a4
d
1
is equal to
a1
(b)
an
(d) none of these
cot
a1 cot
1
an
If three positive unequal numbers a, b, c, are in H.P., then
(a)
a c
(c)
a2
(b)
2b
c2
2ac
a2
c2
2b 2
(d) none of these
If a, b, and c are three terms of an A.P. such that a
b, then
b c
may be equal to
a b
...., then
(a)
(a)
Sn
tan
1
4n
2n 5
(b)
Sn
cot
1
2n 5
4n
(c)
Sn
cot
1
4n
2n 5
(d)
S
cot
1
1
2
MARK YOUR
RESPONSE
( 1)r 1 3r
(r 1)!
... cot
In a G.P., the product of first four terms is 4 and the second
term is the reciprocal of the fourth term. The sum of infinite
terms of the G.P. is
(a) – 8
( 1)r 1 3r 1
(r 1)!
Sn
cot
m
2
( 1)r 3r
r!
(c)
If (m 1)th, (n 1)th and ( r 1)th terms of an A.P. are in G.P..
n
2
Tr
(b) Tr
All the terms of an A.P. are natural numbers and the sum of
the first 20 terms is greater than 1072 and less than 1162. If
the sixth term is 32 then
(a) first term is 12
(b) first terms is 7
(c) common difference is 4 (d) common difference is 5
(a)
11.
(a)
the following values of n is the area of Sn less than 1 sq.
cm.?
(a) 10
(b) 9
(c) 8
(d) 7
If x, y, z are positive numbers in A.P., then
( 1)r 1 (r 2 r 9)3r 1
, then
(r 1)!
If Tr
S1 is 10 cm. then for which of
and m, n, r are in H.P., then the ratio of the first term of the
A.P. to its common difference is
10.
12.
17.
(b)
2
3
(c) 1
(d) 3
If 3 positive real number a, b, c are in A.P with abc = 4 then
[b] can be equal to (where [.] represents the integral part)
(a) 1
(b) 2
(c) 3
(d) 4
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
1.
Column-I
(A) If three real number a, b, c are in A.P. and (1 a),(2 b),(1 c)
p.
are in G.P. then ac can be equal to
(B) Let x be the arithmetic mean and y, z be two geometric means
q.
1
r.
3
s.
4
n
(D)
lim tan
n
tan
r 1
1
1
2r 2
x
3
3
z
is equal to
2 xyz
(C) If a, b, c be three positive number which form three successive
terms of a G.P. and c > 4b – 3a, then the common ratio of
the G.P. can be equal to
between any two positive numbers, then
2.
is equal to
Column-I
Column-II
1 1 1
(A) If a, b, c be positive numbers then (a + b + c)
a b c
must be greater than or equal to
(B) If h be the H.M. and g be the G.M. of two positive
a
numbers a and b such that h : g = 4 : 5, then
can be equal to
b
(C) If S
1
n
and Sn 1
1
r
r
r 02
r 02
then n is greater than or equal to
and S – Sn + 1 < 10–3
n
(D) If (1 + x) (1 + x2) (1 + x4) (1 + x8)..........(1 + x128)
3.
Column-II
1
3
then n is equal to
Column 1
(A) If pth, qth , rth and sth terms of an A. P. are in G..P.
then p q, q r , r s
xr
(D) If the arithmetic mean of (b c) , (c d )
2 and
4
q.
9
r.
10
s.
255
p.
Column 2
are all equal
q.
are in A.P.
r.
are in G.P.
s.
are in H.P.
r 0
(B) If ln x, ln y, ln z ( x, y , z 1) are in G.P. then 2x + ln (ln x),
3x + ln (ln y), 4x + ln (ln z)
(C) If n !, 3 n ! and (n 1)! are in G.P. then n !, 5 n ! and (n 1)!
2
p.
2
(a b) is
the same as the arithmetic mean of (b c 2a)2 , (c a 2b)2 ,
(a b 2c) 2 then a, b, c
1.
MARK YOUR
RESPONSE
2.
3
4.
Column-1
(A) If a, b, c are the sides of any triangle ABC, s is
its semiperimeter and is its area then the minimum value of
s
2
2
2
s a
s a 1
s b
s b 1
s c
2
s c
p.
Column-2
1210
q.
60
r.
1529
s.
1681
1 is
(B) Let {a1 , a2 ,....} be a sequence such that
10
a1
1 and an
an
n2 n
1
2 then
i 1
ai is equal to
(C) If the largest term of the sequence
1
4
9 16
T
,
,
,
,... is T then
is equal to
503 524 581 692
49
3 5
7
9
1
... to n term then
(D) If Sn
is equal to
4 36 144 400
1 S40
4.
MARK YOUR
RESPONSE
1.
2.
3.
The integers x, y and z each are perfect squares and
x > y > z > 0. If x, y, z form an A.P. then the smallest possible
value of x is
Two consecutive numbers from 1, 2, 3,.........., n are removed.
If the arithmetic mean of the remaining numbers is 105/4
then n is equal to
The odd positive integers are arranged in a triangle as
follows
The sum of the numbers in the 10th row of this triangle is
equal to
4.
the limit lim
1
3
7
13
21
1.
MARK
YOUR
RESPONSE
23
x
5
9
15
11
17
25
The sum of infinite terms of the G.P. a ar ar 2 ...
,
where a is the value of x for which the function
7 2 x ln 25 5 x 1 52 x has the grteatest value and r is
5.
19
27
2.
29
3.
x
0 0
t 2 dt
x 2 tan(
x)
is equal to
The number of values of x such that x, [x] and {x} are in H.P.
(where [.] denotes the greatest integer function and {.}
denotes fractional part of x), is equal to
4.
5.
(c)
(a)
(c)
(b)
(c)
(b)
(c)
(b)
(c)
(c)
(c)
1
2
3
4
5
6
7
8
9
10
11
(d)
(c)
1
2
1.
12
13
14
15
16
17
18
19
20
21
22
(b)
(d)
(a)
(a)
(a)
(d)
(a)
(c)
(b)
(b)
(a)
(a)
(b)
3
4
23
24
25
26
27
28
29
30
31
32
33
(b)
(a)
(d)
(c)
(c)
(d)
(a)
(d)
(c)
(b)
(c)
(d)
(d)
5
6
(d)
1
2
3
1.
3.
4
5
6
(b,c)
(a,b,c,d)
(a,b,c)
7
8
9
(a,b,c)
(b,d)
(a, d)
A - r, s; B - q; C - p,s; D - q
A - r; B - q; C - q; D - p
1
(a)
(c)
7
8
2.
(a,b,c,d)
(b,d)
(b,c)
49
2
50
10
11
12
2.
4.
3
1000
(b)
(a)
(a)
(d)
(d)
(b)
(a)
(c)
(c)
(c)
(c)
34
35
36
37
38
39
40
41
42
43
44
4
(c)
(c)
(c)
(c)
(c)
(c)
(b)
(b)
(c)
(d)
(b)
56
57
58
59
60
61
62
(c)
(c)
(a)
(a)
(d)
(b)
(a)
(c, d)
(a, c)
(a, b, c)
16
17
(c, d)
(a,b,c,d)
45
46
47
48
49
50
51
52
53
54
55
(b)
9
(b)
(a,d)
(a,b,d)
(b,c)
13
14
15
A - p, q; B - p; C - r, s; D - s
A - q; B - p; C - q; D - s
3
5
2
1.
(c) Let the common difference of the given A. P. be t.
Then
d
a2
b2
c2
a + 3t = a 2
5t 2
Un =
(a t )2
(a 2t ) 2
3(2a –1)t 3a 2 – a
=
...(1)
0
t is real
D
(r – 1) U n =
9(2a –1)2 – 4(5)(3a 2 – a)
0
24a 2 16a – 9
1
3
0
1
3
70
<a<
12
4.
70
12
2
(b) We have (OM n –1 )2 = (OPn )2 + ( Pn M n –1 )
2
2
Also, (OPn –1 )2 = (OM n –1 )2 + ( Pn –1M n – 1 )
2
n –1
(rejecting both these values since t must be non-zero
integer)
(a)
4
5
t=1
0
( p r ) 2 – 16pr
r
p
2
r
–7
p
3.
(c)
Sn =
n
a(r –1)
r –1
=
1
2
n –1
=
1
22
=
1
2
n –1
n 2
...(1)
y
2
– 4pr
0 [from (1)]
p 2 + r 2 – 14pr
0
r
– 14
p
n
n
n
1
1
= .......... = n =
2
2
roots of p x 2 + qx + r = 0 are all real then
q 2 – 4pr
1
2
= 2 n2 +
n –1
2
OPn =
a+b+c+d =–1+0+1+2=2
p, q, r are in A. P.
2q = p + r
p r
2
r Sn + (1 – r) U n = an
= 2(OPn )2 = 2 n (say)
3
5
When a = – 1, from (1) : t = 1,
ar (r n –1)
– an
r –1
a = – 1, 0
a is integer]
When a = 0, from (1) : t = 0,
r (r n –1)
–n
r –1
a
r –1
(r – 1) U n = r S n – an
0
[
2.
a
n
{r + r 2 + ............ + r – n}
r –1
+1
r
–7
p
0
B
0
P
2
48
M1
M2
4 3
O
n
Un =
n
a
=
r –1
a(r n –1)
r –1
1
n
(r n –1)
n 1
5.
P3 P2
P1
A
x
(c) Given A = a2b + ab2 – a2c – ac2 = a2(b – c)
+ a(b2 – c2) = a(b – c) (a + b + c)
Similarly, B = b(c – a) (a + b + c);
C = c(a – b) (a + b + c)
Now, Ax2 + Bx + C = 0
2
(a + b + c) {a(b – c) x + b(c – a)x
+ c(a – b)} = 0 has equal roots
...(i)
Clearly 1 is a root of equation (i) so, other root is
also 1.
product of root of equation (i) = 1
c(a – b)
=1
a(b – c)
6.
b=
–1
(b) Given a –1 , b , c –1 are in A. P..
a, b, c are in H. P..
a101 c101
> ( ac )101 > b101
2
2ac
a c
a, b, c are in H. P..
(b) Given b2 = ac ( a, b, c are in G. P.) and
2(log2b – log3c) = loga – log2b + log3c – loga
[ given terms are in A. P.]
2b
3c
log
Now, a =
2
= log
cos A =
=
A > 90°
(c)
3c
2b
b=
3b
9c
b2
=
=
2
4
c
Now,
7.
8.
b
2
c –a
2bc
9c 2
4
c2 –
Product of the roots of given equation
=
9.
z
(c) Given a x = b y = c = k
2
a=
81 2
c
16
3c
2
c
2
b
a
= negative
1 1
,
,
a1 a2
1 , ..........., 1 are in A. P..
a3
an
a3 ..... an a1 a2
,
a1
a a
........... , 1 2
1+
......... , 1 +
a2
a1
a3 ..... an ,
a2
10.
(c)
1
,b= k ,c= kz
an2
1
x
= k
1
z
1
y
1
1
1
1
–
=
–
x
z
y
y
x, y, z are in H. P..
y = H. M. of x and z
xz > y
x3
z3
2
3
> ( xz ) > y3
3
x 3 + z > 2 y3
2
ai2 – ai –1 = 2 ai –1 + 1
1
– a12 = 2[ a1 + a2 + ......... + a n ] + n
a1 = 0 and
a a .... an
A= 1 2
]
n
A=
11.
a1
a2
,
, ............. ,
a3 ..... an a1 a3 ..... an
k
1
y
an2 1 = 2(nA) + n [
a1 a2 .... an –1
are in A. P..
an
are in A. P..
c
b
Put i = 2, 3, 4, ......., n + 1 and add. We get
a3 ..... an
a a ..... an
,
,1+ 1 3
a1
a2
an
are in H. P..
a2 ..... an –1
1
y
| ai | = | ai –1 1|
a3 ..... an a1 a3 ..... an
,
, ............. ,
a1
a2
an
a2
and
a3 ..... an
are in A.P..
an
a1 a2 ..... an –1
1
kx
a, b, c are in G. P..
a1 , a2 , a3 , ..................., an are in H. P..
a2
2b101 – a101 – c101
<0
1
triangle is obtuse.
a1 a2
A. M. > G. M. > H. M.)
2b101 – a101 – c101 < 0
3c
2
a is the largest side.
2
(
(c)
y
an2 1 – n
an2 1 1
–
=
2n
2n
2
–
1
2
2 x , being in the first quadrant. The sequence
of x-coordinate 1, 2, 4, 8, ................
the sequence of y-coordinate
2, 2 2 , 2 4 , 2 8 , ................ is a G.P. where the
common ratio is
2.
yn = 2( 2)n –1 = ( 2)n
1
12.
(b)
AL12 + L1M12 = (a 2 12 ) + {(a – 1)2 12 }
2
2
AL2 + L2 M 2 = (a
2
2
2 ) + {(a – 2)
2
=
2
2 }
1 x2
(1 x 2
n –1
n 1
–1
) (1 – x 2 n 1 )
......................................................................................
ALa –12 + La –1M a –12 = a 2 + (a –1) 2
2
a
Now, Sn =
La
M1
M2
1
+
+
+ 2 {12
1– x
= a(a – 1) a
2a – 1
1
= a(a – 1) (4a – 1)
2
2
15.
1
1– x 2
=
n
1
1
–
1 – x 1 – x 2n
3
3
2
2
M
1
0<M 1
Alternatively, Let a + b = x, then c + d = 2 – x
2
M = (a + b) (c + d) = x (2 – x) = 1 – ( x –1)
3
M
3
.......... + m }
16.
1
1.
(a) Given a1 a2 a3 ......... an = c
a1a2 a3 .....an –1 (2an ) = 2c
m2 (m 1)2
= m (2m 1) + 16.
4
2
A. M.
n
]
2
(a) The general term of the given series is tn =
( a b) (c d )
2
M = (a + b) (c + d) > 0.
+ 8 × 2{ 1 + 2 + 3 +
[Put m =
n
Lim x 2 = 0, as | x | < 1]
n
Also a, b, c, d are positive. Therefore
= {13 + 23 + 33 + 43 + ...... + (2m – 1)3 + (2m)3}
– {23 + 43 + ....+(2m)3} + 3{23 + 43 + 63 + ..... + (2m)3}
4
–
(a b)(c d )
M
+ 3(23 43 63 ...to m terms)
6n 5)
n 1
A. M.
We get
Then S2m = ( 13 + 33 + 5 + ...... to m terms)
n2 (2n2
+ .................
1 – x4
(a) Consider two real numbers (a + b) and (c + d). Using
G. M.
3
=
1
[
(d) We have Sn = 13 + 3. 23 + 33 + 3. 43 + 53 + ............
Let n = 2 m
2
n 1
1
1 – x2
1
1
–
1 – x 1 – x2
tn =
22 ..... (a –1) 2 }
(a –1) a (2a –1)
6
2
–
2
n
1
x
= Lim S n =
–1=
1– x
n
1– x
= (a – 1) a 2 + 3
2m(2m 1)
=
2
1 – x2
The sum to infinite terms
= (a – 1) a 2 + {12 22 ..... (a –1)2 }
14.
1
1
C
the required sum
13.
1
–
n
B
L1
Ma
2n 1
(a – 1) }
L2
D
1– x
2
+ {1
A
1
tn =
G. M.
a1 a2
... (1)
a3 ... an –1 2an
n
(a1a2 a3 ......2an )1/ n
x2
n 1
1 – x2
n
a1 a2
a3 + ...................
+ an –1 2an
n(2c)1/ n
17.
(d) Let a = b – d and c = b + d, then a + b + c =
b=
3
2
1
1
1
1
1
1
–
=
=
–
;
–
a2
a1
a2
a4
a3
a3
1
2
1
1 1
–d , ,
d
2
2 2
The numbers are
a1 , a2 , a3 , a4 are in H. P..
20.
[d > 0 as a < b < c]
Now a 2 , b 2 , c 2 are in G. P.
1
2
1
1
1
1
,
,
and
are in A. P..
a1 a2 a3
a4
so
4
=
2
1
–d
2
1
2
2 2
(b2 )2 = a c
2
(b) Let the A. P. be x, x + y, x + 2y, ........... Then
a = x + (p – 1)y; b = x + (q – 1)y; c = x + (r – 1)y
(b – c) = (q – r)y, (c – a) = (r – p)y, (a – b)
= (p – q)y
Let the G. P. be u, uv, uv 2 , .......Then a = uv p –1 ,
d
b = uv q –1 , c = uv r –1
2
1
1
– d2
=
16
4
1
d=
18.
1
– d2 =
4
1
d=
2
(
2
1
4
Now, log{ ab – c bc – a c a – b } = (b – c)log a
+ (c – a)log b + (a – b)log c
= (q – r)y log ( uv p –1 )+ (r – p)y log ( uv q –1 )
d>0)
+ (p – q)y log ( uv r –1 ) = 0. log u + 0. log v = 0
(a) The general term can be given by
tr
1
a
= 2n
a2n
1– r
1– r
– ar
ar
1
ab – c . a c – a . c a – b = 1
, r = 0, 1, 2, ............... n – 1
a b bc c a = a c b a c b
1
21.
a (2n – r )d – {a1 rd } (n – r )d
=
= 1
a1 (2n – r )d {a1 rd }
a1 nd
The required sum is
n –1
Sn =
=
=
19.
(c)
r
x 2 – (a + b) x + ab = 0
n –1
tr
1
=
0
r
0
(n – r )d
a1 nd
2
or, x –
[
m –1
m
n –1
x + (m – 1)( n –1) = 0
n
mn
2
n(n 1)d
n (n –1) (n – 2) .... 1
d =
2an 1
a1 nd
n(n 1) a2 – a1
.
2
an 1
n –1
1
m –1
a=
. Similarly, b =
.
n
1– a
m
Required quadratic equation is
(b) m =
22.
d = a2 – a1 ]
or, mnx – (2mn – m – n)x + mn – m – n + 1 = 0
(a) Since the line joining the mid points of two sides of
triangle is parallel to the third side and half in length.
Therefore each of the subtriangles is equilateral with
side lengths 12, 6, 3, ......etc.
a a3
a1 a4
1
1
1
1
+
=
+
or,,
= 2
; so
a1a4
a2 a3
a4
a1
a3
a2
1
1
1
1
–
=
–
a4
a3
a2
a1
24 cm
...(1)
12 cm
12 cm
3(a2 – a3 )
a –a
Also
= 1 4 ; so
a2 a3
a1a4
B
24 cm
12 cm
C
24 cm
The sum of perimeters of of all triangles
1
1
–
3
a3 a2
1
1
=
–
a4
a1
From (1) and (2)
... (2)
= 3[24 + 12 + 6 + 3 + .......
] = 3.
24
1
1–
2
= 144 cm
23.
(b)
c de y
b ce y
=
=
c – de y
a – be y
b – ce y
a be y
2a – (a – be y )
a – be
2a
a – be
y
2a
a – be y
y
–1=
=
=
Again,
2b – (b – ce y )
b – ce
2b
b – ce
2b
b – ce y
=
c – de
c – de
y
y
a
x
–1
26.
2c
(c)
b
y
a+b+c+d
27.
4
(a b)(c d ) = 2 2
8.
2b a
n 1
a and 2b, then A m = a + m
Again, let B 1 , B 2 , ........ , Bn be airthmetic means
between 2a and b , then Bm = 2a + m
(2n 1)th term of the A. P, i.e. a + 4n .
Now, Am = Bm
G. P. is (a + 4n), (a + 4n) (0.5),
(a + 4n) (0.5) 2 ............... (a + 4n) (0.5)2 n .
m
Its middle term is (a + 4n) (0.5) n
28.
(d)
a + 2n = (a + 4n) (0.5) n
b 2a
n 1
Sk = 1 +
a+m
m
1
k 1
2n – 4n(0.5)
4n =
n
(0.5) – 1
(k 1)2
+ ......... to
2n
1
1–
2
n
=
n.2
n
2n –1
ab
ac
b2
bc )
bc
=2
k 1
29.
[
b 2 = ac]
=
k 1
k
k 1
kSk =
n
k 1
(k+1)
= 2 + 3 + ...... + (n + 1) =
a
c
2a
2c
Now,
+
=
+
x
y
a b b c
ac
1
k Sk = k + 1
n 1
a b
b c
(d) Given, b = ac, x =
,y=
2
2
ac
1
=
1
2
2(ab
a =
m
.
b
n m 1
1
+
(0.5) n – 1
n
b 2a
n 1
2b a
= 2a +
n 1
b a
=a
n 1
2n – 4n(0.5) n
Required middle term = a + 4n =
=
4
(c) Let A1 , A2 , .........., An be airthmetic means between
Again for the last (2n + 1) terms the first term will be
According to the given condition,
=2
G. M.
( a b) (c d )
2
b = c = d
a, b, c, d are in G.. P.
a
b
c
(a) Let the first term is a, then first (2n + 1) terms are a, a
+ 2, a + 4, ...., a + 2.2n. Clearly the middle term of the
sequence of 4n + 1 terms is (2n + 1)th term,
ab ac b 2 bc
24
(a + b) (c + d)
But A. M
2b(b c a b)
= 4.
log 2 {(a + b) (c + d)}
c – de y
i. e. a + 4n. Also the middle term of the A. P.of
(2n + 1) terms is (n + 1)th term i.e., a + 2n.
25.
b
x
c
y
log 2 (a + b) + log 2 (c + d)
1 – b ey = 1 – c ey = 1 – d ey
a
b
c
a=
1
b c
=
a – be y = b – ce y = c – de y
a
b
c
24.
b
1
= 2b
y
a b
2c – (c – de y )
2c
–1=
y
=
y
b
x
(a)
tn =
1
n 2
.
n(n 1) 2
n
=
2(n 1) n 1
.
2
n(n 1)
=
1
n
n (n 3)
2
n
1 n 1
1
1
–
.
2
n 1 2
n
.
n
1 1
Sn =
tn =
1 2
n 1
1
1 1
2 2
+
+
30.
1 1
3 2
2
1 1
n 2
n 1
1 1
2 2
32.
1
(b) The general term is
tn =
+ ..........
n
1
1
n 1 2
=1–
1
(n 1)2 n
=
1
1
2 1.3.5.....(2n 1)
1
1.3.5....(2 n 1)
=
1
T
2 n
where Tn =
(d) Length of line segments on one side of the diagonal
are
Sn =
2 , 2 2 , 3 2 , ..............., (n – 1) 2
=
[Since the distance between the lines is
1
A0 A1 = A1 A2 = ........... = 1.
2
33.
=
Therefore, the side of square is divided into n points
equidistant at 1 units].
So , the required sum =
2
(c)
tn = 1 T0 T1 T1 T2 ..... Tn
2
n 1
1
n
16
n
1
=
=
+
+
4
4n 2 1
16
(2n 1)(2n 1)
[Perform actual division ]
2
4
1
4n 2 1
+
32
16
1
2n 1
AL1 L1M1
=
;
AB
BC
n
Sum, Sn =
1
LM
= 1 1;
n 1
a
L1M1 =
a
n 1
AL2
L2 M 2
=
;
AB
BC
L2 M 2 =
............ +
=
n 1
n
tn =
+
=
Ln
B
a
n 1
Mn
C
4n 2 1
16
1 n
32 n 1
1
2n 1
1
2n 1
1
1
.....
5
2n 1
1
2n 1
1
1 n(n 1)(2n 1)
+
n
16
6
4
M1
M2
1
1
1
32
3
1
3
1
n
1
2
( 4 n + 6n + 5 ) +
1
32
48
2n 1
L2 M 2
=
;
n 1
a
2
=
n(4n2
6n 5)
n
+
48
16(2n 1)
2
2a
etc.
n 1
The required sum =
2
A
L1
L2
1
2n 1
+
n(n 1)
2
+n 2 =n
2
1
1.3.5.....(2n 1)
n
2 2 3 2 ..... ( n 1) 2 + n 2
=2 2
31.
Tn
Sum to infinite terms, S = Lim Sn = 1 .
n
2
(c) The general term is
A3
A2
A1
A0
1
1
1
1
(T T ) =
1
2 0 n
2
1.3.5....(2n 1)
tn
2
1
n
(2n 1) 1
=
2
1.3.5....(2n 1)
1.3.5....(2n 1)(2n 1)
34.
a
2a
3a
+
+
+
n 1 n 1 n 1
na
a
n(n 1)
an
=
.
=
n 1 n 1
2
2
f (n) = n(4 n + 6n + 5 )
(b) We have two successive terms of the series
tr =
1
r (r 1)(r 2)( r 3)
and tr 1 =
1
(r 1)(r 2)( r 3)( r 4)
Þ rt r = (r + 4) t r +1
1
a + nx - a
[ a + nx - a ] =
x
x( a + nx + a )
=
Þ rt r – ( r + 1 ) t r +1 = 3 t r +1
Put r = 1 , 2 , 3 , ...... , n – 1, we have
=
1. t1 – 2 t2 = 3 t2
2. t2 – 3 t3 = 3 t3
n
3. t3 – 4 t4 = 3 t4
36.
(a)
........................................
4
å (2r – 1) = 14 + 34 + 54 + ............. + (2n – 1)
4
4
+ (2n) – [ 24 + 44 + ............... + (2n) ]
t1 – n tn = 3 [t 2 + t 3 + ...... + t n ]
4
= f (2n) – 16{ 14 + 24 + ........... + n } = f(2n) – 16 f
(n) for all n Î N; whether n is even on odd.
Þ 4t1 – n tn = 3 ëét1 + t2 + ...... + tn ûù
\ Sn = t1 + t2 + .............. + tn = 4 t1 – n tn
3
3
37.
(d)
æ
n
n
å
k =1
1
\ f (n) =
3(n + 1)(n + 2)(n + 3)
=
ALTERNATIVE METHOD :
1
(r + 3) - r
1
=
r (r + 1)(r + 2)(r + 3) 3 r (r + 1)(r + 2)(r + 3)
a + rx + a + (r - 1) x
=
a + rx - a + (r - 1) x
a + rx - a - (r - 1) x
1
= [ a + rx - a + (r - 1) x ]
x
\ t1 + t 2 + t3 + .......... tn
1
[ { a + x - a} + { a + 2x - a + x } +
x
.......................... + { a + nx - a + (n - 1) x } ]
k =1
k (k + 1)(2k + 1)
1 n
=
å (2k 3 + 3k 2 + k )
6 k =1
6
2
1 ì n(n + 1)(2n + 1) ü
1 ì n(n + 1) ü
í
ý +
ý
2 íî
3 î 2 þ
6
þ
a =
1
3
2
4
1 ì n(n + 1) ü
í
ý = 12 {n + 4n + 5n + 2 n}
6 î 2 þ
1
1
1
5
and e = 0.
,b= ,c=
,d=
3
6
12
12
So, a + b + c + d + e = 1
ù
1
1
18 3(n + 1)(n + 2)(n + 3)
1
=
\
tr = ê
-å
( n + 1)( n + 2)( n + 3) úû
3 ë1.2.3
r =1
=
35. (a)
1
m =1
+
ù
1é
1
1
3 êë r ( r + 1)( r + 2) ( r + 1)( r + 2)( r + 3) úû
1é 1
n
............ + k 2 )
1
= 1 –
3(n + 1)(n + 2)( n + 3)
18
Now
ö
k
2
2
2
å ç å m2 ÷ = å ( 1 + 2 + 3 +
è
ø
k =1
n
= 4. 1
–
3 1.2.3.4 3.n(n + 1)(n + 2)(n + 3)
n
4
r =1
On adding, we get
=
a + a + nx
= 14 + 24 + 3 4 + ...........
(n – 1) t n-1 – ntn = 3 tn
tr =
n
38.
a b
+
(d) As A. M ³ G. M., b c ³
2
³
c d
+
a b and d e
.
b c
2
c d
.
d e
æ a bö æ c d ö
a
\ çè b + c ÷ø çè d + e ÷ø ³ 4
c
c
a
= 4
e
e
a
æa cö æb dö
Similarly, ç + ÷ ç + ÷ ³ 4
è b dø è c eø
e
a b c d e
+ + + +
a b c d e
Also b c d e a ³ 5 . . . .
= 1
b c d e f
5
39.
1 1 1 1
(b) Since, a1, a2 , a3 , a4 are in H.P., a , a , a , a are
1
2
3
4
42. (c)
D ABC = D PBC + D PAC + D PAB
AB
1
1
1
1
.a. 3h = a. h +
a h1 + a. h2
2
2
2
2
A
in A.P.
1
1
–
= d Þ a1 - a2 = d a1a2
a1
a2
Similarly, a2 - a3 = da 2a 3
Add
a3 - a4 = da3a4
h1
a1 - a4 = d[a1a2 + a2 a3 + a3a4 ]
1
a1 - a4
a1 - a4
1
also
= 3a1a4
=
+ 3d Þ
a1
d
d
a4
Þ a1a2 + a2 a3 + a3a4 = 3a1a4
40.
33
20
14
6
(3 + 99) +
(5 + 100) +
(7 + 98) –
2
2
2
2
41.
2
1- 5
1+ 5
1+ 5
(Q r > 1)
<r<
Þ 1<r<
2
2
2
\
[r] = 1. Also –
1+ 5
<–r<–1
2
\ [– r] = – 2
[r] + [– r] = 1 – 2 = – 1.
= x ( x8 + 1) ( x 4 + 1) ( x 2 - 1)
Now x16 – 1 = ( x8 + 1) ( x 4 + 1) ( x 2 + 1) ( x 2 - 1)
=
7
( x 2 + 1) = 7
x
2
éæ
ù
1 ö
êç x + 2 ú > 14
÷
êè
ú
xø
ë
û
16
\ x > 15
44. (c) We have
f (k ) =
n
f ( k ) Sn
=
-1
å ar - ak = Sn - ak Þ
ak
r =1
ak
" k = 1, 2,......., n
Given a1, a2 ,..., an are in H.P. Þ
1 1
1
, ,.....,..
an
a1 a2
are in A.P.
Þ
Sn
S
S
- 1, n - 1....., n - 1
a1
a2
an
are in A.P..
f (1) f (2)
f (n) are in A.P..
,....,.
,
a2
an
a1
45. (c) We have
(c) Let sides of triangle be a, ar, ar 2
Since r > 1 \ ar 2 is greatest side \ a + ar > ar
Þ r2 – r – 1 < 0
h1 + h2
2
3
x15 - x13 + x11 - x9 + x 7 - x5 + x - x
Þ
4
(15 + 90) – (21 + 84) – (35 + 70) = 2838.
2
C
h1 + h2 = 2h Þ h =
43. (c)
\ given expression = 3.
It is a root of x2 + 2x – 15 = 0
(a) The integers divisible by 3 are 33 in number and are
3, 6, ,......., 99.
The integer divisible by 5 are 20 in number and are
5, 10, ..............., 100.
The integers divisible by 7are 14 in number and are
7, 14, ..............98.
The integers divisible by both 3 and 5 are 6 in number
and are 15, 30, ............, 90.
The integers divisible by both 3 and 7 are 4 in number
and are 21, 42, 63 and 84.
The integers divisible by both 5 and 7 are 2 in number
and are 35 and 70.
There are no integers divisible by all three.
Hence the sum of the numbers divisible by 3 or 5
or 7 is
G
h
B
Þ a1a2 + a2 a3 + a3 a4
=
h2
P
1 + x + x 2 + ...... + x n = ( x - x1 )( x - x2 )...( x - xn ) "x
\
Put x = 1 Þ
(1 - x1 )(1 - x2 )......(1 - xn ) = 1 + 1..... + 1 = n + 1
Again, 1 + x + x 2 + ...... + x n + x n +1
= ( x - y1 )( x - y2 )...( x - yn )( x - yn+1 )
Take log and differentiate, we have
1 + 2 x + 3x 2 + ...... + nx n -1 + (n + 1) x n
1 + x + x 2 + ...... + x n + x n+1
... (i)
1
1
1
1
+
+ ..... +
+
x - yn x - yn +1
x - y1 x - y2
=
49.
1
1
1
1
and
+
+
=
ab bc ca abc
(c)
Put x = 1 Þ
a+b+c
1
1
³ (abc )1/ 3 Þ abc £
Þ
³ 27
abc
3
27
1 + 2 + 3 + ...... + n + ( n + 1)
1
1
=
+
1 - y1 1 - y2
1 + 1 + ........ + 1 + 1
50.
1
1
+
+ ..... +
1 - yn 1 - yn +1
(c) We have Tr =
5r - ( r - 1)
=
1
1
(n + 1)(n + 2)
Þ
=
+
2(n + 2)
1 - y1 1 - y2
\
...(ii)
51.
, where r ³ 2
1
-
5r -1 ( r - 1)
éæ 1
1 ö
1
5r .r
æ 1
1 ö
æ 1
1 ö
å Tr = êçè 51.1 - 52.2 ÷ø + çè 52.2 - 53.3 ÷ø + çè 53.3 - 54.4 ÷ø
ë
r=2
+......to infinity ] =
From (i) and (ii), (1 - x1 )(1 - x2 ).......(1 - xn ) = 2
é 1
1
1
1 ù
+
+ ..... +
+
ê
ú
1
y
1
y
1
y
1
yn +1 û
n
2
1
ë
5 .r ( r - 1)
=
5r .r ( r - 1)
¥
1
1
+ ..... +
+
1 - yn 1 - yn +1
4r + 1
r
(b) We have, Tr =
1
5
8r
4
4r + 1
[Q terms tend to zero as n ® ¥]
8r
=
2
(2r + 2r + 1)(2 r 2 - 2 r + 1)
46. (c) We have 2b = a + c and a, p, b, q, c are in A.P.
é (2r 2 + 2r + 1) - (2r 2 - 2r + 1) ù
= 2ê
ú=
2
2
ëê (2r + 2r + 1)(2r - 2r + 1) ûú
a+b
b+c
,q =
2
2
Þ p=
Again, p ' = ab and q ' = bc
\ p2 - q2 =
2
(a + b) - (b + c )
4
é
ù
1
1
2ê 2
ú
2
ëê (2r - 2r + 1) (2r + 2r + 1) ûú
2
\
(a - c )(a + c + 2b)
=
= (a - c )b = p '2 - q '2
4
2r .r
2r [2(r + 1) - (r + 2)]
47. (c) We have t r =
=
(r + 1)(r + 2)
(r + 1)(r + 2)
¥
52.
(b)
L
=
13
13
k =1
k =1
k =1
k =1
å (7 + 3k )2 = 9 å k 2 + 42 å k + 49 å (1) = 11830
L
L
1ö
ù
L
20
ai = å (n 2 + n)
å
i =1
i =1
=
48. (c) We have
13
æ 1
û
= an- 3 + 2(n - 2 + n - 1 + n)
20
Þ
æ 2n +1
2n ö 2n +1
+....... + ç
-1
÷=
è n + 2 n + 1ø n + 2
13
1ö
= a1 + 2(2 + 3 + ... + n) = 2(1 + 2 + ...n) = n( n + 1)
é æ 22 2 ö æ 23 22 ö æ 24 23 ö ù
- ÷ +ç - ÷ +ç
- ÷ú
Sn = å t r = ê ç
3 3ø è 4
3ø è 5
4 øú
r =1
ëê è
û
Hence, 49 = 7 + 14d Þ d = 3
13 A.M.‘s are given by 7 + 3k, k = 1, 2, ........., 13
\ Desired sum
æ1
an = an -1 + 2n = an -2 + 2(n - 1 + n)
n
æ 7 + 49 ö
= 364 Þ n = 13
A1 + A2 + ........ + An = n ç
è 2 ÷ø
1ö
ë
r =1
2r +1
2r
=
r + 2 r +1
\
éæ 1
å Tr = 2 ê çè 1 - 5 ø÷ + èç 5 - 13 ø÷ + èç 13 - 25 ø÷ + ....ú = 2
53.
(c)
20 ´ 21 ´ 41 20 ´ 21
+
= 3080.
6
2
( w + x + y + z )4 £ 256 xywz
but A.M. ³ G.M.
Þ A.M. = G.M.
\ x= y=z=w
Also, 3w + 5x + 7 y + 9 z = 24.
\
x = y = z = w =1
So, the equation is 2t 2 + t + 2 = 0
\ D<0
Roots are imaginary.
n
54.
(d)
å
r =0
57. (c)
....[ ab + ( n - 1)( a + b) + ( n - 1) 2 ]
n -1
(-1)r ( n + 2r ) 2 = n 2 - ( n + 2) 2 + ( n + 4) 2 - ... - (3n) 2
= nab + (a + b)
= (n - n - 2)(n + n + 2) + (n + 4 - n - 6)
n(n - 1) (n - 1))(n)(2n - 1)
+
2
6
n n(n - 1) n(n - 1)(2n - 1)
+
+
6
6
6
n
= (-2)[2 n. + 2 + 10 + 18 + ...n / 2 terms]
2
=
n
[1 + (n - 1){1 + 2n - 1}]
6
é
næ
æn ö öù
= (-2) ê n2 + ç 4 + ç - 1÷ 8÷ ú
è 2 ø øû
è
4
ë
=
n
n
[1 + 2 n( n - 1)] = (1 - 2 n + 2n 2 ).
6
6
= ( -2 ) é n2 + n + n2 - 2n ù = -4n2 + 2n
ë
û
(b) Series can be summed as
58. (a)
21 + 2 2 + 23 + ... + 2n = 2n+1 - 2
22 + 23 + ... + 2 n = 2 n+1 - 2 2
L
59. (a)
n - 1 = 29
Þ
Þ
n = 2 + 1 = 513
Þ
2ab
Þ 2ab = c (a + b)
a+b
2ab + 2c (a + b) + 2c 2 = 2(a + b)2
...(2)
c (a + b) + 2c(a + b) + 2c 2 = 2(a + b) 2
2(a + b) 2 - 3c (a + b) - 2c 2 = 0
x 2 + 9 y 2 + 25 z 2 - 15 yz - 5xz - 3xy = 0
-2.5 xz - 2.3xy = 0
Þ
...(1)
From (1) and (2),
Þ 2 x 2 + 2.9 y 2 + 2.25 z 2 - 2.5 z.3 y
Þ
ab + c(a + b) + c 2 = (a + b)2
also, c =
9
2
n +1
-1
n!
log(a + c ) + log(b + c) = 2 log(a + b)
Þ
æ 15 5 3 ö
x 2 + 9 y 2 + 25 z 2 = xyz ç + + ÷
è x y zø
Þ
r +1
r
- (-1)r -1
r!
(r - 1)!
(a + c)(b + c) = (a + b) 2
Given that 2n +1 (n - 1) + 2 = 2n +10 + 2
Þ
Tr = (-1)r
Tr = Vn - V0 = (-1)n
å
r= 1
= 2n +1 (n - 1) + 2
(n - 1)2n +1 = 2n +10
é r
r2 + r +1
r + 1ù
= (-1)r ê
+
ú
r!
ë (r - 1)! r ! û
n
= n(2n +1 ) - (2n +1 - 2)
Þ
Tr = (-1)r
Tr = Vr - Vr -1
23 + ... + 2 n = 2 n+1 - 23
L
L
L
+2n = 2n+1 - 2n
(c)
r+
=
= (-2)[(2n + 2) + (2n + 10) + (2n + 18) + ... + n / 2 terms]
56.
n -1
r2
å
å
r= 1
r= 1
= nab + (a + b)
(n + 4 + n + 6) + ...
55.
S = ab + [ab + (a + b) + 1] + [ab + 2(a + b) + 22 ] +
2
a +b =
\
a + b = 2c
2
( x - 3 y ) + (3 y - 5 z ) + (5z - x ) = 0
x = 3 y = 5z
x y z
= =
Þ x, y, z are in H.P..
1 1 1
3 5
60. (d)
3c ± 9c 2 + 16c 2 3c ± 5c
c
=
= 2c or 4
2
4
\
Tr = (-1)r .
(Q
a, b, c > 0 )
r2 + r +1
r!
é r
1
1ù
= ( -1) r ê
+
+ ú
(
r
1)!
(
r
1)!
r
!û
ë
é (-1)r (-1)r ù é (-1)r
(-1)r ù
=ê
+
+
ú+ê
ú
( r - 1)!ûú ëê (r - 1)! (r - 2)!ûú
ëê r !
n
é (-1)r (-1)r -1 ù é (-1)r -1 (-1) r - 2 ù
Sn =
ú-ê
ú
ê
r!
(r - 1)! ûú ëê (r - 1)! ( r - 2)! ûú
r =1 ëê
å
é (-1)n (-1)n -1 ù
=ê
ú - 1.
(n - 1)! ûú
ëê n!
61. (b)
62.
(a)
tn
Þ
3n-1
tn
3n-1
2
2 æ2ö
æ2ö
+ ç ÷ + ... + ç ÷
3 è3ø
è3ø
Hence it is a G.P. from second term onwards.
(c) As obtained in (1), tn = a.2n – 1, n > 1
1
1 1 ¥ 1
1 2
1
\ å = å
=
=
r -1
1
t
a
a
a
2
r =2 r
r =2
12
n
(a) Given Sn =
1
1
= n(n + 1)(n + 2)(n + 3) - (n - 1)n (n + 1)(n + 2)
4
4
=
1
n(n + 1)(n + 2) {n + 3 - n + 1} = n(n + 1)(n + 2)
4
å
é
æ a + a öù
= 2n -3 ê 2na1 + 2n ç 1 n ÷ ú
è 2 øû
ë
= 2n -2 [na1 + Sn ].
p
2
4.
(b)
p
2
I1 = ò cos x cos xdx = ò cos 2 x dx =
0
0
We have I n+1 =
p
2
ò cos
n +1
p
4
x cos(n + 1) x dx
0
Integrating by parts taking cos(n + 1)x as second
function
I n +1 = cos
n +1
sin( n + 1) x
x
n +1
p
2
0
p
2
- ò (n + 1) cos n x (- sin x ).
1
n(n + 1) (n + 2) (n + 3)
4
\ tn = Sn - Sn -1
k
n
Ck [2a + (k - 1)d ]
å
2
k =0
¥
3.
2 -1
= n.2n -3 [4a1 + an - a1 ] = n.2n -3 [3a1 + an ]
n -1
tn +1
= 2, n ³ 2
tn
3 -1
= a1.n.2n -1 + dn(n - 1)2n - 3.
å
-1 =
å tn =
Ck Sk =
) - 2 ( 2n - 1)
dö
d
æ
= ç a1 - ÷ n.2n-1 + [n.2n-1 + n(n - 1)2n- 2 ]
è
2ø
2
n -1
(d) Given Sn = a.2n - b Þ t1 = S1 = 2a - b
Now tn = Sn – Sn – 1 = a.2n – a.2n – 1 = a.2n – 1, n > 1
Now
2.
3n - 2
æ 2ö
=ç ÷
è 3ø
n
å
Þ
1.
-
tn -1
å
k =0
n
(
ú
úû
3 3n - 1
å
r -1
tr -1 ö
æ tr
æ2ö
=
ç r -1
÷
ç ÷
3r - 2 ø r = 2 è 3 ø
r =2 è 3
n
\
t n = 3n - 2n , n ³ 1
n -1 ù
n
n
éæ
ù
dö
d
= êç a1 - ÷
k n Ck +
k 2 n Ck ú
2 ø k =0
2 k =0
êëè
úû
t n = 3tn -1 + 2n -1 , n > 1 and t1 = 1
tn
2
- tn -1 =
3
3
\
n
+... + (3tn -1 + 2n -1 )
n -1
é æ2ö
- 1 = 2 ê1 - ç ÷
n -1
êë è 3 ø
3
tn
Þ S=
Sn = 1 + (3 ´ 1 + 2) + (3 ´ 5 + 2 2 ) + (3 ´ 19 + 23 )
\
Þ
0
sin(n + 1) x
dx
n +1
p
2
= 0 + ò cos n x sin(n + 1)sin x dx
0
=
p
2
ò cos
0
n
x {cos nx - cos( n + 1) x cos x} dx
A.P. with common difference lnt. Also, x, y, z are in
A.P. (say with common difference d).
[Q cos n x = cos (n + 1 – 1) x = cos (n +1)x cos x +
sin(n+1)x sin x]
I
1
\ I n +1 = I n - I n +1 Þ n +1 = " n Î N
In
2
Hence x - y = ( p - q)d etc.
and l n x - l n y = ( p - q)ln t. etc.
Hence, I1, I2, I3,........, In,.......are in G.P., with common
Let E = x y - z . y z - x .z x - y so that
1
ratio
2
5.
(d) We have I1 =
l n E = ( y - z )l n x + ( z - x ) l n y + ( x - y ) l n z
= (q - r )d l n x + (r - p)d l n y + ( p - q )d l n z
1
p
and common ratio = , so the sum of
2
4
pæ
1ö
14 çè 2n ÷ø p
n terms =
= (1 - 2n )
1
2
12
= d [ p(l n z - l n y) + q (l n x - l n z ) + r (l n y - l n x)]
= d l n t[ p(r - q) + q( p - r ) + r (q - p)] = 0 Þ E = 1
8.
lna, lnb, lnc, lnd are in A.P. with common difference
ln r. Also a x = b y = c z = d v
p
6.
sin(2 n + 1) x - sin(2n - 1) x
(d) We have an + 1 – an = ò
sin x
Þ
0
p
0
x ln a = ylnb = z lnc = v lnd
Þ xlna = y (lna + lnr ) = z (lna + 2lnr )
[Note that denominator is free from n]
=ò
(c) Since a, b, c, d are in G.P. (say with common ratio r),
= v(lna + 3lnr )
p
2 cos 2 nx sin x
dx = 2ò cos 2n x dx = 0
sin x
0
\ an+1 = an " n Î N
So, a1 = a2 = a3 = ...........
p
We have a1 =
sin x
ò sin x dx = p
0
\ a1 = a2 = a3 = ........ = an = p
9.
Þ
x-y
y-v
y-z
x- z
=2
=2
and
y
v
z
z
Þ
1 1 2
1 1 2
+ = and + = Þ x, y , z , v are in H.P..
x z y
y v z
(b) Here a, b, c are in H.P..
Þ a -1, b -1, c-1 are in A.P..
or a1 + a2 + a3 + ....... + an = np
7.
(a) Let the base be taken as e. Since x, y , z are terms of a
G.P. (say with common ratio t), l n x, l n y, l n z are in
1.
(d) Solving first two equations we get
x = -2 Þ b = - m and g = m
Now,
1
1
1
+
=
x+ p x+q r
Þ x 2 + ( p + q - 2r ) x + pq - r ( p + q) = 0
\
2.
b + g = p + q - 2r = 0
Þ p, r , q are in A.P..
(b) Statment-2 is correct as
a 2 + b 2 + c 2 - ab - bc - ca
æ1ö
Þ ç ÷
è4ø
a -1
-1
b -1
æ1ö
, ç ÷
è4ø
-1
Þ 4- a , 4 - b , 4 - c
æ1ö
, ç ÷
è4ø
-1
c -1
are in G.P..
are in G.P..
1
= [(a - b)2 + (b - c)2 + (c - a) 2 ] ³ 0
2
Also,
f ( x) = (a1 x - a2 )2 + (a2 x - a3 )2 + ... + (an -1 x - an )2
Thus f ( x ) £ 0
Þ
a1x - a2 = a2 x - a3 = ...... = an -1 x - an = 0
Þ
x=
Þ
a1, a2 ,..., an are in G.P..
a
a2 a3
=
= ... = n
a1 a2
an -1
1.
(a,b,c,d) Given
3(a2 + b2 + c2 + 1) – 2(a + b + c + ab + bc + ca) = 0
Þ [(a –b)2 + (b – c)2 + (c – a)2] + [(a – 1)2
+ (b – 1)2 + (c – 1)2] = 0
Þ a – b = b – c = c – a = 0 and
a 2 ¹ 1 [If a 2 = 1 Þ r = 0, which is not possible]
Þ (3a 2 – 1) (a 2 – 3) £ 0 and a 2 ¹ 1
1
Þ a 2 Î éê , 3ùú and a 2 ¹ 1.
ë3 û
a–1=b–1=c–1=0 Þ a=b=c=1
2.
(b,d)
c c
a+b+ +
2 2 ³
4
or, a + b + c ³
4
\
Þ
(a + b + c )4
4
4
c c
a.b. .
2 2
4.
1
(a + b + c )4
64
For the greatest value of abc 2 the numbers have
c
1
=
2
4
Also, given the greatest value =
Let G. P. be A, AR, A R 2 , ......................, then
1
t 2n - 1 = A R 2 n - 2 = x Þ R n - 1
\ b = tn = A R
1
.
64
Let H. P. be A,
Let the three numbers in G. P. be a, ar and ar 2 , then
a + ar + ar 2 = a S
...(1)
2 4
and a 2 + a 2 r 2 + a r = S 2
... (2)
Þ
Þ
(1 + r + r 2 )2
(1 + r 2 – r ) (1 + r 2 + r )
1+ r + r2
1– r + r 2
(1 + r + r 2 )2
= a2
2
4
1+ r + r
=a
x- A
A+ x
=
2
2
æ xö 2
=ç ÷
è Aø
1
So a + b + c = 1
Solving (1) and (2), we have
x-A
2
\ a = tn = A + (n – 1)d = A +
1
\ the greatest value of abc =
(a + b + c )4 .
64
(b,c)
Let A. P. be A, A + d, A + 2d, ................... then
(n – 1)d =
2
3.
(b,c)
t2n - 1 = A + (2n – 2)d = x (say), then
abc 2
4
to be equal, i.e. a = b =
1
or 3 for which r = –1, or 1
3
2
æ1 ö
\ a Î ç , 1÷ U (1, 3)
è3 ø
abc 2
4
³
4
abc 2 £
4
2
Also, a ¹
2
= a2
2
2
2
2
Þ (a –1)r – (a + 1)r + (a – 1) = 0
Q r Î R Þ (a 2 + 1) 2 – 4(a 2 – 1) 2 ³ 0 and
t 2n –1 =
n -1
= A æç x ö÷ 2 Þ
è Aø
Ax
1
1
,
,....................... then
1
1
+D
+ 2D
A
A
1
= x then
1
+ (2n – 2) D
A
(n – 1)D =
1æ1 1ö
ç – ÷
2 è x Aø
1
1
\ c = tn =
=
1
1 1æ1 1ö
+ (n – 1) D
+ ç – ÷
A
A 2 è x Aø
=
1
1æ 1 1ö
ç + ÷
2 è x Aø
Clearly a, b and c are A. M., G. M. and H. M. between
the numbers, x and A respectively.
Hence a ³ b ³ c also b 2 = ac
5.
(a,b,c,d) a(n) = 1 +
1
1 1 1
+ + + ...... +
=1+
2 3 4
(2n ) – 1
6.
(a,b,c)
Then an = length of diagonal of S n + 1 = 2 an + 1
1ö
æ 1 1ö æ 1 1 1 1ö æ1 1
ç + ÷ + ç + + + ÷ + ç + + ... + ÷
15 ø
è 2 3ø è 4 5 6 7 ø è 8 9
Þ
æ 1
1
1 ö
+ ....... +
+ .......... + ç n –1 + n –1
÷ <1
n
è2
2
1
2
1ø
+
-3
14444442444444
an + 1
an
=
1
2
Þ a1, a2 , a3 , .......... from a G.. P.
with common ratio
2n -1 terms
æ 1 ö
\ an = a1 ç
è 2 ÷ø
æ 1 1ö æ 1 1 1 1ö
+ç + ÷ +ç + + + ÷ +
è 2 2ø è 4 4 4 4ø
1
424
3 1442443
2 terms
Let an be the length of a side of the square Sn .
4 terms
1
2
n –1
.
æ 1 ö
= 10. ç
è 2 ÷ø
n –1
\ Area of square
1ö
æ1 1
çè + + .... + ÷ø
8 8
8
1442443
2
æ 1 ö
Sn = an = 100 ç
è 2 ÷ø
8 terms
2( n –1)
=
100
2n –1
Sn
æ 1
1
1 ö
+ ......... + ç
+
+ ... +
÷
n
n
n
–1
–1
è2
24244444
2 –13ø
14444
2n -1 terms
S n +1
= 114
+ 14
+ 1244
+ .... +
31 = n
n times
\ a (n) < n. In particular a (100) < 100 and a (200)
< 200. So, (a) and (b) are correct.
Given that area of Sn < 1 Þ
1 æ1 1ö
Again, a(n) = 1 +
+ç + ÷
2 è3 4ø
æ1 1 1 1ö
+ ç + + + ÷ + ............. +
è5 6 7 8ø
1
1
1ö
1
æ
1
+ ... +
+ ÷ –
+
n
nø
èç 2n –1 + 1 2n –1 + 2
2 –1 2 3
2n
1444444424444444
2n -1 terms
> 1+
(a,b,c)
A. M. of x, z = y, G. M. of x, z = xz and
2
A. M. ³ G. M. So, y ³ xz.
Also,
2 xz x + y
x+ y
=
x + z 2y - x
x+ z - x
=
4 terms
2 n -1 terms
=1+
7.
AM ³ HM Þ y ³
1
1
1ö
æ 1
çè n + n + .... + n ÷ø – n
2
2 2444
23 2
1444
1
1 1 1
1
1ö n
æ
+ + + .... + – n = ç1 – ÷ +
è
2
2
2
2
2
2n ø 2
144
42444
3
y+z
x+ y
+
2y - x 2y - z
³
\
2
= 1+
x+ y y+ z
y+ z
,
=
z
x
2y - z
x+ y y+ z
.
z
x
y( x + y + z)
3y2
= 1+
xz
xz
[Q x + z = 2y ]
n terms
æ
1 ö 200
> 100
\ a (200) > çç1 – 200 ÷÷ +
2
è 2
ø
Þ (c) is correct. Similarly (d) is also correct.
<1
Þ 2n –1 > 100 Þ n – 1 ³ 7 or,, n ³ 8.
\ n = 8, 9, 10 satisfy the required condition.
1 æ 1 1ö æ1 1 1 1ö
+ ç + ÷ + ç + + + ÷ + .......... +
2 è 4 4ø è8 8 8 8ø
1
424
3 1442443
2 terms
100
2n –1
\
y2
x+ y
y+z
³ 2 1 + 3.
³4
+
xz
2y - x
2y - z
[Q y 2 ³ xz ]
8.
(b,d)
We have 1072 < 10 (2a + 19d) < 1162 and
a + 5d = 32
\ 1072 < 20 (a + 5d) + 90d < 1162
Þ 1072 < 640 + 90d < 1162
\
11.
(a,b,d)
1ö æ
1ö
æ
çr + ÷-çr - ÷
2
2
ø è
ø
= tan -1 è
1 öæ
1ö
æ
1 + ç r + ÷ç r - ÷
2 øè
2ø
è
432
522
and d is natural number,,
<d <
90
90
so d = 5 Þ a = 7
9.
(a, d)
Given (a + nd ) 2 = (a + md )(a + rd )
2
a
a
a
Þ æç + nö÷ = æç + mö÷ æç + r ö÷
èd
ø
èd
øèd
ø
1ö
1ö
æ
æ
= tan -1 ç r + ÷ - tan -1 ç r - ÷
è
è
2ø
2ø
...(i)
(m + r )n
2mr
Also n =
Þ mr =
m+r
2
Sn =
...(ii)
\
10. (a,d)
12.
(b,c)
(m + r )n
2
from (ii)
m + r - 2n
n2 -
Let the first four terms be
1
(-1)r -1 (r 2 + r - 9)3r -1
(r - 1)!
Tr =
=
a
n
mr
=- =d
2
m+r
(-1)r -1 3r -1 (-1)r +1 3r +1
= Vr -1 - Vr +1
(r - 1)!
(r + 1)!
n
Sn =
a a
, , ar , ar 3 , then
r3 r
a
1
1
Þ r2 = .
=
r ar 3
2
3
So, the sum to infinite terms, S = r
1 - r2
(-1)n .3n (-1)n +1.3n +1
(n + 1)!
n!
= -2 - (-1)n
13.
a
Tr = V0 + V1 - Vn - Vn +1
å
r =1
= 1- 3-
æ a ö æ aö
3
4
çè 3 ÷ø çè r ÷ø ( ar ) (ar ) = 4 Þ a = 4 Þ a = ± 2
r
Also,
1
2
a
æ aö
æ an ö
2 æ aö
çè ÷ø + 2 çè ÷ø + n = çè ÷ø + (m + r ) + mr
d
d
d
d
n2 - mr
a
Þ
=
=
d m + r - 2n
æ 4n ö
åTr = tan-1 æçè n + 2 ö÷ø - tan -1 2 = tan-1 çè 2n + 5÷ø
1
S ¥ = lim S n = tan -1 2 = cot -1 .
n®¥
2
Now from (i),
2
æ 4r 2 + 3 ö
1
-1
Tr = cot -1 ç
÷ = tan
1
è 4 ø
1+ r2 4
(c, d)
3n (n - 2)
.
(n + 1)!
Let b = a + p, c = a + 2 p, d = a + 3 p
1
1
1 1
+
+
a d = a a + 3 p = (a + p )(a + 2 p)
1
1
1 1
a(a + 3 p)
+
+
b c a + p a + 2p
a
(Q first term = 3 and common ratio = r 2 )
r
1
a
2a
2
or S =
= 4 (Q r = )
3
2
r
r
Where
=
\
a 2 + 3ap + 2 p 2
a 2 + 3ap
>1
1 1 1 1
+ > +
a d b c
1 ö æ
1 ö æ
1 ö
æ
(a, r) = ç 2,
÷ , ç - 2,
÷ , ç - 2, ÷
2
2ø
2
è
ø è
ø è
æ 1
1 ö
æ1 1ö
+
÷ (a + a + 3 p)
ç + ÷ (a + d ) = ç
èb cø
è a + p a + 2p ø
æ
1 ö
and çç 2 ,÷÷
2ø
è
=
(2 a + 3 p) 2
a 2 + 3ap + 2 p 2
= 4+
p2
a 2 + 3ap + 2 p 2
> 4.
14.
é1 + a1a2 ù
-1 é1 + a2 a3 ù
cot -1 ê
ú
ú + cot ê
a
a
ë 2 1û
ë a3 - a2 û
(a, c)
+ cot
Also,
2
2
2
Þ a + c > 2 ac > 2b (G.M. > H.M.)
-1 é1 + a3 a4
ù
-1 é1 + an an -1 ù
ê
ú +... + cot ê
ú
ë a4 - a3 û
ë an - an -1 û
= cot -1 a1 - cot -1 a2 + cot -1 a2 - cot -1 a3 + ...
a 2 + c2
> a 2 c2
2
Þ a 2 + c 2 > 2b 2
If a, b, c are pth, qth and rth terms of A.P. then
16. (c, d)
+ cot -1 an -1 - cot -1 an
is always a rational number.
= cot -1 a1 - cot -1 an = tan -1 an - tan -1 a1
17. (a,b,c,d) b ³ ac Þ b3 ³ abc
Þ b3 ³ 4 or b ³ (4)1/ 3 Þ [b] ³ 1
a+b+c
> b (H .M) Þ a + c > 2b
3
15.
(a, b, c)
1.
A - r, s; B - q; C - p,s; D - q
(A) 2b = a + c and(2 + b)2 = (1 + a)(1 + c)
Þ
4 + 4b + b 2 = 1 + a + c + ac
Þ
4 + 4b + b 2 = 1 + 2b + ac
Þ
ac = b 2 + 2b + 3 = (b + 1)2 + 2 Þ ac > 2
( Q b ¹ -1, otherwise a and c are non-real)
(B) If the numbers are a and b, then
x=
1
æ bö 3
a+b
and b = ar3 Þ r = ç ÷
2
è aø
y 3 + z3 a 3 r 3 + a 3 r 6 a (1 + r 3 ) a + b
=
=
=2
=
a+b
xyz
x
x (ar ) (ar 2 )
2
2
(C) c > 4b – 3a Þ ar > 4ar – 3a
Þ r2 – 4r + 3 > 0 Þ r < 1 or r > 3. But the terms are
positive so r Î (0, 1) È (3, ¥)
2.
A - p, q; B - p; C - r, s; D - s
æ 1 1 1ö
Þ (a + b + c ) ç + + ÷ ³ 9
è a b cø
2ab
2ab
h
, g = ab Þ =
a+b
g (a + b) ab
(B) h =
Þ
4 2 ab
=
5 a+b
Now,
-1 æ 1 ö
-1 æ 2 ö
-1 (2 r + 1) - (2 r - 1)
(D) tan ç 2 ÷ = tan ç 2 ÷ = tan
è 2r ø
è 4r ø
1 + (2r + 1)(2r - 1)
a+b+c
3
³
1 1 1
3
+ +
a b c
(A) A.M ³ H .M Þ
9
Þ =
1
or
(
(
) Þ
2
b)
a+ b
a-
a+ b
a- b
2
a+ b
= ±3
or a : b = 1 : 4 or 4 : 1
= tan -1 (2 r + 1) - tan -1 (2r - 1)
\
n
å tan
-1 æ
r =1
1 ö
-1
-1
çè 2 ÷ø = tan (2n + 1) - tan (1)
2r
= tan -1 (2 n + 1) -
p
4
(2 n + 1) - 1
æ 1 ö ïü
n
ïì
=
\ tan í å tan -1 ç 2 ÷ ý =
è 2r ø ïþ 1 + (2n + 1).1 n + 1
ïî r =1
n
é n
æ 1 öù
n
=1
\ lim tan ê å tan -1 ç 2 ÷ ú = lim
è
ø
n
+1
n®¥
n
®¥
r
2
ëê r =1
ûú
=3
a- b
(C) S =
1-
1
1-
1
2
= 2 and Sn+1 =
\ S - Sn+1 =
1
n
2
<
1
2
1-
n +1
1
2
= 2-
1
2n
1
Þ 2n > 1000 .
1000
But 29 < 1000 < 210
\ n ³ 10
(D) (1 + x) (1 + x2) (1 + x4) (1 + x8)...........(1 + x128)
=
1 - x 256 255 r
= å x Þ n = 255
1- x
r =0
b-c
a-b
3.
A - r; B - q; C - q; D - p
(A) Ap = a + ( p - 1)d
...(1)
Aq = a + (q - 1)d
...(2)
Ar = a + (r - 1)d
...(3)
As = a + ( s - 1)d
...(4)
Aq = kAp
³5
As = k 3 Ap
.)
( Q Ap , Aq , Ar , As in G.P.)
Ap - Aq
(1 - k )
= Ap
from (1) and (2)
d
d
( q - r ) = Ap k
(1 - k )
from (2) and (3)
d
(1 - k )
d
p - q, q - r , r - s are in G.P..
( r - s ) = Ap k 2
Þ
Þ
ln(ln x), ln(ln y ), ln(ln z ) are in A.P..
Þ
2 x + ln(ln x ), 3x + ln(ln y), 4 x + ln(ln z ) are in A.P..
(C) n !,3 ´ n ! and (n + 1)! are in G.P..
Þ
9(n !)2 = n !(n + 1)!
Þ
(n + 1) = 9 Þ n = 8
\
n ! = 8!
5 ´ n ! = 5 ´ 8!
n !, 5 ´ n ! and (n + 1) ! are in A.P..
(b + c - 2a)2 + (c + a - 2b)2 + (a + b - 2c) 2
=
3
4.
2
K
K
K
ai = 12 + 2 2 + ...i 2 =
S=
10
(C) Tn =
i (i + 1)(2i + 1)
6
i (i + 1)(2i + 1)
= 1210.
6
ai = å
å
i =1
i =1
n2
500 + 3n 2
1 500
=
+ 3n
Tn
n2
æ 1000 ö
Þ n=ç
è 3 ÷ø
1/ 3
.
[( s - a)2 + (s - a) + 1] [( s - b) 2 + ( s - b) + 1]
( s - b)
(s - a )
1/ 3
< 7.
Hence T7 is the largest term, so largest term in the
given sequence is
(D) Tr =
a=b=c
A - q; B - p; C - q; D - s
(A)
a3 = a2 + 32 = 12 + 22 + 32
(b + c - 2a)2 - (b - c )2 + (c + a - 2b)2 - (c - a) 2
+ ( a + b - 2c ) - ( a - b ) = 0
Þ
2
2
2
(B) a2 = a1 + 2 = 1 + 2
æ 1000 ö
Now, 6 < ç
è 3 ÷ø
(b - c ) 2 + ( a - b ) 2 + (c - a ) 2
3
2
1
2
dTn (500 + 3n 2 ).2n - n 2 9n 2
n(1000 - 3n3 )
=
=
=0
dn
(500 + 3n 2 ) 2
(500 + 3n 2 ) 2
9!+ 8! = 5 ´ 9!
Þ
Equality holds if a = b = c =
Let U n =
(n + 1)! = 9!
(D)
the minimum value is 3 × 4 × 5 = 60.
10
(B) ln x, ln y, ln z are in G.P..
Þ
1
1
öæ
ö
æ
+ 1÷ ç ( s - b) +
+ 2÷
çè ( s - a) +
ø
s-a øè
s -b
³3
³4
1
æ
ö
+ 3÷
çè (s - c) +
s-c ø
Ar = k 2 Ap
( p - q) =
[( s - c)2 + (s - c ) + 1]
=
( s - c)
2r + 1
2
r (r + 1)
n
Sn =
å
r =1
2
=
49
.
1529
(r + 1) 2 - r 2
r 2 (r + 1) 2
n
Tr =
[Vr - Vr +1 ]
å
r =1
= V1 - Vn +1 = 1 -
1
( n + 1) 2
=
1
r2
-
1
(r + 1) 2
1.
2.
Ans : 49
Let x = a2, y = b2, z = c2. Then 2b2 = a2 + c2
Since b is at least 2, consider the values of b from 2 onwards.
The first such value occurs when b = 5 (and then a = 7, c = 1)
So, the least possible value of x = a2 = 49.
Ans : 50
Let k and k + 1 be removed. Then
\
r = lim
3.
4m 2 + 103(1 - m)
. Clearly (1 – m) must be divisible
4
x®0 3x
\
5.
Þ 1 £ 16t 2 - 95t + 1 < 8t + 2 Þ t = 6 and so, n = 50
Ans : 1000
nth row will contain n elements.
Þ
I=
Let f ( x ) = 7 + 2 x ln 25 - 5x -1 - 52 - x
ln 5
f '( x ) =
[20 - 5 x + 53- x ] = 0 Þ x = 2
5
f "( x ) < 0 Þ f ( x) is max at x = 2
2
1-
1
3
= 3.
2 x{x}
x + {x}
2 f (I + f )
I +2f
I2 + 2If = 2If + 2 f 2
Þ
I2 = 2 f 2 Þ I = ± 2 f
1
f = 0,
n
[first term + last term]
2
n
= [n 2 - (n - 1) + n2 + (n + 1)] = n3 .
2
Ans : 3
a
=
1- r
Let x = I + f
[ x] =
Sum of elements in
tan(p + x )
1
3
=
Ans : 2
Q
the nth row =
2
Sum of G.P. =
by 4.
Let m = 1 + 4t, then we get k = 16t2 – 95t + 1 and 1 £ k < n
2
Last element of nth row will be n + (n + 1)
{x 2 tan( p + x )}
x3
r = lim
1st element of nth row will be n2 - (n - 1)
4.
ò
t 2 dt
x
x ®0 0
n
(n + 1) - 2k - 1
105
2
Þ 2n 2 - 103n - 8k + 206 = 0
=
4
n-2
Since n and k are integers, so n must be even, say n = 2m
then k =
a=2
2
If f = 0, I = 0 Þ x = 0 which is not possible
1
\
f =
\
I=± 2
\
x = 1+
2
1
2
1
2
= ±1
, -1 +
1
2
.