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PATHANIA INSTITUTE OF MATHEMATICS
S.C.F 13 PHASE: 2 MOHALI, PH– 98145– 06093
SEQUENCE AND SERIES
M.M.: __________
1.
2.
If the altitude of a triangle are in A.P., then the
sides of the triangle are
(a) A.P.
(b) H.P.
(c) G.P.
(d) A.G. progression
Date: _________
3.
13  2  33  ....  123
is equal to
12  22  32  ....  122
243
234
(a)
(b)
25
25
263
(c)
(d) none of these
27
4.
Let x be the A.M. and y, z be two G.M.’s
y3  z3
between any two +ve numbers. Then

xyz
(a) 2
(b) 1
(c) 3
(d) 4
5.
If a, b, c are in A.P., then 10ax+10, 10bx+10,
10cx+10, x  0 are in
(a) A.P.
(b) G.P. only when x > 0
(c) G.P. for all x
(d) G.P. only when x < 0
In the sequence [1], [2, 3], [4, 5, 6],
[7, 8, 9, 10], … of sets, the sum of elements in
the 50th set is
(a) 62525
(b) 65255
(c) 56255
(d) 55625
(c) 3
(d) none of these
Sol.
6.
If Tn denotes the nth term of the series
2 + 3 + 6 + 11 + 18 + … , then T50 is equal to
(a) 492 + 2
(b) 492
2
(c) 50 + 1
(d) 492 – 1
9.
Every term of a G.P. is positive and also every
term is the sum of two preceding terms. Then
the common ratio of the G.P. is
5 1
(a) 1
(b)
2
5 1
1 5
(c)
(d)
2
2
Sol.
7.
Let the sequence a1, a2, a3, … an, form an A.P.,
then a12 – a22 + a32 – a42 + … + a2n-12 – a2n2 is
equal to
n
2n 2
2
(a)
(b)
(a12  a 2n
)
(a 2n  a12 )
2n  1
n 1
n
2
(c)
(d) none of these
(a12  a 2n
)
n 1
10.
Sol.
11.
Sol.
8.
Let S1, S2, S3 be the sum of n terms of three
series in A.P., the first term of each being 1 and
the common differences 1, 2, 3 respectively. If
S1 + S3 = S2, then the value of  is
(a) 1
(b) 2
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 is
(a) 9
(b) 5/4
(c) 0
(d) 18
Thee consecutive terms of a progression are 30,
24, 20. The next term of the progression is
1
(a) 18
(b) 17
7
(c) 16
(d) none of these
12.
{an}| and {bn} be two sequences given by
1
1
1
2 n (y)2
a n  (x)  (y) and bn  (x)
n  N. Then a1 a2 a3 …. An is equal to
xy
(a) x – y
(b)
bn
xy
xy
(c)
(d)
bn
bn
2n
2n
1
n
15.
for all
Let an be the nth term of a G.P. of positive
100
integers. Let
a
n 1
2n
  and
100
a
n 1
2n 1
  such
that   . Then the common ratio is


(a)
(b)


1
  2
(c)  

Sol.
1
  2
(d)  

Sol.
16.
13.
If a1, a2, …. An are positive real numbers whose
product is a fixed number c, then the minimum
value of a1 + a2 + … + an-1 + 2an is
(a) n (2c)1/n
(b) (n + 1) c1/n
(c) 2n c1/n
(d) (n + 1) (2c)1/n
If the first two terms of harmonic progression
1 1
be , , then the harmonic mean of first four
2 4
numbers is
(a) 5
(b) 1/5
(c) 10
(d) 1/10
Sol.
Sol.
17.
Sum of the n terms of the
2  8  18  32  .... is
n(n  1)
(a)
(b) 2n (n + 1)
2
n(n  1)
(c)
(d) 1
2
series
Sol.
14.
Sol.
If the arithmetic and geometric means of two
distinct, positive numbers are A and G
respectively, then their harmonic mean is
(a) A/G
(b) G/A
(c) G2/A
(d) A2/G
2
G
Since G2 = AH  H 
A
18.
Three numbers are in G.P. If we double the
middle term, we get an A.P. Then the common
ratio of the G.P. equals
(a) 2  3
(b)_ 3  2
(c) 3  5
(d) 5  3
22.
Sol.
Sol.
23.
19.
If x > 0, then the sum of the series
e x  e2x  e3x .... is
1
1
(a)
(b) x
x
1 e
e 1
1
1
(c)
(d)
x
1 e
1  ex
If A, G and H are respectively the A.M., the
G.M. and the H.M. between two positive
numbers ‘a’ and ‘b’, then the correct
relationship is
(a) A = G2 H
(b) G2 = AH
2
(c) A = GH
(d) A = GH2
2
G = AH
The length of a side of a square is “a” metre. A
second square is formed by joining the middle
points of this square. Then a third square is
formed by joining the middle point of the
second square and so on. Then the
sum of the area of squares which carried upto
infinity is
(a) a2
(b) 2a2
2
(c) 3a
(d) 4a2
Sol.
Sol.
49
20.
If (1.05)50 = 11.658, then
 (1  05)
n
equals:
n 1
(a) 208  34
(c) 212  16
(b) 212  12
(d) 213  16
Sol.
21.
Sol.
The product of n positive numbers is 1, then
their sum is a positive integer, that is
(a) equal to 1
(b) equal to n + n2
(c) divisible by n
(d) never less than n
24.
Sol.
Consider an infinite series with first term ‘a’
and common ration ‘r’. If its sum is 4 and the
second term is 3/4, then
7
3
3
(a) a  , r 
(b) a  2, r 
4
7
8
3
1
1
(c) a  , r 
(d) a  3, r 
2
2
4
28.
25.
Two A.M.’s A1 and A2, two G.M.’s G1 and G2
and two H.M.’s H1 and H2 are inserted
between any two numbers, then H1-1 + H2-1
equals
(a) A1-1 + A2-1
(b) G1-1 + G2-1
G1G 2
A  A2
(c)
(d) 1
A1  A 2
G1G 2
The sum of three numbers which are consecutive
terms of an A.P. is 21. If the second number is
reduced by 1 while the third is increased by 1,
three consecutive terms of a G.P. result. The three
numbers are
(a) 3, 7, 11
(b) 12, 7, 2
(c) 1, 7, 11
(d) none of these
Sol. A,B
Sol.D
26.
The
number
of
terms
in
the
series
1
2
20, 19 , 18 ,... of which the sum is 300, is
3
3
Sol.
(a) 25
(a) (b)
(b) 36
(c) 31
29.
(d) none
Sol.
27.
Sol.
If the sum of four numbers in G.P. is 60 and the
A.M. of the first and the last is 18, then the
numbers are
(a) 8, 12, 16, 20
(b) 32, 16, 8,4
(c) 4, 8, 16, 32
(d) none of these
(b) (c)
The sum of each of two sets of three terms in A.P.
is 15. The common difference of the first set is
greater than that of the second by 1 and the ratio
of the products of the terms in the first set and
that of the second set is 7 : 8. The two sets of
numbers are
(a) 3, 5, 7 and 4, 5, 6
(b) 3, 5, 7 and 7, 8, 9
(c) 2, 4, 6 and 4, 5, 6
(d) 21, 5, -11 and 22, 5, - 12
(a) (d)
30.
The sum of three numbers in A.P. is 15. If 1, 3, 9
are added to them respectively, the resulting
numbers are in G.P. The numbers are
(a) 3, 5, 7
(b) 2, 5, 8
(c) 15, 5, -5
(d) none of these
Sol.
(a) (c)
31.
The sum of three numbers in A.P. is 15. If 1, 4,
19 are added to them respectively, the resulting
series is in G.P, then the numbers are
(a) 2, 5, 8
(b) 26, 5, - 16
(c) 3, 5, 7
(d) none of these
(a) (b)
Sol.
32.
Sol.
If the H.M. of two numbers is to their G.M. as 12
: 13, then the numbers are in the ratio
(a) 4 : 9
(b) 9 : 4
(c) 2 : 9
(d) 9 : 2
(a) (b)
Column matching
33. Sum of the series upto n terms
Column I
(a) + (1.5)3 + 23
+ (2.5)3 + …
13
(b) 1(22) + 2(32)
+ 3(42) + …
(c) (n2 – 1)2 + 2(n2– 22)
+ 3(n2 – 32) + …
(d) (2) (5) + (5) (8)
+ (8) (11) + …
Sol.
Column II
1
n(n  1)
12
× (n  2)(3n  5)
(p)
(q) n(3n2 + 6n + 1)
1
(n  1)2
32
1
(n  2)2 
8
1
(s) n 2 (n  1)2
4
(r)
Passage type questions
The sum of three terms of a strictly increasing G.P. is S
and sum of the squares of three terms is S2.
34. 2 lies
(a) (1/3, 2)
(b) (1, 2)
(c) (1/3, 3)
(d) none of these
Sol.
(c)
1
( 5  3)
2
(d)
1
( 3  5)
3
Sol.
36.
If r  2, then 2 equals
1
(b)
(2  3)
7
1
(c) (11  6 2)
(d)
7
Sol. C Put r  2 in (4) and solve.
(a)
35.
If 2 = 2, then value of r equals
(a)
1
(5  3)
2
(b)
Set 2
Column Matching Same
33.
34. Put r  2 in (4) and solve.
35.
36.
1
(3  5)
2
1
(11  7)
4
1
(11  6 2)
5