SRI CHAITANYA IIT ACADEMY, INDIA

Sri Chaitanya IIT Academy., India.

A.P, TELANGANA, KARNATAKA, TAMILNADU, MAHARASHTRA, DELHI, RANCHI

A right Choice for the Real Aspirant

ICON Central Office – Madhapur – Hyderabad
MATHS : SEQUENCE & SERIES
1. Suppose that the number of terms in an A.P. is 2k, k  N. If the sum of all odd terms of
theA.P. is 40, the sum of all even terms is 55 and the last term of the A.P. exceeds the first
term by 27, then k is equal to
1) 6 2) 4 3) 8 4) 5
[22-01-25, shift-2]
Topic: Sequences & Series
Sub Topic: A.P / Sum of n Terms
Type:1
Ans: 4
Sol: Given a1 , a2 , a3 ,.....a2 k are in A.P
k k
But  a2r 1  40,  a2r  55
r 1 r 1

40 55
 a1    k  1 d   kd
k k
k  5

1(a) Let a1.a2 , a3 ,     be in an Arithmetic progression of positive terms. Let

Ak  a12  a22  a32  a42      a22k 1  a22k if A3  153, A5  435, and a12  a22  a32  66, then a17  A77 
1) 910 2) 920 3) -910 4) -920
Ans: 1
Sol: Ak  kd 2a   2k  1 d   A3  153,
A5  435, d  3, a  1
k
1(b) For x  0, the least value of k, for which 41 x  41 x , ,16 x  16 x are three consecutive terms of
2
A.P
1) 8 2) 4 3) 10 4) 16
Ans: 3
 1  1 
Sol: K  4  4 x  x    42 x  2 x 
 4   4 
 4  2  2  10
SRI CHAITANYA IIT ACADEMY, INDIA
100
2. Let  an  be a sequence such that a0  0, a1  1 / 2 and 2an2  5an1  3an , n  0,1,2,3,...... Then  ak
k 1

is equal to : [28-01-25, SHIFT-1]

a) 3a99  100 b) 3a100  100 c) 3a100  100 d) 3a99  100
Topic : Series & Sequences
Sub Topic : Series & Sequences
Type: 1
Key: b
1
Sol: a0  0, a1 
2
2an2  5an1  3an
2 x2  5x  3  0  x  1,3 / 2
n
3
 an  A1n  B  
2

n
3
 an  1   
2
100 
3 
100 k

 k 
a   ( 1)   2  
  
k 1 k 1 

 3  3  
100

 2    2   1
   
 100 
3
1
2
  3 100 
 100  3     1 
 2  
 
 3.(a100 )  100

2a. Let  an  be a sequence such that a0  0, a1  2 and 4an2  16an1  15 , n=0,1,2,3,……..Then

16a5 
a)544 b)-544 c)2882 d)-2882
Key: c
Sol: 4an2  16an1  15
4 x2  16 x  15
4 x2  16 x  15  0
4 x2  10 x  6 x  15  0
2x(2x  5)  3(2 x  5)  0
(2x  5)(2x  3)  0
x 3/ 2 & x 5/ 2

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SRI CHAITANYA IIT ACADEMY, INDIA
n n
3 5
 an  A    B  
2 2
a0  A  B  0
3 5
a1  A B  2
2 2
3 A  5B  4
 A  2, B  2
n n
3 5
an  2    2  
2 2
5 5
3 5
a5  2    2  
2 2
35 55
a5   
24 24
16a5  243  3125  2882

2(b) Let  an  be a sequence such that a1  1, a2  4 and an2  3  4an1 , n=0,1,2,3,……. Then
50
2 an  50 
n 1

a) 3a50 b) 2a50 c) 3a51 d) 2a51

Key: a
Sol: an2  3  4an1
x2  3  4 x
x2  4 x  3  0
x2  3 x  x  3  0
x( x  3)  1( x  3)  0
x  1,3
an  A(1)n  B(3)n
a1  1  A  3B  1
a2  4  A  9B  4
1 1
A ,B 
2 2
1 n 1 n
an  (1)  (3)
2 2
50
1 1 3(350  1)
  an  (50) 
n 1 2 2 3 1
3  350  1 
 25   
2  2 
3
 25  .a50
2
50  3a50

2

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SRI CHAITANYA IIT ACADEMY, INDIA

3. The roots of the quadratic equation 3x 2  px  q  0 and 10th and 11th terms of an arithmetic
3
progression with common difference . If the sum of the first 11 terms of this arithmetic
2
progression is 88,then q-2p is equal to________

[23-01-25, SHIFT-02]

Topic: Series & Sequences

Sub topic: AP
Type: 1

Key:474
Sol: S11 
11
2
 2a  11  1 d   88  a  5d  8
3 1
a  8  5  
2 2
1 3 1  27
T10  a  9d   9.   14
2 2 2
1 3 31
T11  a  10d   10. 
2 2 2
31 p
S .O.R  14  
2 3
31 q
P.O.R  14. 
2 3
p 59
 , q  7.31.3
3 2
59.3
p q  651
2
q-2p=651-59  3=474

3(a). Roots of a quadratic equation x2  9dx  20d 2  0 (d  0) are r th ,(r  1)th terms (r>1) of an
A.P with common difference  the sum of 1st (r+1) terms going A.P is ad then r
Key : 8
Sol: Givne 4d,5d ARE r th ,(r  1)th terms of an A.P with common difference d
Tr  a  (r  1)d  4d
Tr  1  a  rd  5d
a  d (5  r )

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SRI CHAITANYA IIT ACADEMY, INDIA
r 1
 2a  rd   9d
Sr  1 
2
Given (r  1)[2d (5  r )  rd ]  18d
(10  r )(r  1)  18
r 8

4(O). Let Tr be the r th term of an A.P. If for some m, Tm = 1/25, T25 = 1/20 and
2m
 Tr
25
20 Tr  13 then 5m is [28-01-25, SHIFT-02]
r 1
r m
a) 112 b) 126 c) 98 d) 142
Topic : Series & Sequences
Sub Topic : Arthematic progression
Type: 1
Key: b
25
1 1
Sol: Tm  ,T25  ,20 Tr  13
25 20 r 1
1
Tm  a  (m  1)d  ........(1)
25
1
T25  a  24d 
20
25  1 1
20.  a    13  a 
2  20  500
25 1
also,20 S25  20. 2a  24d   13  d 
2 500
1 m 1 1
from(1)    m  20
500 500 25
Now,
2m 40
5m  Tr  100  Tr  126
r m r 20

50
4(a).Let Tr be the r th term of an A.P. If for some m, Tm = 1/100, T50 = 1/20 and 40 Tr  51
r 1

then
a) 30 b) 31 c) 35 d) 40
Key: b
1 1
Sol.: Tm  a  (m  1)d  , T50  a  49d 
100 20
50
50  1
40 Tr  51  40.  a    51
r 1 2  20 

 50 20a  1  51
51 1 1
 20a  1   a
50 50 1000

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SRI CHAITANYA IIT ACADEMY, INDIA
1 1 1 49 1
Since a  49d   49d     d
20 20 1000 1000 1000
1 1 1 1 1 1
Since a  (m  1)d    (m  1)    1  m  1   m 10
100 1000 1000 100 1000 100
2m
20m  Tr
r m
20
= 20(10)  Tr
r 10

 20  2 19  10  2 9 
 200       
 2 1000 1000  2 1000 1000 
10   21  11   31 
200   2   = 200  5   31 .
2   1000  1000  1000 

1 1 1 1
4(b). Let , , ,...., ,  xi  0,for i  1, 2,3,.... be an A.P., such that x1  4 and x21  20 . If n is the
x1 x2 x3 xn
least +ve integers, for which xn  50 , then n=
a) 25 b) 3 c) 23 d) 24
Key: 1
1 1 1 1 1 1 1  5 4 1
Sol.:   20d    20d  20d     
x21 x1 20 4 20 4 20 20 5
1 1 1 1
Since xn  50     ( x  1)d 
xn 50 x1 50
1 (n  1) 1
  
4 (100) 50
25  n  1 1
 
100 50
 26 – n < 2  n > 24
 The least value of n is 25.

5. Let a1 , a2 , a3 ............. be a GP of increasing positive terms. if a1a5  28 and a2  a4  29

Then a6 is

1) 526 2) 812 3) 628 4) 784

[22-01-25, SHIFT-1]
Topic: SEQUENCE & SERIES
Sub Topic: G.P.
Type:1
Key : 4
Sol: r > 1
a1a5  28  a1 (a1r 4 )  28  a 21r 4  28  a1r 2  2 7

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a2  a4  29  a1r  a1r 3  29  a1r (1  r 2 )  29
a1r  a2  28, a2  1  a1r
1
 
5
a6  a1r 5  28  784
28

5(A). In a G.P the ratio of the sum of the first 5 terms to the sum of their reciprocals is 49
and sum of the first and third term is 35. The fifth term of the G.P is
7 7 7
1)7 2) 3) 4)
2 4 8
Key : 3
a a 2
Sol : Let five terms in G.P be 2 , , a, ar , ar
r r
a(r 2  r 1  1  r  r 2 )
Given  49
1 2 1 2
(r  r  1  r  r )
a
a 2  49 a  7
a
Also  a  35
r2
 a =-7 is not possible
a
A = 7 and 2  28
r
1
r2 
4
Now ,fifth term = ar 2
7
=
4
5(B). Let a1 , a2 , a3 ... be a G.P. such that a1  0, a1  a2  4 and a3  a4  16 then a5 
1) 16 2) -16 3) 64 4) -64
Key : 4
Sol: a1  a2  4
a1  a1r  4 -------------(1)
a3  a4  16
a1r 2  a1r 3  16 ------------(2)
Solving (1) & (2) r 2  4
r  2
If r = -2
(1)  a1  2a1  4
 a1  4
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SRI CHAITANYA IIT ACADEMY, INDIA
a1  4
a5  ar 4
 4(2)4
= - 64
25
5(C) . In an increasing G.P , a1 , a2 , a3 ..... if a2  a6  and a3a5  25 then a 4 =
2
1) 2 2)3 3)4 4) 5
Key : 4
25
Sol : a2  a6 
2
25
ar  ar 5 
2
25
ar (1  r 4 ) 
2
a3a5  25
ar 2 ar 4  25
ar 3  5
a4  5
n
(2n  1)(2n  1)(2n  1)(2n  5) n
1 
6. If Tr  , then lim    is equal to
n
r 1 64 r 1  Tr 
2 1
1) 2) 1 3) 4) 0
3 3
[22-01-25, shift-01]
Topic: SEQUENCE & SERIES
Sub Topic: general term
Type:1
Key : 3
(2n  1)(2n  1)(2n  3)(2n  5) (2n  3)(2n  1)(2n  1)(2n  3)
Sol: Tn  Sn  Sn1  
64 64
8(2n  1)(2n  1)(2n  3) (2n  1)(2n  1)(2n  3)
 
64 8
1 8

Tr (2r  1)(2r  1)(2r  3)
n
1  n
 8  2
lim    n 
 lim  3
n
r 1  Tr r  1  (2r  1)(2r  1)(2n  3) 
n
 1 1   1 1  2
 2lim      2lim   
n
r 1  (2r  1)(2r  1) (2r  1)(2r  3)  
n (1)(3) (2n  1)(2n  3)  3
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n  n  1 n  2  n
6(A). If Tr 
2026
then lim  =?
3 n
r 1 Tr
1) 0 2) 1013 3) 2026 4) 4052
Key: 3
Sol: tn  sn  sn1
n n 1
n(n  1)(n  2) (n  1)(n)(n  1)
  tr   tr  
r 1 r 1 3 3
n(n  1) n(n  1)
  n  2   (n  1)   (3)
3 3
 n(n  1)
1 1 1 1
   
tn n(n  1) n n  1
2026  n 1 1   1 
  2026        2026 1  
Tr  r 1  r r  1    n 1
 n 
 2026  
 n 1
2026  n 
 lim   lim(2026)    2026(1)
n Tr n
 n 1
=2026

7. Consider an A. P. of positive integers, whose sum of the first three terms is 54 and the
sum of the first twenty terms lies between 1600 and 1800. Then its 11th term is:
1) 90 2)108 3) 84 4)122
[29-01-25-SHIFT-01]

Topic: Sequence & Series

Sub Topic: A.P
Type: 1
Key: 1
Solution: a  a  d  a  2d  54  a  d  18
20
S 20   2a  19d   20a  190d  20  a  d   170d
2
 360 170d
1600  360  170d  1800
1240  170d  1440
7.29  d  8.47  d  8
T11  a  10d  10  80  90

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7(a). Consider an A. P. of positive integers, whose sum of the first three terms is 30 and the
sum of the first 24 terms lies between 1200 and 1400. Then its 13th term is:
1) 50 2)54 3) 46 4) 58
Key: 2
Sol: a  a  d  a  2d  30
a  d  10 ……….(1)
S24  122a  23d 
= 12 2 10  d   23d  …….From eq (1)
S24  240  12  21d
1200  S24  1400
1200  240  12  21d  1400
960  252d  1160
3.8  d  4.6
d 4
a6
T13  a  12d
 6  12  4 = 54

7(b). Consider an A. P. of positive integers, whose sum of the first 5 terms is 40 and the
sum of the first 16 terms lies between 250 and 350. Then its 9th term is:
1) 16 2)19 3) 20 4) 22
Key: 3
Sol: a  a  d  a  2d  a  3d  a  4d  40
5a  10d  40
a  2d  8 ………(1)
s16  82a  15d 
S16  16 8  2d   120d
S16  128  88d
250  128  88d  350
1.38  d  2.52
d 2
a4
t9  4  8  2  20

7(C). Let a1 , a2 ,.... a2024 be an Arithmetic Progression such that

a1   a5  a10  a15  ...  a2020   a2024  2233 . Then a1  a2  a3  ....  a2024 is equal to
Key:11132
Sol.: a1  a3  a10  ....  a2020  a2024  2233

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In an A.P. the sum of terms equidistant from ends is equal
a1  a2024  a5  a2020  a10  a2015 ....
 203 pairs
 203  a1  a2024   2233
Hence
2024
s2024  (a1  a2024 )
2
= 1012 × 11
= 11132.

8. Let a1 , a2 ,.... a2024 be an Arithmetic Progression such that

a1   a5  a10  a15  ...  a2020   a2024  2233 . Then a1  a2  a3  ....  a2024 is equal to
[29-01-25-SHIFT-02]

TOPIC: SEQUENCE & SERIES

SUB TOPIC: A.P.
TYPE:1
Key:11132
Sol.: a1  a3  a10  ....  a2020  a2024  2233
In an A.P. the sum of terms equidistant from ends is equal
a1  a2024  a5  a2020  a10  a2015 ....
 203 pairs
 203  a1  a2024   2233
Hence
2024
s2024  (a1  a2024 )
2
= 1012 × 11
= 11132.

8a) If a1 , a2 , a3 .........an is sequence of positive number which are in A.P with common difference d
16
 A
and a1  a4  a7  ......  a16  147 then maximum value of a1.a2 ......a16 is   then the value of
B
 
B
A  ______
Key : 49
Sol : a1  a4  a7  ......  a16  147
3 a1  a16   147

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SRI CHAITANYA IIT ACADEMY, INDIA
Now using
a1  a2  ......  a16
  a1........a16 
1/16

16
8  a1  a16 
  a1.......a16 
1/16

16
8........49
  a1.......a16 
1/16

16
1/16 16
 A  49 
       a1a2 .......a16 
B  2 
A  49,   2

 a  
2 2
49  49

8b) Let a1 , a2 , a3 .........a2029 will be an arithmetic progression such that

a6   a10  a20  a30  ........a2020   a2024  1020 . Then a1  a2  a3  .........a2029 is equal to
Key : 10145
Sol : a6   a10  a20  ........a2020   a2024  1020
a6  a2024  a10  a2020  a20  a2010  ........
 102 pairs
 102  a6  a2024   1020
a6  a2024  10
a1  a2024  a6  a2024
2029
s2029   a1  a2029 
2
 2029  5
=10145

1 1 1 1
9. Let S n      .... upto ‘n’ terms. If the sum of the first six terms of an A.P
2 6 12 20
with first term –p and common difference ‘p’ is 2026S 2025, , then the absolute
difference between 20th and 15th terms of the A.P is

1)20 2)45 3)90 4)25

[24-01-25-SHIFT-01]

Topic: SEQUENCE & SERIES

Sub Topic: Telescopic Sum
Type:2
Key:4
Solution:

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SRI CHAITANYA IIT ACADEMY, INDIA

1 1 1
9A. Let 𝑆𝑛 = + + + ⋯ . . + upto n terms (d>0) and Sm' be the sum of first
1+𝑑 1+3𝑑+2𝑑 2 1+5𝑑+6𝑑 2
𝑆2025 ′
m terms of an A.P with first term k and common difference 2k (k>0) such that −
2025
𝑆2024 ′ 𝑑
= 8𝑑, then [√ ] = ([.] denotes GIF)
2024 𝑘
1)2 2)3 3)4 4)1
Key:1
Solution:
𝑛
Telescopically adding the terms of 𝑆𝑛 𝑔𝑖𝑣𝑒𝑠𝑆𝑛 = and
1+𝑛𝑑
𝑚
𝑆𝑚 ′ = (2𝑘 + (𝑚 − 1)(2𝑘)) = 𝑘𝑚2
2
′
𝑆𝑚 𝑆𝑚+1 ′ 𝑆𝑚 ′
⟹ = 𝑘𝑚 ⟹ − = 𝑘 = 8𝑑
𝑚 𝑚+1 𝑚

10. If the first term of an A.P. is 3 and the sum of its first four terms is equal to one-fifth of
the next four terms, then the sum of the first 20 terms is equal to
1) -1020 2) – 1080 3) – 1200 4) – 120
[23-02-25-SHIFT-01]

Topic: SEQUENCE & SERIES

Sub Topic: A.P.
Type:1

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SRI CHAITANYA IIT ACADEMY, INDIA

10A. If the first term of an A.P. is 4 and the sum of its first seven terms is half of the sum of the
next seven terms, then the sum of the first 28 terms is equal to
1) 490 2) 480 3) 390 4) 590
Key : (1)
1
Sol : Let a  4 , Given  28  21d    28  70 d   d  1
2
S28   2  4    28  11   14  35  490
28
2

10B. If the first term of an A.P. is 2, the common difference is 1 and the sum of the cubes of the
first n terms is equal to 11 times the sum of the first n terms, then ‘n’ is equal to
1) 2 2) 5 3) 4 4) 3
Key : (4)

Sol : tn  2   n  11  n  1

 tn    n  1
3 3

Sn 
n
2
 2  2    n  11   n  3
n
2
n
  n  1  11.  n  3
3

2
n3
10C. The sum of the first 10 terms of an A.P. is 300, while the sum of the next 10 term is 600,
then the 50th term in equal to
1) 163.5 2) 153.5 3) 169.5 4) 163.25
Key : (1)

10
Sol : S10   2a  9d   300  10a  45d  2a  9d  60...1
2
S20  S10  600

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SRI CHAITANYA IIT ACADEMY, INDIA
20 10
 2a  19d    2a  9d   600
2 2

10  2a  19d   5 2a  9d   600

2a  29d  120 …. (2)

(2) – (1) 20d  60  d  3

33
(1)  2a  9  3  60  2a  33  a 
2
33 33
t50  a   n  1 d    50  1 3   147  163.5
2 2
10D. If the common difference of A.P. is 4 and the sum of its first six terms is equal to one-third
of the sum of the next six terms, then the sum of first 30 terms is equal to
1) 1600 2) 1700 3) 1900 4) 1800
Key : (4)

1
Sol : Given S6   S12  S6 
3

1
 3  2a  20   6  2a  44   3  2a  20    a  2
3

S30 
30
2
 2  2    30  1 4   15  4  116   15  120  1800
10E. Let S n denotes the sum of the first n terms of an A.P. If S 2 n  3S n , then the value of
S3n

Sn
1) 3 2) 9 3) 6 4) 4
Key : (3)

Solution : S2 n  3.Sn

2n n
 2a   2n  1 d   3.  2a   n  1 d 
2 2

2a   n  1 d

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SRI CHAITANYA IIT ACADEMY, INDIA
3n
 2a   3n  1 d  3  n  1 d   3n  1 d 
    3.4 n d  6
S3 n 2

Sn n
 2a   n  1 d   n  1 d   n  1 d 2nd
2

1
For positive integers n, if 4an   n2  5n  6 and, Sn   
n
11.  then the value of 507 S2025 is:
k 1  ak 

1) 540 2) 675 3)1350 4) 135

[28-01-25, shift-02]
Topic: SEQUENCES & SERIES
Sub Topic: TELESCOPIC SUM
Type:3

ANS: 2
SOL: 4an  n 2  5n  6
n
1
Sn  
k 1 ak

n2  5n  6  n  2  n  3
an  
4 4
1 4

an  n  2  n  3
2025
 1   1  1 1 
S2025  4     , S2025  4  
n 1  n  2   n  3   3 2028 
 675  675
 4 , ,507 S2025675
 2028  507

11(i) For every positive integer n  2, t1  t2   tn  2n 2  9n  13 . If tk  103, then what is the value of
k?
1) 20 2) 21 3) 22 4) 24
ANS: 4
Sol: tn  Sn  Sn1
 ,   4 

11(ii) If A  1  r a  r 2a  r 3a  , a  0, and B  a  r 2b  r 4b  r 6b  , a  0, for r  1, then 4,0

A

P
2

 x, y 
B
1
is
Y 40
B   ,4

equal to
A 1 
2

1) log B A 2) log1B 1  A 3) log B1   4) 2log A B

B  A 

ANS: 3

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SRI CHAITANYA IIT ACADEMY, INDIA
1 1
Sol: A  ,B  a
1 r a
1  r 2b

11(iii) . If  Tr  n  2n2  9n  13 and Sn   Tr , then

n n
2
S 20 is equal to
r 1 r 1 3
1) 405 2) 460 3) 355 4) 604
ANS: 2
n
1
(iv). For positive integer n, if 2an  4n2  1 and Sn   , then the value of 2025 S1012 is
k 1 ak

1) 2024 2) 2025 3) 1012 4) 1013

ANS: 3
1 1 1
Sol: use  
an 2n  1 2n  1

12. In an arithmetic progression, if S 40  1030 and S12  57 , then S30  S10 is equal to
1) 515 2) 505 3) 510 4) 525
[24-01-25-SHIFT-02]

Topic: Sequences of series

Sub topic: n-th terms
Type: 1

Key: 1
Sol: S40  1030  20[2a  39d ]

 0,2 

A ,  2 

 2,0 
P
1

y  2 B   , 2 

12(a). If S n denotes the sum of n terms of AP and S n  16 and S6  48 then S10 =
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SRI CHAITANYA IIT ACADEMY, INDIA
1) -320 2) -260 3) -380 4) -410
Key: 1
4
Sol: S4  16   2a  3d   16 ----------(1)
2
6
S6  48   2a  5d   48 ___________(2)
2
By using a = 22, b = -12
S10  320

12(b). If S n denotes the sum of n terms of an A.P then Sn3  3S n 2  3S n1 - S n 

1) 0 2) 1 3) 3 4) 2

Key : 1
Sol: ( Sn3  Sn 2 )  2( Sn 2  Sn1 )  ( Sn1  Sn )
Tn3  2Tn2  Tn1  (Tn3  Tn 2 )(Tn 2  Tn1 )  0
1 1 1
13. If 7  5   5     2  5  2   3  5  3   ....... , then the value of  is
7 7 7
6 1
1) 1 2) 6 3) 4)
7 7
[24-01-25-SHIFT-02]

Topic: Sequences and series

Sub topic: sum of the terms
Type: 1

Key: 2
1 1 1
Sol: 7  5   5     2  5  2   3  5  3   .......
7 7 7

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SRI CHAITANYA IIT ACADEMY, INDIA
5  5 5 3
75   2  3  2  ......
7 7 7 7 7
 1 1 1  1 2 3 
7  5 1   2  3  ........      2  3  ....... 
 7 7 7  7 7 7 
  2
 1   1  1 
7  5 5      1  
 1 
1  7  7 
 7
7  1 49 
7  5     . 
6  7 36 
35  7  7 7
7     
6  36  6 36
 6

1 1 1
13(a). 9  7   7     2  7  2   3 (7  3 )  .................. , then the values of  is
9 9 9
1) 8 2) 9 3) 10 4) 11
Key: 1
 7 7 7  1 2 3 
Sol: 9   7   2  3  .........     2  3  ......
 9 9 9  9 9 9 
 1 1 1  1 2 3 
 7 1   2  3        . 1   2      
 9 9 9  9 9 9 
  2
 1    1
=7  1
1  9  9 
1  
 9
 9   9
2

= 7    
 8  9 8
63  81
 9 
8 9 64
9 9

8 64
 8

1 1
13(b). 11  9   9     2  9  2   ........ , then the values of  is
11 11
1) 10 2) 11 3)8 4) 9

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SRI CHAITANYA IIT ACADEMY, INDIA
Key : 1
9  9 2
Sol: 11  9    2  2  ........
11 11 11 11
 1 1 1  1 2 3 
11  9 1   2  3  .....     2  3 ....
 11 11 11  11 11 11 
 
 1  1 2 3 
11  9      1   2  .....
1  
1 11  11 11 
 11 
2
 11    1
11  9    1  
10  11  11 
2
99  10 
11  
10 11  11 
11   11 
2


10 11 10 
11  11 
  
10  100 
  10

14. The interior angles of a polygon with n sides, are in an A.P with common difference 60. If
the largest interior angle of the polygon is 2190, then n is equal to ______.
[28-01-25, shift-02]
Topic: SEQUENCES & SERIES OR PROGRESSIONS
Sub Topic: SUM OF N TERMS
Type:1
ANS: 20
SOL:
n
2
 2a   n  1 6  =  n  2 .1800
 a n  3n2  3n   n  2 .1800 ---------------------(1)
From given a   n 1 60  2190
a  225  6n --------------------(2)
1 and 2
n2  14n  120  0
n  20, 6  Re jected 
n = 20

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SRI CHAITANYA IIT ACADEMY, INDIA

14(i). If each interior angle is equal to 144̊, then how many sides does a regular polygon have?
1) 6 2) 9 3) 8 4) 10
ANS: 4
n2
SOL:    180
0

 n 

14(ii) . What is the sum of interior angles of a 10 sides polygon?

1) 720degrees 2) 360 degrees 3) 240 degrees 4) 120 degrees
ANS: 1
n2
Sol:   
 180
0

 n 

14(iii) . What is the sum of interior angles of a kite?

1) 120̊ 2) 360̊ 3) 240̊ 4) 540̊
ANS: 2
n2
SOL;    180
0

 n 

14(iv) . The interior angle of a convex polygon are in A.P. The smallest angle is 120̊ and the
common difference is 5̊. The number of sides is
1) 16 2) 9
3) 9 or 16, both values being permissible 4) 7

ANS: 3
n2
Sol:    180
0

 n 

14(v). What is the value of the interior angle of a regular octagon?

1) 135̊ 2) 270̊ 3) 360̊ 4) 90̊

ANS: 1
n2
Sol:   
 180
0

 n 

n
k 3  6k 2  11k  5
15 . The value of lim  is
n
k 1 (k  3)!
1) 4 2) 2 3) 7 4) 5
3 3 3
Key : 4
Sol:

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SRI CHAITANYA IIT ACADEMY, INDIA
n
K 3  6 K 2  11K  6  1
lim 
n
1  K  3 !
 lim 
 K  1 K  2  ( K  3)  1
n

n
1  K  3 !
n
 lim 
 K  1 K  2  ( K  3)  1
1  K  3  K  2  K  1 K !  K  3!
n

n
1 1
 lim  
n
1 K !  K  3 !
1 1 1 1 1 1 1 
 lim    .........     .....  
n 1!
 2! n ! 4! 5! 6!  n  3 ! 
1 1 1 5
   
1 2 6 3

 k 3  9k 2  26 K  23  m
n
15A. The value of lim    is  where gcd of  m, n  is1.Then m  n 
n
k 1   k  4 !  n
1) 41 2) 31 3) 51 4) 21

Key: 1
Sol:

lim 
 k  4  k  3 k  2   1
n

n
k 1  k  4 !
n
1 1
lim  
k 1  k  1  k  4 !
n

1 1 1 1  1 1 1 
lim             
n 2! 3! 4!
  n  1!  5! 6!  n  4 !
1 1 1 17
   
2 6 24 24

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