DPP Sequences and Series

Answer Keys and Solutions BITSAT Crash Course

ANSWER KEYS
1. (1) 2. (2) 3. (4) 4. (4) 5. (1) 6. (2) 7. (1) 8. (1)
9. (2) 10. (4) 11. (1) 12. (1) 13. (3) 14. (4) 15. (1) 16. (4)
17. (2) 18. (1) 19. (2) 20. (2) 21. (2) 22. (2) 23. (1) 24. (1)
25. (3) 26. (2) 27. (2) 28. (1) 29. (1) 30. (2) 31. (3) 32. (2)
33. (3) 34. (3) 35. (1) 36. (3) 37. (4) 38. (4) 39. (2) 40. (3)

1. (1) Let the numbers be a − d, a, a + d

∴ a + d + a + a − d = 21

⇒ 3a = 21

⇒ a = 7

Again, (a − d)(a + d)= 45

⇒ (a − d) a (a + d)= 45 × a

= 45 × 7

= 315

2. (2)
The numbers between 100 and 500 that are divisible by 7 are, 105, 112, 119, 126, 133, 140, 147, … , 483, 490, 497 .
Let such numbers be n.
Then, 497 = 105 +(n − 1)×7
⇒ n = 57

The number between 100 and 500 that are divisible by 21 are, 105, 126, 147, … , 483 .
Let such numbers be m.
Then, 483 = 105 +(m − 1)×21
⇒ m = 19

∴ Required number = n − m = 57 − 19 = 38
3. (4)
∴ We know that sum of equidistant term from end & beginning is same in any A. P.⇒ a1 + a16 = a2 + a15 = a3 + a14 = a4 + a13 = a5 + a12 = … = a7 + a10

∵ a1 + a4 +a7 + a10 + a13 + a16 = 114

⇒ 3(a1 + a16 )= 114 ⇒ a1 + a16 = 38

Then,
a1 + a6 +a11 + a16

= 2(a1 + a16 )= 76

4. (4)
We know that n term T th
n = Sn − Sn−1 , and difference of any two consecutive terms is common difference d.
Sn+3 − 3Sn+2 + 3Sn+1 − Sn

=(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 )

= d − d = 0 .
5. (1)
If a, b, c are in AP, then b is A.M of a & c.
∴ 2 b = a + c

2 sin 2α−1 4−2 sin 2α

⇒ 28 = 3 + 3

Putting, 3 2 sin 2α
= x we get,
x 81 2
28 = + ⇒ x − 84x + 243 = 0
3 x

⇒ (x − 3)(x − 81) = 0

∴ 3
2 sin 2α
= 3 or 3 4

sin 2α =
1

2
, ∵ sin 2α ≠ 2
Terms are 1, 14, 27, … … then
T6 = 1 + 5(13) = 66

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6. (2)
Now, 1

a1 a2
+
1

a2 a3
+ ….+
1

a4000 a4001

a2 −a1 a3 −a2 a4001 −a4000

1
= ( + +. . . . + )
d a1 a2 a2 a3 a4000 a40001

1 1 1 1 1 1 1
= ( − + − +. . . . + − )
d a1 a2 a2 a3 a4000 a4001

=
1

d
(
1

a1
−
1

a4001
)=
4000

a1 a4001
= 10 (given)
⇒ a1 a4001 = 400 . . . . . .(i)

a1 + a4001 = a2 + a4000 = 50 . . . .(ii)

( as sum of terms equidistant from beginning is always equal)

2 2
∴ (a1 − a4001 ) = (a1 + a4001 ) − 4a1 a4001

2
= (50) − 1600

⇒ |a1 − a4001 |= 30 .
7. (1)
Let, the first 3 terms of the A. P . are a − d, a, a + d

⇒ sum = 3a = 33 ⇒ a = 11

And product (a − d) a (a + d)= a(a 2 2

− d )= 1155

2 2
⇒ 11(11 − d )= 1155

2 1155
⇒ 121 − d = = 105
11

2
⇒ d = 121 − 105 = 16

⇒ d = ±4

For a = 11, d = 4, the A. P . is 7, 11, 15, …

The n term of an A. P . , with first term A and common difference d is T

th
n
= A +(n − 1)d

So, T 11 = 7 + 10 × 4 = 47

For a = 11, d = −4, the A. P . is 15, 11, 7, …

So, T 11 = 15 + 10(−4)= −25.

8. (1) Since,
( Sn )
1
=
2n+3
...(i)
( Sn ) 6n+5
2
n
[ 2a1 + ( n−1 ) d1 ]
2 2n+3
⇒ n =
[ 2a2 + ( n−1 ) d2 ] 6n+5
2
( n−1 )
a1 + d1
2 2n+3
⇒ =
( n−1 ) 6n+5
a2 + d2
2

Put n−1

2
= 12 ⇒ n = 25
a1 +12d1 53
∴ =
a2 +12d2 155

( T13 )
1 53
⇒ =
( T13 ) 155
2

9. (2) We have a 1 + a2 + ... + an = 4500

⇒ a11 + a12 + ... + an = 4500 − 10 × 150 = 3000

⇒ 148 + 146 + ... = 3000

n−10
⇒ ⋅(2 × 148 +(n − 10 − 1)(−2))= 3000
2

Let n - 10 = m
⇒ m × 148 − m(m − 1)= 3000

2
⇒ m − 149m + 3000 = 0

⇒(m − 24)(m − 125)= 0

∴ m = 24, 125

giving n = 34, 135

But for n = 135, we have
a135 = 148 −(135 − 1)(2)= 148 − 268 < 0

But a34 is positive.

Hence, n = 34 is the only answer.

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10. (4)
It is given
2
a1 +a2 +.....+ap p
=
a1 +a2 +.....+aq 2
q

Replacing p = 6 and then 5

a1 +a2 +.....+a6 2
6
=
a1 +a2 +.....+aq q
2

a1 +a2 +.....+a5 2
5
=
a1 +a2 +.....+aq 2
q

on substracting
2 2
a6 6 −5

a1 +.....+q
=
2
.....(i)
q q

similarly replacing p = 21 and then 20 and subtracting we get

2 2
a21 ( 21 ) − ( 20 )

a1 +......+q
=
2
.....(ii)
q q

Divide equation (2) by (1)

a6 / a21 = 11/41.

11. (1) Given log (7

1/2
)
x + log
(7
1/3
)
x + log
(7
1/4
)
x + … 20 times = 460

⇒ (2 + 3 + 4 + … . . +21) log x = 460

7

20
⇒ (2 + 21) log x = 460
2 7

⇒ log x = 2
7

⇒ x = 49

12. (1) x = 1 + a + a 2 1 1 1 x−1

+. . . = ⇒ = 1 − a ⇒ a = 1 − =
1−a x x x

y−1
2 1 1 1
y = 1 + b + b +. . . = ⇒ = 1 − b ⇒ b = 1 − =
1−b y y y

1 1 xy xy
2 2
1 + ab + a b +. . . = = = =
x−1 y−1
1−ab xy− ( x−1 ) ( y−1 ) x+y−1
1− ( ) ( )
x y

13. (3) We know that some of infinite geometric progression a + ar + ar 2

+ ……..=
1−r
a
, |r|< 1

1 1 1
1/6 1/36
Hence, (32)(32) (32) … = 32
1+
6
+
36
+...
= 32
1−1/6

6/5
5
= (2 ) = 64

14. (4)
2

Since,
a+ar+ar 125
=
2 3 4 5 152
a+ar+ar +ar +ar +ar
2
1+r+r 125
⇒ =
( 1+r+r
2
) ( 1+r
3
) 152

3 152
⇒ 1 + r =
125
3
3 27 3
⇒ r = = ( )
125 5

3
⇒ r =
5

15. (1)
SInce , Three numbers in a G.P is given by a , ar , ar
2

Given that : a + ar + ar
2
= 26 ⇒ a(1 + r + r )
2
= 26 and a ⋅ ar + ar ⋅ ar
2
+ a ⋅ ar
2
= 156 ⇒
2 2
a r ( 1 + r + r ) = 156
2 2
a r ( 1+r+r )
156 6
= ⇒ ar = 6 ⇒ a =
a(1+r+r )
2 26 r

6 2 2 2 1
⇒ (1 + r + r ) = 26 ⇒ 6 + 6r + 6r = 26r ⇒ 3r − 10r + 3 = 0 ⇒ r = 3 ,
r 3

1
If r = , then Required terms are 18 , 6 , 2
3

If r = 3 , then Required terms are 2 , 6 , 18

16. (4)
Let the GP be a, ar, ar ,
2 3
ar , . . . . . . . . ar
n−1

Where a = Frist term and r = Common ratio

According to question
We have t1 + t2 = 12 ⇒ a + ar = 12 . . .(i)

2 3
t3 + t4 = 48 ⇒ ar + ar = 48 . . . . .(ii)

Divide the equations (i) & (ii)

2 3
ar +ar 48
=
a+ar 12
2
ar ( 1+r ) 48 2
⇒ = ⇒ r = 4
a ( 1+r ) 12

∴ r = ±2

But the terms are alternately positive and negative,

∴ r = −2

Now using equation (i)a = 1+r

12
=
1−2
12
=
12

−1
= −12

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17. (2)
Let a be the first term and r the common ratio of the G. P . Then
sum= 57 ⇒ 1−r
a
= 57 .......(i)
Sum of the cubes = 9747
3 3 3 3 6
⇒ a + a r + a r + … = 9747
3

= 9747 .......(ii)
a
⇒
3
1−r

Dividing the cube of (i) by (ii), we

3
3
3 ( 1−r ) (57)
get a

3
⋅
3
=
9747
(1−r) a

3
1−r
⇒ = 19
3
(1−r)

2
1+r+r
⇒ = 19
2
(1−r)

2
⇒ 18r − 39r + 18 = 0

⇒ (3r − 2)(6r − 9) = 0

2 3
⇒ r = or r =
3 2

2
⇒ r =
3

because −1 < r < 1 for an infinite G. P )

3
(∵ r ≠
2

2 3 4

18. (1) 1
a+ar+ar +ar +ar

1 1 1 1
= 49
+ + + +
a ar
ar2 ar3 ar4
2 3 4
a [ 1+r+r +r +r ]

⇒ = 49
1 4 3 2
[ r +r +r +r+1 ]
ar4

2 4
⇒ a r = 49

2
⇒ ar = 7

⇒ T3 = 7

T1 + T3 = 35

T1 + 7 = 35

T1 = 28

19. (2) We know, the sum of infinite terms of G. P. S∞ =

a

1−r
.

Here, a =first term, r =common ratio

Then, the sum of infinite terms of G. P. is equal to 20 ⇒ 1−r
a
= 20

⇒ a = 20(1 − r) . . .(i)
2

and the sum of the square of infinite terms of G. P. is equal to 100 ⇒

a
= 100
2
1−r
2 2
( 20 ) ( 1−r )
⇒
( 1−r ) ( 1+r )
= 100 (from (i))
⇒ 4(1 − r)= 1 + r

3
⇒ r =
5

20. (2)
a + 3c = 2b and b 2
= 4ac
2
2b−a b
⇒ c = =
3 4a

2 2 2 2
∴ 8ab– 4a = 3b ⇒ 4a – 8ab + 3b = 0

2 8±√64−4×4×3
4
a

2
–8
a

b
+ 3= 0 ⇒ a

b
=
2×4
b

a 2±√4−3 2±1 3 1
= = = , .
b 2 2 2 2

21. (2) We know, the sum of infinite terms of G. P. S∞ =

a

1−r
.

Here, a =first term, r =common ratio

Then, the sum of infinite terms of G. P. is equal to 20 ⇒ 1−r
a
= 20

⇒ a = 20(1 − r) . . .(i)
2

and the sum of the square of infinite terms of G. P. is equal to 100 ⇒

a
= 100
2
1−r
2 2
( 20 ) ( 1−r )
⇒
( 1−r ) ( 1+r )
= 100 (from (i))
⇒ 4(1 − r)= 1 + r

3
⇒ r =
5

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22. (2) Let the three numbers of the GP be a, ar and ar where r is the common ratio.
2

According to the given condition, a, 2ar and ar are in AP.

2

2
a+ar
∴ 2ar =
2

2
∴ a + ar = 4ar

2
∴ r + 1 = 4r

2
∴ r − 4r + 1 = 0

4±√16−4
∴ r =
2

4±2√3
= = 2 ± √3
2

As it is an 'increasing' GP, r > 1 .

∴ r = 2 + √3 .
23. (1) b = ar
2
c = ar

3a, 7b and 15c are in A.P.

⇒ 14b = 3a + 15c

2
⇒ 14(ar)= 3a + 15(ar )

2
⇒ 14r = 3 + 15r

2
⇒ 15r − 14r + 3 = 0 ⇒(3r − 1)(5r − 3)= 0

1 3
r = , .
3 5

only acceptable value is r = 1

3
, because
1
r ∈(0, ]
2

∴ common difference = 7b − 3a = 7ar − 3a =

7

3
a − 3a = −
2

3
a

2
∴ T4 = 3a +(4 − 1)(− a)= 3a − 2a = a
3

24. (1)
Let the first term of the given sequence be a
First 11 terms are: a, a + 2, . . . a + 20

Since, eleventh term of the sequence is the first of last 11 terms of this sequence.
Last 11 terms are:a + 20, (a + 20). 2, . . . (a + 20) ⋅ 2
10

Thus, the sequence obtained is as follows:

2 10
a, a + 2, … , a + 20, (a + 20)2, (a + 20)2 , … … , (a + 20)2

th
(11+1)
Now middle term of the AP is the ( 2
) term, i.e., 6 term and the same goes for GP.
th

So, (a + 10) = (a + 20)2 5

6
( 1−2 )
5 5
⇒ 10 − 20 ⋅ 2 = a(2 − 1)⇒ 10 = a
5
2 −1

6
( 1−2 )

Middle term of the whole sequence = T 11 = a + 20 = 10

5
+ 20
2 −1

−1 10
= 10[ ]=
5 5
2 −1 1−2

25. (3) Since, a+b

2
,
1
,
2

b+c
are in AP.
b

2 1 1
⇒ = 2( + )
b a+b b+c

1 2b+a+c
⇒ =( )
2
b ab+ac+b +bc

2 2
⇒ ab + ac + b + bc = 2b + ab + bc

2
⇒ b = ac

⇒ a, b, c are in GP.
26. (2) Let the first term of an AP be a and common difference be d.
Since, a + 3d = 5

3
. . . . .(i)

and a + 7d = 3. . . . . .(ii)
On solving an equation (i) and (ii), we get
2 1
a = , d =
3 3

2 5 7
∴ T6 = a + 5d = + =
3 3 3

⇒ 6th term of HP is 3

7
.

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27. (2) S = 1 + 4x + 7x
2
+ 10x +. … … .
3

2 3
xS = x + 4x + 7x +. … … . .

On subtracting, we get,
2 3
S(1 − x)= 1 + 3x + 3x + 3x +. … … . .

1
= 1 + 3x( ), |x|< 1
1−x

1−x+3x 1+2x
(1 − x)S = =
1−x 1−x

(given)
1+2x 35
S = =
2 16
( 1−x )

2
16 + 32x = 35 + 35x − 70x

2
⇒ 35x − 102x + 19 = 0

⇒(5x − 1)(7x − 19)= 0

1 19
x = ,
5 7

But, |x|< 1 ⇒ x = 1

28. (1)
Let,S = (10)
9
+ 2(11)
1
(10)
8
+ 3(11)
2
(10)
7
+ ...... + 10(11)
9
= k(10)
9

2 9
9 11 11 11
S = (10) {1 + 2( )+3( ) +. .... +10( ) }
10 10 10

2 9

So,k = 1 + 2(
11 11 11
)+3( ) +. .... +10( )
10 10 10

2 3 10
11 11 11 11 11
( )k = + 2( ) + 3( ) +. .... +10( )
10 10 10 10 10

Subtracting
2 9 10
k 11 11 11 11
−( )= 1 +( )+( ) +. … … + ( ) − 10( )
10 10 10 10 10

Here,
First term = 1
Common ratio= 11

10

Number of terms= 10
10
11
1{ ( ) −1 }
10 ⎡ 10 ⎤
k 11
∴ = 10( ) −⎢ ⎥
10 10 11
{ −1 }
⎣ 10 ⎦

10 10
11 11
k = 10[10( ) − 10{( ) − 1}]
10 10

10 10
11 11
k =[100( ) − 100( ) + 100]
10 10

k = 100

29. (1) We know that geometric mean of numbers a 1, a2 … . . an = (a1 . a2 … . . an )

n

1/n
Hence, Geometric mean = (7.7 2
.7
3
…7 )
n

1/n
1+2+3+…+n
= (7 )

1/n
n ( n+1 ) n+1
( )
2
= (7 2 ) = 7

30. (2)
a+b
A. M. of a and b = 2

So we have
n n
a +b a+b
⇒ =
n−1 n−1 2
a +b

n n n n−1 n−1 n
⇒ 2a + 2b = a + a. b + b. a + b

n n n−1 n−1
⇒ a + b − a. b − b. a = 0

n−1 n−1
⇒ a (a − b)+b (b − a)= 0

n−1 n−1
⇒(a − b)(a − b )= 0

Since a ≠ b
n−1 n−1
⇒(a − b )= 0

n−1
a
⇒ ( ) = 1
b

n−1 0
a a
⇒ ( ) = ( )
b b

⇒ n − 1 = 0

⇒ n = 1

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31. (3) Let d be the common difference. Then

83 7
(− )−( )
2 2 −5
d = =
18 2

So, the 17 mean is given by th

83 5 83 5 78
− −(− )= − + = − = −39
2 2 2 2 2

32. (2)
Let S n
=
1

1⋅2
+
1

2⋅3
+
3⋅4
1
+. . . +
1

n⋅ ( n+1 )

2−1 3−2 4−3 ( n+1 ) −n

⇒ Sn = + + +. . . +
1⋅2 2⋅3 3⋅4 n⋅ ( n+1 )

1 1 1 1 1 1 1
⇒ Sn = 1 − + − + − +. . . + −
2 2 3 3 4 n n+1

1
⇒ Sn = 1 −
n+1

⇒ Sn =
n+1
n
.
33. (3)
1 1 1
tr = (1 + 2 + 3 … . . + r)= . r (r + 1)
r r 2

1 n
S = Σtr = ∑ (r + 1)
2 r=1

20
1 20 1
S = ∑ r + ∑1
2 r=1 2
r=1

1 1
S = . 20(20 + 1)+ . 20
4 2

S = 115

34. (3)
We have, t n = n(n !)

={(n + 1)−1}n !

=(n + 1) ! − n !

Now, ∑ 15

n=1
tn = 16! − 15! + 15! − 14!+. . . . . . . +2! − 1

= 16! − 1

35. (1) Let, I(r)= 4

r

2
=
2
r

2
r +r +1 ( r +r+1 ) ( r −r+1 )

2 2
( r +r+1 ) − ( r −r+1 )
1
= { }
2 2 2
( r +r+1 ) ( r −r+1 )

1 1 1
= − ( − )
2 2 2
r +r+1 r −r+1

1
= − (V (r)−V (r − 1))
2
n 1
⇒ ∑ I(r)= − (V (n)−V (0))
r=1 2

1 1
= − ( − 1)
2 2
n +n+1
2
( n +n )
1 1 1
= (1 − )=
2 2
n +n+1 2 2
( n +n+1 )

36. (3) t r
=
1

( r+1 ) √r+r√r+1

( √r+1−√r )
1
= ×
√r ( r+1 ) { √r+1+√r } ( √r+1−√r )

√r+1−√r
1 1
= = −
√r ( r+1 ) √r √r+1

99 1 9
P = ∑ tr = 1 − =
r=1 10
√100

37. (4) We have,

5 + 55 + 555 + . . . . . . . . . to n terms
= 5[1 + 11 + 111 + . . . . . . . . . to n terms]
5
= [9 + 99 + 999 + ... n terms]
9
5 2 3 n
= [(10 − 1)+(10 − 1)+(10 − 1)+ ... +(10 − 1)]
9
5 2 3 n
= [(10 + 10 + 10 + ... +10 )−(1 + 1 + 1 + ... n times)]
9
n
( 10 −1 )
5
= [10 × − n]
9 10−1

5 10 n
= [ (10 − 1)−n]
9 9

5 n+1
= [10 − 10 − 9n]
81
5 101
S 100 = [10 − 910]
81

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38. (4)
2 2 2 2
3n ( 1 +2 +3 +…+n )
3n×n ( n+1 ) ( 2n+1 )
tn = =
2n+1 6 ( 2n+1 )

1 3 2
= (n + n )
2
15 15 1 3 2
∴ S15 = ∑ tn = ∑ (n + n )
n=1 n=1 2

1 15 3 15 2
= [∑ n + ∑ n ]
2 n=1 n=1

2 2
n ( n+1 ) n ( n+1 ) ( n+2 )
1 15×16 15×16×31 3 2
= ×[( ) + ] {∑ n = ( ) , ∑n =( )}
2 2 6 2 6

= 7200 + 620

= 7820

39. (2) AM ≥ GM for positive numbers. So,

x 4
4 +
4x x 4
≥ √4 ⋅ x
= 2.
2 4

x 4
4 + x
≥ 4
4

40. (3)
AM ≥ GM

a+b+b+c+c+c 6
2 3
≥ √ab c
6

a+2b+3c 6
2 3
⇒ ≥ √ab c
6

⇒ a b c
2 3
≤ 2
6
.

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