# Sequence & Series MCQ

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Sequence & Series MCQ

1. Determine 25th term of an A.P. whose 9th term is 6 and common difference is

(a) 21

(b) 17

(c) 20

(d) 14.

d= , tn=- 6, t9=a+(9 -1) X (- ,), or – 6= a+8 X (- ,), Or, – 6 = a + 10, Or, a = – 16

t25 = – 16 + (25 – 1) x =- 16 + (24x ,) = – 16 + 30 = 14. So, option (a) is correct.

2. Which term of the A.P. 5, 13, 21,                             is 181 ?

(a)    31st

(b)    22nd

(c)    23rd

(d)    25th

a = 5   d = 13 – 5 = 8, tn = a + (n – 1) X d,  or, 181 = 5 + (n – 1)  X 8, Or, 8 (n – 1) = 181 – 5,

or, 8 (n – 1)  =176, Or, (n -1) = , Or, (n -1) =22, Or, n = 22 + 1, or n= 23. Hence, 23rd  term is 181

So, option (c) is correct.

3. The ratio of the 7th to the 3rd term of an A.P. is 12:5. Find the ratio of 13th to the 4th term

(a)     6 : 5

(b)     12 : 7

(c)     9 : 5

(d)     10 : 3

t7 = a + (7 – 1)d  = a + 6d, t3 = a + (3 – 1)d  = a + 2d, t13 = a + 12d , t4 = a + 3d

Now  = = . Or, 5a + 30d = 12a + 24d, Or, 30d – 24d = 12a – 5a, Or, 6d = 7a, or d=  a.

4. If the 9th term of an A.P. is 99 and 99th term is 9, find 108th term

(a)    0

(b)    3

(c)    5

(d)    9.

t9 =  a + 8d = 99

t99 =  a + 98d = 9

So, -90d= 90. Or d = – 1, or a = 99 – 8 x (- 1), or a = 99 + 8 = 107.

So, t108 =  107 + 107 (- 1) = 107 – 107 = 0. So, option (a) is correct.

5. Determine the sum of the first 35 terms of an A.P. If t2 = 2 and t7 = 22

(a)    2,210

(b)    2,310

(c)    2,620

(d)    2,810.

t2 = a + (2 – 1)d, Or, a + d = 2   or, a = 2 – d, t7 = a + (7 – 1)d =22, Or, a + 6d = 22. Or, 2 – d + 6d = 22

Or, 5d = 22 – 2 = 20, Or, d = 4, Or,  a = 2 – 4 = – 2

Sum of First 35 terms= s35 =  X {2 X(- 2) + (35 – 1) X 4} =  X {(- 4) + (34 X 4)}

=  X {(- 4) + 136)}=   X 132= 2310. So, option (b) is correct.

6. If the 12th term of an A.P. is 13 and the sum of the first four terms is 24, what is the sum of the first 10 terms ?

(a)    0

(b)    7

(c)    3

(d)    6

t12 =a + (12 – 1) X d, or – 13 = a + 11d, or  a = – (11d + 13)

Now, s4=  X {2a + (4 – 1) Xd}, Or, 24 = 2  X (2a + 3d), Or, 2a + 3d = 12, Or, 2 X (- 11d – 13) + 3d = 12 ,

Or, – 22d – 26 + 3d = 12 , Or, – 19d – 26 = 12, Or, -19d = 12 + 26 = 38. Or d=-2.

So, s10=  X {(2×9)+(10 -1)x (- 2)}= 5 X [18 +{ 9 x( – 2)}] = 5 X (18 – 18) = 5 x 0 = 0. So, option (a) is correct.

7. Write down the 20th term of the G.P. 1, 1, 1, 1,                      .

(a)    -10

(b)    -1

(c)    1

(d)    -2.

Common ratio =  = -1. so, t20=arn1  or, t20=(1) X (-1) 201, or, t20= (1) X (-1) 19 = 1 X (-1) = -1. So, option (b) is correct.

8. The 3rd term of a G.P. is the square of the first term. If the second term is 8, determine the 6th term

(a)    140

(b)    134

(c)    128

(d)    120.

tn= arn1  So, t3= ar31 = ar2.  t2= ar21 = ar. Now, ar2 = a2, or a = r2 . again ar=8. So, r2 X r = 8, or r3=8, or r=2. Again, ar=8, So, a= =4. Now 6th term = t6= ar61 = ar5. = 4X 25 = 4 X 32= 128. So, option (c) is correct.

9. A progression 2, 8, 32,                       is given the 10th number will be

(a)    218

(b)    219

(c)    220

(d)    1.

Here a=2, r= =4, 10th term t10= ar101 = ar9 = = 2 X 49 = 2 X (22)9 = 2 . 218 = 219. So, option (b) is correct.

10. The sum of 40 terms of an A.P. whose first term is 2 and common difference 4, will be

(a)    3,200

(b)    2,100

(c)    2,600

(d)    3,800.

Here a = 2, d = 4. S40=  X {(2 X 2) + (40 -1) X 4}= 20 X {4 + (39 x 4)}= 20 X (4 + 156) = 20 x 160 =3200. So, option (a) is correct.

11. If 15th term of an A.P. is 20 and 20th term is 15, then its 7th term is equal to

(a)    32

(b)    40

(c)    22

(d)    30.

t15 = a + (15 -1)d = 20, Or, a + 14d = 20 ….(1)

t20 = a + (20 -1)d = 15, Or, a + 19d = 15 ….(2)

Subtracting equation (2) from equation (1), we get, -5d=5, or d=-1.

So, a = 20 – 14 x (- 1) = 20 + 14 = 34. So, t7 = 34 + {(7 -1) X (- 1)} = 34 + 6 = 40. So, option (d) is correct.

12. Which term of the A.P. 64, 60, 56, 52,                       is zero ?

(a)    20th

(b)    17th

(c)    12th

(d)    11th.

It is an AP. Here a = 64, d = 60 – 64 = – 4. Let nth term = 0.

Now nth term tn = a + (n -1) d, or 0 = 64 + (n -1) x( – 4), Or,  0 = 64 – 4n +4, Or, 4n = 68, or, n=17. So, option (b) is correct.

13. Find the sum of 20 terms of the sequence 1, 4, 7, 10                      .

(a)    590

(b)    550

(c)    310

(d)    470.

By observation, we see it is an AP. Here a = 1, d = 4-1 = 3. We  know, nth term tn = a X {(n -1)  X d }

So, S20 =  X {2 . 1 + (20 – 1) X 3} = 10 X {2 + (19 x 3)} = 10 (2 + 57) = 10 x 59 = 590. So, option (a) is correct.

14. Which term of the A.P.  5, 8, 11, 24,                       is 320?

(a)    104th

(b)    109th

(c)    106th

(d)    98th.

By observation, you see it is an AP. Here, a = 5    d = 8 – 5 = 3. We know, tn = a + (n – 1)d

So, 320 = 5 + (n – 1)X 3, Or, 320 = 5 + 3n – 3, Or, 320 = 2 + 3n, Or, 3n = 320 – 2 = 318. or, n=106. So, option (c) is correct.

15. Calculate arithmetic mean between 14 and 18

(a)    16

(b)    24

(c)    20

(d)    10.

Let arithmetic mean be x. So, x= = =16. So, option (a) is correct.

16. The 3rd and 5th terms of a G.P. are 12 and 48. Its second term is

(a)    6

(b)    10

(c)    16

(d)    24.

3rd term in G.P. = ar31 = ar2 = 12, 5th term in G.P. = ar51 = ar4 = 48

=  =4, or r2=4, or r=2. So, First Term =a=  = =3. second term = ar21 = ar = 3X2=6. So, option (a) is correct.

17. The last term of the series 5, 7, 9, ……to 21 terms is

(a)    40

(b)    42

(c)    45

(d)    38

We observe, the series is an AP. Here, a = 5, d = 7 – 5 = 2.

21st term= t21  = a + (n – 1) d = 5 + (21 – 1) X 2 = 5 + (20 x 2) = 5 + 40 = 45. So, option (c) is correct.

18. The two arithmetic means between 6 and 14 is

(a) ,

(b) , 7

(c ) – , – 7

(d)

Let two arithmetic mean between -6 and 14 be x1, x2 . So, total terms n=4

Here, a=-6, t4 = 14, t4  = -6  + (4 – 1) X d, Or, 14 = – 6 + 3d, Or, 3d = 20, or d= .

So, x1 = t2= – 6  + (2 – 1) X  = – 6+  = . X2 = t3= – 6  + (3 – 1) X  = – 6  +  =  = 7 . So, the two mean terms are : , 7  . So, option (b) is correct.

19. The sum of 3 consecutive terms in AP is 15 and their product is 80. The numbers are

(a)    2, 8, 5

(b)    8, 2, 5

(c)    2, 5, 8

(d)    None of the above

Let three integer be a – b, a, a + b (this form will make computation easier). So, (a – b) + a + (a + b) = 15

Or, 3a = 15   or, a = 5. Also, (a – b) x a x (a + b ) = 80, Or, (5 – b)x  5 x (5 + b) = 80,

Or, (5 – b) (5 + b) = 16, Or, 52 – b2 = 16, Or, b2 = 25 – 16 = 9, or b= +3,

So, the numbers are (5 – 3), 5, (5 + 3) (considering b=3, numbers in ascending order), or 2.5.8

and {5-(-3)}, 5 , {5+(-3)}  (considering b=-3, numbers in descending order), or 8,5,2, So, option (c) is correct.

20. The arithmetic mean between 33 and 77 is

(a)    30

(b)    40

(c)    55

(d)    66

Let arithmetic mean between 33 and 77 be x1. So, 3rd term t3 =77, 1st term a= 33.

So, t3 = 33 + (3 – 1) X d, Or, 77 = 33 + 2 X d, Or, 2d = 77 – 33 = 44, Or, d = 22

So, x1 =  2nd term= 33 + (2 – 1) X 22 = 33 + 22 = 55

You may also compute single  A.M between 2 terms of a,b = (a+b)/2=(33+77) / 2=55. So, option (c) is correct.

21. The second term of a GP is 243 and the fifth term is 81. The series is

(a)    16, 36, 24, 54, ….

(b)    12, 36, 53, …..

(c)    16, 24, 36, 54, ….

(d)    12, 24, 36, 72

tn = ar n–1, t2 = 24 = ar 2–1 = ar, t5 = ar 5–1 = ar4= 81

=  = . Or r3 =  = 3. So, r= .  So, t2=ar = 24x   = 16.

So, the series is 16, (16 x 3/2), (16 x 3/2 x 3/2), (16 x 3/2 x 3/2 x 3/2) Or, 16, 24, 36, 54. So, option (a) is correct.

22. The sum of 3 numbers of a G. P. is 39 and their product is 729. The numbers are

(a)    18, 27, 9

(b)    81,27,9

(c)    3, 9, 27

(d)    9, 27, 81

Let a/r, a, ar be 3 numbers (this form will make computation easier)

So, (a / r) x a x (ar) = a3 = 729 = 93 or a=9. Now   + a + (ar) = 39. Or,  + 9 + (9Xr) = 39.

+ 9 + (9Xr) = 39.  + (9Xr) =30, or {(9+9r2)} / r  =30, Or, 9r2 + 9 = 30r, Or, 9r2 – 30r + 9 = 0

Or, 9r2 – 3r – 27r + 9 = 0, Or, 3r (3r – 1) – 9 (3r – 1) = 0, Or, (3r – 1) (3r – 9) = 0,

or either 3r – 1 = 0, i.e 3r = 1, So, r = ,

Or, 3r – 9 = 0, or r= =3.

So, when r=3 the numbers are  9, 9X3, or 3,9,27

when r= , the numbers are, or 9/ ( , 9, 9 X   or, 27,9,3,

You may cross verify 3+9+27=39, 3x9x27=729. So, option (c) is correct.

23. If the terms 2x, (x + 10) and (3x + 2) be in A. P. , the value of x is

(a)    8

(b)    12

(c)    6

(d)    4

2x, (x + 10), (3x + 2) are in A.P. So, (x + 10) – 2x = (3x + 2) – (x + 10), Or, x + 10 – 2x = 3x + 2 – x – 10

Or, 10 – x = 2x – 8, Or, 3x = 18, Or, x =  6

So, the AP series is 2×6, 6+10, 3×6+2, or 12, 16, 20. So, option (c) is correct.

24. The A.M. of two positive numbers is 40 and their G.M. is 24. The numbers are

(a)    (72, 8)

(b)    (64, 8)

(c)    (60, 10)

(d)    (50, 30)

Let two numbers be a & b. So,  =40, or a+b=80. Also, (ab)1/2=24, or ab=(24)2=576

Now, (a – b)2 = (a + b)2 – 4 a b, = (80)2 – 4x(24)2 = 6400 – 4x 576 = 6400 – 2304 = 4096 = (64)2

So, a-b=64. a+b=80. So, 2a=(80+64)=144, or a=72. So, b=80-72=8. So, the numbers are 72,8

You mya cross check 72+0=90, (72X8)1/2 = 24 = (576)1/2=24. So, option (a) is correct.

25. Three numbers are in A.P. and their sum is 15. If 8, 6, 4, be added to them respectively, the numbers are in G.P. The numbers are

(a)    1, 6, 9

(b)    3, 5, 6

(c)    3, 5, 7

(d)    2, 5, 7

Let numbers be a – b , a, a + b (this makes computation easy). So, (a – b) + a +( a + b) = 15, Or, 3a = 15, a = 5. So, the Three numbers are 5 – b, 5, 5 + b.

On adding 8, 6, 4, to the respective numbers, the numbers would be  (5 – b + 8), 5+6, (5 + b + 4).

As the numbers would be in GP, So, )} = {(5+6)/(5+b+4)}. Or,  =

Or, (13 – b) (9 + b) = 121, Or, 117 – 9b + 13b – b2 = 121, Or, 4b – b2 = 121 – 117 = 4

Or, 4b – b2 = 4, Or, b2 – 4b + 4 = 0, Or, b2 – 2. b. 2 + 22 = 0, Or, (b – 2)2 = 0, Or, b – 2 = 0, Or, b = 2

So, Three numbers are : 5 – 2, 5, 5 + 2, Or, 3, 5, 7

Cross Check 3 +5 +7 =15. and (3+8), (5+6), (7+4), or 11, 11, 11 . These numbers are in GP. So, option (c) is correct.

26. Find three numbers in A.P. whose sum is 6 and the product is 24

(a)    -2, 2, 6

(b)    -2, -3, 3

(c)    1, 3, 6

(d)    2, 4, 9

Let three numbers be a – b, a, a + b. So, (a – b) + a + (a + b) = 6, Or, 3a = 6, Or, a = 2,

So, (2 – b) X (2) X (2 + b) =  – 24. Or, (2 – b) (2 + b) =  – 12, Or, 22 – b2 = – 12, Or, 4 – b2 = – 12

Or, b2 = 12 + 4 = 16, Or, b = ± 4, So, the Numbers are 2 – 4, 2, 2 + 4, Or, – 2, 2, 6. So, option (a) is correct.

27 An A.P. consists of 60 terms. If the first and last term 7 and 125 respectively, Find its 32nd term.

(a)    30

(b)    91

(c)    73

(d)    69

1st term (a) = 7, t60 = 7 + (60 -1) X d = 125.

So, 125 = 7 +  59d = 125, Or, 59d = 125 – 7, Or, 59d = 118, Or, d = 2

So, t32 = 7 + (32 -1) . 2 = 7 + (31X 2)= 7 +  62 = 69. So, option (d) is correct.

28. The 4 terms in A P in  between 2 and 23 are

(a)    3, 13, 8, 18

(b)    18, 3, 8, 13

(c)    3, 8, 13, 18

(d)    4, 6, 8, 10

There are 4 terms between -2 and 23. So, first term is -2, and 6th term is 23.

t6 = – 2 + (6 -1) X d  =- 2+5d =23, So, -2+5d =23, Or, 5d = 23 + 2 = 25, Or, d = 5,

So, The terms are : – 2, (- 2 + 5=3), (3 + 5=8), (8 + 5=13), (13 + 5=18), 23 = – 2, 3, 8, 13, 18, 23.

So, 4 terms in between are 3, 8, 13, 18. So, option (c) is correct.

29. The 1st and the last term of an AP are 4 and 146. The sum of the terms is 7171. The number of terms is

(a)    101

(b)    103

(c)    96

(d)    98

We know, in AP, Sn=  X (a+l),where a = 1st term,  l = last term, n = number of terms, sn = sum of n terms. Here, sn = 7171,  a = – 4,  l= 146. sum of n terms sn = 7171. So, 7171=  X {(-4)+146}.

Or, 7171 =  X 142}= 71n. or n= = 101. So, number of terms = 101. So, option (a) is correct.

30. Of a series nth term is 2n + 5. Find the nature of the series.

(a)    G.P

(b)    A.P

(c)    H.P

(d)    Any of these

For n = 1  2n + 5 = 2 . 1 +  5 = 7

For n = 2    2n + 5 = 2. 2 +  5  = 9

For n = 3    2n + 5 = 2 . 3 +  5 = 11

For n = 4     2n + 5 = 2 . 4 + 5 = 13

So, the series is 7,9,11,13. From inspection, it appears that the common difference between the numbers = (9 – 7) = (11-9) = (13-11) =2. So, the Series is in A . P. So, option (b) is correct.