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The correct option is indicated by (√) mark with explanation in the rightmost column

**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= _{, }t_{n}=- 6, t_{9}=a+(9 -1) X (- _{,}), or – 6= a+8 X (- _{,}), Or, – 6 = a + 10, Or, a = – 16

t_{25} = – 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, t_{n} = 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

t_{7} = a + (7 – 1)d = a + 6d, t_{3} = a + (3 – 1)d = a + 2d, t_{13} = a + 12d , t_{4} = a + 3d

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

t_{13}/t_{4}= = = = = = 10:3. So, option (d) is correct. |

**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.

t_{9} = a + 8d = 99

t_{99} = a + 98d = 9

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

So, t_{108} = 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 t**_{2}** = 2 and t**_{7}** = 22**

(a) 2,210

(b) 2,310

(c) 2,620

(d) 2,810.

t_{2} = a + (2 – 1)d, Or, a + d = 2 or, a = 2 – d, t_{7} = 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= s_{35} = 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

t_{12} =a + (12 – 1) X d, or – 13 = a + 11d, or a = – (11d + 13)

Now, s_{4}= 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, s_{10}= 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, }t_{20}=ar^{n}^{–}^{1 } or, t_{20}=(1) X (-1) ^{20}^{–}^{1}, or, t_{20}= (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.

t_{n}= ar^{n}^{–}^{1} So, t_{3}= ar^{3}^{–}^{1 }= ar^{2}. t_{2}= ar^{2}^{–}^{1 }= ar. Now, ar^{2} = a^{2}, or a = r^{2 }. again ar=8. So, r^{2} X r = 8, or r^{3}=8, or r=2. Again, ar=8, So, a= =4. Now 6^{th} term = t^{6}= ar^{6}^{–}^{1 }= ar^{5.} = 4X 2^{5} = 4 X 32= 128. So, option **(c)** is correct.

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

(a) 2^{18}

(b) 2^{19}

(c) 2^{20}

(d) 1.

Here a=2, r= =4, 10^{th} term t_{10}= ar^{10}^{–}^{1 }= ar^{9} = = 2 X 4^{9} = 2 X (2^{2})^{9} = 2 . 2^{18 }= 2^{19}. 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. S_{40}= 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.

t_{15} = a + (15 -1)d = 20, Or, a + 14d = 20 ….(1)

t_{20} = 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, t_{7} = 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 t_{n} = 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 t_{n} = a X {(n -1) X d }

So, S_{20} = 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, t_{n} = 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. = ar^{3}^{–}^{1 }= ar^{2} = 12, 5th term in G.P. = ar^{5}^{–}^{1 }= ar^{4} = 48

= =4, or r^{2}=4, or r=2. So, First Term =a= = =3. second term = ar^{2}^{–}^{1 }= 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.

21^{st} term= t_{21} = 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 x_{1}, x_{2 }. So, total terms n=4

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

**So, **x_{1} = t_{2}= – 6 + (2 – 1) X = – 6+ = . X_{2} = t_{3}= – 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, 5^{2} – b^{2} = 16, Or, b^{2} = 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 x_{1}. So, 3^{rd} term t_{3 }=77, 1^{st} term a= 33.

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

So, x_{1} = 2^{nd} 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

t_{n} = ar ^{n}^{–1}, t_{2} = 24 = ar ^{2–1} = ar, t_{5} = ar ^{5–1} = ar^{4}= 81

= = . Or r^{3} = = ^{3}. So, r= . So, t_{2}=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) = a^{3} = 729 = 9^{3} or a=9. Now + a + (ar) = 39. Or, + 9 + (9Xr) = 39.

+ 9 + (9Xr) = 39. + (9Xr) =30, or {(9+9r^{2})} / r =30, Or, 9r^{2} + 9 = 30r, Or, 9r^{2} – 30r + 9 = 0

Or, 9r^{2} – 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 – b^{2} = 121, Or, 4b – b^{2} = 121 – 117 = 4

Or, 4b – b^{2} = 4, Or, b^{2} – 4b + 4 = 0, Or, b^{2} – 2. b. 2 + 2^{2} = 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, 2^{2} – b^{2} = – 12, Or, 4 – b^{2} = – 12

Or, b^{2} = 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, t_{60} = 7 + (60 -1) X d = 125.

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

So, t_{32} = 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.

t_{6} = – 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, S_{n}= X (a+l),where a = 1st term, l = last term, n = number of terms, s_{n} = sum of n terms. Here, s_{n} = 7171, a = – 4, l= 146. sum of n terms s_{n} = 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.