PROGRESSION SERIES SOLVED QUESTIONS

PROGRESSION SERIES QUESTIONS

         1. Locate the ninth term and sixteenth term of the A.P. 5,8,11, 14, 17...

     A -40

     B - 50

     C - 60

     D - 70

 Solution : In the given A.P. we have a=5, d= (8-5) = 3

 ∴ Tn= a+ (n-1) d= 5+ (n-1)³ = 3n+2

 T16= (3*16+2) = 50

        2.   What numbers of term arrive in the A.P. 7, 13, 19, 25... 205?

        A - 34

        B - 35

        C - 36

        D - 37

  Solution : Let the given A.P contain A.P. contain n terms. At that point,

   A=7, d = (13-7)= 6andTn = 205

    ∴ a+ (n-1) d =205 ⇒ 7+ (n-1)*6 = 198 ⇒ (n-1) =33 ⇒ n = 34

    Given A.P contains 34 terms.

 3.   Locate the ninth term and the nth term of the G.P. 3,6,12, 24 ...

     A - 738, 4^n-1

     B - 748, 5^n-1

     C - 758, 6^n-1

      D - 768, 6^n-1

  Solution : Given numbers are in G.P in which a= 3 and r =6/3 = 2.

  ∴ Tn = ar^n-1 ⇒ T9= 3*28 = (3*256) = 768

  Tn = 3*2^n-1 = 6^n-1

 4.  (1+2+3+4....+99+100+99+98+.....+3+2=1) =?

     A - 200

     B - 500

     C - 1000

     D - 10000

  Solution : Give series = 2 (1+2+3+4+?..+99) + 100

   = 2 * 99/2 * (1+99) + 100=2* 4950 +100 = 9900 + 100 = 10000

 

 5.  If the sixth term of an H.P. is 10 and the 11th term is 18 Find the 16th term.

     A. 30

     B. 75                   

     C. 90                    

     D. 80

  Solution : Option C

   Here a+5d=1/10

   And a+10d=1/18

   Solving these two equations, we get a=13/90 and d=-2/225

   we have a+15d=(13/90)-(2/15)=1/90

    Hence the 16th term is 90.

 

  6. The first and third terms of an A.P. {A1} are A1 = a and A3 = b, and a1 = a and a5 = b   respectively be the first and fifth terms of another A.P. {a1}. Find the ratio of An+1 and    a2n+1.

       A. 2

       B. 4

       C. 1

       D. b/a

   Solution : Option C
    For the first A.P. a + 2D1 = b i.e. D1= [b – a]/2.
    For the second A.P. a + 4D2 = b i.e. D2 = [b – a]/4.
    Now the value of An+1will be a + (n + 1 – 1) * [b – a]/2.
    Similarly the value of a2n+1 will be a + (2n + 1 – 1) * [b – a]/4. Taking the ratio of these      two, we get:
  Thus 3rd option is the answer.

 

  7.  An Arithmetic Progression has 23 terms, the sum of the middle three terms of this     arithmetic progression is 720, and the sum of the last three terms of this Arithmetic     Progression is 1320. What is the 18th term of this Arithmetic Progression?

      A. 240

      B. 360

      C. 340

      D. 440

  Solution : Option B
    The middle three terms of this Arithmetic Progression represent the eleventh, twelth &       thirteenth term.
   The average of these three terms will represent the twelth term i.e. the 12th term will be     720/3 = 240. The average of the last three terms will be the twenty second term i.e. 1320/3 = 440
 The difference of twenty second term and the twelth term will give us ten times the  difference i.e. 440 – 240 = 200/10 = 20 will be the difference. If the twelth term is 240, the  18th term will be 240 + 6d i.e. 240 + 6 * 20 = 360 will be the 18th term. If you want to apply  the Arithmetic Progression formula to solve, you can also do it like that. Therefore,  2nd option is the correct answer.

 

 8.  The first term of an AP is 10 and the last term is 28. If the sum of all terms is 190, what   is the common difference?

       A. 5

       B. 3

       C. 1

       D. 2

 Solution : Option D
 The sum of all terms is (n/2) (a + l) = 190. This gives us n = 10.
 The 10th term is (a + 9d) = 28. Substituting for a, we get d = 2.

 

 9. The sum of three numbers in Arithmetic Progression is 72 and their product is 11880.   What are the numbers?

       A. 21, 24, 27

       B. 12, 24, 36

       C. 18, 24, 30

       D. 15, 24, 33

Solution : Option D
 Assuming that the numbers in the Arithmetic Progression are (a – d), a, (a + d) and their   sum is 72, we get the middle number as 24. Now, the product (a – d) (a + d) = 495. Solving, we get d = 9. Therefore, the numbers are 15, 24 and 33.

 

 10. In an AP, the ratio of the 2nd term to the 7th term is 1/3. If the 5th term is 11, what is  the 15th term?

        A. 28

        B. 31

        C. 33

        D. 36

 Solution : Option B
  The 2nd and the 7th terms are (a + d) and (a + 6d) respectively. The ratio of these terms is  1/3. Solving this ratio, we get 2a = 3d. The 5th term is (a + 4d) = 11. Substituting for a, we  get a = 3 and d = 2. Therefore, the 15th term is (a + 14d) = 31.

  

  11. The sum of the first 3 terms in an AP is 51 and that of the last 3 is 99. If the AP has 11      terms, what is the arithmetic mean of the middle 3 terms? 

         A. 17

         B. 19

         C. 21

         D. 25

  Solution : Option D
  The AP can be expressed as a, (a + d), ---, (a +10d). The sum of the first 3 terms is (3a + 3d) = 51 and that of the last 3 is (3a + 27d) = 99. Solving these equations, we get a = 15 and d = 2. The arithmetic mean of the 3 terms is the value of the middle term = 25.

 

 12.  The 5th term of an AP is 17/6 and the 9th term is 25/6. What is the 12th term?

        A. 29/6

        B. 31/6

        C. 33/6

        D. 34/6

 Solution : Option B

 The 5th and 9th terms are (a + 4d) and (a + 8d). Equating these terms with their values and solving as simultaneous equations, we get a = 3/2 and d = 1/3. So the 12th term is (a + 11d) = (3/2) + (11/3) = 31/6.

 

 13.  Find the 5th term of the G. P.: 1/7,1/14, 1/28 ...

         A. 1/108

         B. 1/112

         C. 1/128

         D. 2/115

 Solution : Option B
 Here the first term is 1/7 and the common ratio = 1/2
 We have Tn=ar^n-1

 

  14. Find the sum of the following infinite G. P. 

        A. 1/2

        B. 1

        C. 1/3

        D. 1/5

Solution : Option D
Here a = 1/3 and r = -2/3
Hence the required sum =

 

  15. Find the G. M. between 4/9 and 169/9.

       A. 

       B. 

       C. 

       D. 

Solution : Option B
The geometric mean between two terms ‘a’ and ‘b’ are given by  
Hence the G.M. between 4/9 and 169/9 = 

 

 16. The arithmetic mean between two numbers is 75 and their geometric mean is 21. Find     the numbers.

        A. 133 and 17

        B. 63 and 87

        C. 3 and 147

        D. 73 and 77

Solution : Option C
Let the required numbers are ‘a’ and ‘b’.
We have  = 75

a+b = 150……….(i)
Also  = 21 or ab = 441

We know that (a-b)² = (a+b)² - 4ab = 150² - 4 x 441 = 22500 -1764 = 20736
a-b = 144………..(ii)
Adding (i) and (ii), we get 2a = 294

a = 147 , b = 3
So the numbers are 147 and 3.

 

  17. The product of first three terms of a G. P. is 512. If we add 2 to its second term, the     three terms form an A. P. Find the terms of the G. P.

         A. 4, 8, 16

         B. 16, 8, 4

         C. 12, 24, 48

         D. Option A or B

Solution : Option D
Let the first three terms of G.P. are  ,a,ar

Given that  * a * ar = 512

ð  A³ = 512

ð  A = 8
Now ,a+2 are in A.P.

⇒ (a+2)-  = ar - (a-2)

⇒ 10 -  = 8r - 10

⇒ 8r +  = 20

⇒ 8r² - 20r + 8 =0
⇒ 2r² - 5r + 2 =0
⇒ 2r² - 4r - r + 2 =0
⇒ 2r(r-2) - (r-2) =0
⇒(2r-1)(r-2) = 0
⇒ r = 2 or 
When r = 2, the terms are 4, 8, 16
When r = 1/2, then the terms are 16, 8, 4.

 

 18. If 6, 24, 96 are in G. P. then insert two more numbers in this progression so that it again   forms a G. P.

        A. 12 and 48

        B. 16 and 48

        C. 12 and 40

        D. 20 and 32

Solution : Option A
G.M. between 6 & 24 = 
GM between 24 & 96 =  = 48

If we insert 12 between 6 and 24 and 48 between 24 & 96, then 6, 12, 24, 48, 96 from a G.P. So the required numbers are 12 & 48.

 

  19. Find the sum of the geometric series 6, -3, -3/2, -3/4, …. Up to 10 terms

             A. 512/255

             B. 256/1021

             C. 1024/515

             D. 1023/256

Solution : Option D
The given GP is 6, -3,  ---- upto 10 terms

Here a = 6, r = -1/2.

= 

 

 20.  There are three terms x, y, z between 4&40 such that (i) their sum is 37, (ii) 4, x, y are   consecutive terms of an A.P. and (iii) y, z, 40 are the consecutive terms of a G.P. Find value of z.

         A. 20

         B. 10

         C. 12

         D. 15

Solution : Option A
Here 4, x, y are in AP

=> 2x = 4+y ---(1)
Again y, z and 40 are in GP

=>  z² = 40y

=>  y =  ---(2)    

Also x + y + z = 37

=> x = 37 – y – z ---(3)
Put the value of (3) in (1) we get
2 (37 – y – z) = 4 + y

=>  74 – 2y – 2z = 4 +y
=>  3y = 70 – 2z
Putting the value of y from (2), we get = 70 – 2z

=>  3z² = 2800 – 80z
=>  3z² + 80z – 2800 = 0
Solving we get z = 20, - 140/3
Rejecting z =  we get z = 20

∴   y = 10 & x = 7

21. Divide one hundred loaves of bread among five persons so that the second person receives as much more than the first as the third receives more than the second and the fourth more than the third and the fifth more than the fourth. Also, the ratio of loaves received by the first two to those received by the last three is 4 : 21. How much does the fourth person get?

Solution:

It is clear from the question that the number of loaves of bread received by the five people will be in A.P.

Let them be (a - 2d), (a - d), a, (a + d), (a + 2d). Their sum is 5a which has to be equal to 100. So, a = 20.
Loaves received by the first two = (a - 2d) + (a - d) = (2a - 3d) and those received by the last three = a + (a + d) + (a + 2d) = (3a + 3d). Their ratio has been given as 4 : 21.
So, (2a - 3d) / (3a + 3d) = 4/21. Substituting a = 20, we get d = 8.
Hence the fourth person gets (a + d) = 28 loaves

22. Two harmonic means between  ½ , 4/17 are

Solution:
If  x, y are the harmonics means, then 12, x ,y, 417 are in H.P.
Then 2, 1x,1y,174 are in A.P.
From above, d=174−23=943=34
1x=2+34=114
1y=2+64=144=72
Harmonic means are 411,27

23. The sum of infinite terms of the series 1 + 3/4 + 5/42 + 7/43....... equals:

Solution : Let,
S = 1 + 3/4 + 5/42 + 7/43 + ...........upto ∞ - - - (a) 
By dividing S by 4,
S/4 = 1/4 + 3/42 + 5/43 7/44.......upto ∞ - - - (b)
By Subtracting (b) from (a)
S - S/4 = 1 + (3/4 - 1/4) + (5/42 - 3/42) + . . .
3S/4 = 1 + 2/4 + 2/42 + 2/43 + 2/44 + . . . . . upto ∞
3S/4 = 1 + 2[1/4 + 1/42 + 1/43 . . . . upto ∞ ]
Here, 1/4 + 1/42 + 1/43 is in G.P.
We know that, Sum of infinite terms in GP = a1−r
3S/4 = 1 + 2[(1/4) / {1 - (1/4)}]
3S/4 = 1 + 2 x 1/3 = 5/3
S = (5 x 4) / (3 x 3) = 20/9

24. For a fibonacci sequence of positive integers, from the third term onwards, each term in the sequence is the sum of the previous two terms in that sequence. If the difference in squares of seventh and sixth terms of this sequence is 517, what is the tenth term of this sequence?

Solution :

Let sixth and seventh terms are t6, and t7, then:
(t7)2−(t6)2 = 517,
(t6−t7)(t6+t7)= 11 x 47, since both are prime numbers, so we have
t6−t7 = 11 and t6+t7 = 47, from these two equations
t6 = 18 and t7 = 29,
t8= 18 + 29 = 47, t9 = 47 + 29 = 76, t10 = 76 + 47 = 123

 

Directions for questions 6 to 10: Read the instructions and solve the questions accordingly

1. x%y means x+ y
2. x&y means x – y
3. x$y means x² + y²
4. x@y means x² – y²

 

25. Find 3%(4&5)

   A. 5

   B. 18

   C. 3

   D. 2

   Solution : Option D

     3% (4-5) = 3% (-1) = 3+ (-1) = 2

     26. Find the value of (2&5) % (3@7)

        A. -45

        B. -35

        C. -43

        D. -38

    Solution: Option C

      (2-5) % (3² – 7²) = (-3) % (-40) = -3-40 = -43

27. In an arithmetic progression the first term is 10 and its common difference is 8. If the general term is a_n , find a₁₉ - a₁₁.

      A. 65

    B. 64 

      C. 66

      D. 61

      E. 63

Correct Ans:64

Given a = 10 and common difference = 8.
General Term a_n = a +(n-1)d => a_n = 10 + 8(n-1) = 8n + 2
a₁₉ = 152 + 2=154
a₁₁ = 88 + 2 = 90
a₁₉ - a₁₁ = 64

28. In an arithmetic progression the first term is 9 and its common difference is 6. If the general term is a_n , find a₁₇ - a₁₁.

    A. 37

    B. 39

    C. 38

    D. 36

    E. 35

Correct Ans:36

Given a = 9 and common difference = 6.
General Term a_n = a +(n-1)d => a_n = 9 + 6(n-1) = 6n + 3
a₁₇ = 102 + 3=105
a₁₁ = 66 + 3 = 69
a₁₇ - a₁₁ = 36

29. In an arithmetic progression the first term is 9 and its common difference is 7. If the general term is a_n , find a₃₂ - a₂₆.

      A. 42

      B. 45

       C. 44

      D. 39

       E .41

Correct Ans:42

Given a = 9 and common difference = 7.
General Term a_n = a +(n-1)d => a_n = 9 + 7(n-1) = 7n + 2
a₃₂ = 224 + 2=226
a₂₆ = 182 + 2 = 184
a₃₂ - a₂₆ = 42

30. In an arithmetic progression the first term is 5 and its common difference is 3. If the general term is a_n , find a₁₈ - a₁₃.

        A. 15

        B. 18

         C. 17

D. 12

         E. 14

Correct Ans:15

Given a = 5 and common difference = 3.
General Term a_n = a +(n-1)d => a_n = 5 + 3(n-1) = 3n + 2
a₁₈ = 54 + 2=56
a₁₃ = 39 + 2 = 41
a₁₈ - a₁₃ = 15

31. In an arithmetic progression the first term is 12 and its common difference is 5. If the general term is (a_n) , find a₁₇ - a₁₁.

     A. 31

     B. 33

      C. 30

     D. 27

     E. 29

Correct Ans:30

Given a = 12 and common difference = 5.
General Term a_n = a +(n-1)d => a_n = 12 + 5(n-1) = 5n + 7
a₁₇ = 85 + 7=92
a₁₁ = 55 + 7 = 62
a₁₇ - a₁₁ = 30

32. In an arithmetic progression the first term is 7 and its common difference is 6. If the general term is (a_n) , find a₂₁ - a₁₆

      A. 31

      B. 30

      C. 32

      D. 27

      E. 41

Correct Ans:30

Given a = 7 and common difference = 6.
General Term a_n = a +(n-1)d
=> a_n = 7 + 6(n-1) = 6n + 1

a₂₁ = 126 + 1=127
a₁₆ = 96 + 1 = 97
a₂₁ - a₁₆ = 30

33. In an arithmetic progression the first term is 8 and its common difference is 5. If the general term is (a_n) , find a₁₁ - a₅

     A. 31

     B. 33

     C. 32

     D. 30

     E. 29

Correct Ans:30

Given a = 8 and common difference = 5.
General Term a_n= a +( n-1 )d => a_n = 8 + 5(n-1) = 5n + 3
a₁₁ = 55 + 3 = 58 a₅ = 25 + 3 = 28
a₁₁ - a₅ = 30

34. In an arithmetic progression the first term is 9 and its common difference is 7. If the general term is a_n , find (a₁₂ - a₆) / 6 .

    A. 6.5

    B. 7.5

    C. 7

    D. 8

    E. 41

Correct Ans:7

Given a = 9 and common difference = 7.
General Term a_n = a +(n-1)d => a_n = 9 + 7(n-1) = 7n + 2
a₁₂ = 84 + 2=86
a₆ = 42 + 2 = 44
a₁₂ - a₆ = 42

35. In an arithmetic progression the first term is 7 and its common difference is 6. If the general term is a_n , find a₁₀ - a₇.

    A. 19

    B. 21

    C.18

    D. 15

    E. 17

Correct Ans:18

Given a = 7 and common difference = 6.
General Term a_n = a +(n-1)d => a_n = 7 + 6(n-1) = 6n + 1
a₁₀ = 60 + 1=61
a₇ = 42 + 1 = 43
a₁₀ - a₇ = 18

36. In an arithmetic progression the first term is 11 and its common difference is 6. If the general term is a_n , find a₁₈ - a₁₃.

     A. 31

     B. 33

     C. 32

     D. 30

     E. 29

Correct Ans:30

Given a = 11 and common difference = 6.
General Term a_n = a +(n-1)d => a_n = 11 + 6(n-1) = 6n + 5
a₁₈ = 108 + 5=113
a₁₃ = 78 + 5 = 83
a₁₈ - a₁₃ = 30

37. In an arithmetic progression the first term is 7 and its common difference is 3. If the general term is a_n , find a₂₀ - a₁₃.

       A. 22

       B. 24

       C. 21

       D. 18

        E. 20

Correct Ans:21

Given a = 7 and common difference = 3.
General Term a_n = a +(n-1)d => a_n = 7 + 3(n-1) = 3n + 4
a₂₀ = 60 + 4=64
a₁₃ = 39 + 4 = 43
a₂₀ - a₁₃ = 21

38. In an arithmetic progression the first term is 7 and its common difference is 3. If the general term is a_n, find a₂₂ - a₁₄.

      A. 25

      B. 27

      C. 24

      D. 21

      E .23

Correct Ans:24

Given a = 7 and common difference = 3.
General Term a_n = a +(n-1)d => a_n = 7 + 3(n-1) = 3n + 4
a₂₂ = 66 + 4=70
a₁₄ = 42 + 4 = 46
a₂₂ - a₁₄ = 24

39. In an arithmetic progression the first term is 8 and its common difference is 3. If the general term is a_n, find a₄₂ - a₃₁.

       A. 34

       B. 36

       C. 33

       D. 30

       E .32

Correct Ans:33

Given a = 8 and common difference = 3.
General Term a_n = a +(n-1)d => a_n = 8 + 3(n-1) = 3n + 5
a₄₂ = 126 + 5=131
a₃₁ = 93 + 5 = 98
a₄₂ - a₃₁ = 33

40. In an arithmetic progression the first term is 4 and its common difference is 2. If the general term is a_n , find a₂₂ - a₁₆.

A. 13

      B. 15

      C. 14

      D. 12

      E. 11

Correct Ans:12

Given a = 4 and common difference = 2.
General Term a_n = a +(n-1)d => a_n = 4 + 2(n-1) = 2n + 2
a₂₂ = 44 + 2=46
a₁₆ = 32 + 2 = 34
a₂₂ - a₁₆ = 12

41. In an arithmetic progression the first term is 5 and its common difference is 4. If the general term is a_n , find a₆ x a₃.

       A. 325

       B. 415

       C. 335

       D. 445

       E. 11

Correct Ans:325

Given a = 5 and common difference = 4.
General Term a_n = a +(n-1)d => a_n = 5 + 4(n-1) = 4n + 1
a₆ = 24 + 1= 25
a₃ = 12 + 1 = 13
a₆* a₃ = 325

42. In an arithmetic progression the first term is 21 and its common difference is 2. If the general term is a_n , find a₂₁ - a₁₄.

      A. 15

      B. 17

      C. 14

      D. 11

      E. 13

Correct Ans:14

Given a = 21 and common difference = 2.
General Term a_n = a +(n-1)d => a_n = 21 + 2(n-1) = 2n + 19
a₂₁ = 42 + 19 = 61
a₁₄ = 28 + 19 = 47
a₂₁ - a₁₄ = 14

43. In an arithmetic progression the first term is 11 and its common difference is 2. If the general term is a_n , find a₂₁ - a₁₃.

      A. 17

      B. 16

      C. 18

      D. 13

      E. 15

Correct Ans:16

Given a = 11 and common difference = 2.
General Term a_n = a +(n-1)d
=> a_n = 11 + 2(n-1) = 2n + 9
a₂₁ = 42 + 9=51
a₁₃ = 26 + 9 = 35
a₂₁ - a₁₃ = 16

44. In an arithmetic progression the first term is 7 and its common difference is 1. If the general term is a_n , find a₁₁ - a₈.

      A. 4

      B. 6

      C. 3

      D. 0

      E. 2

Correct Ans:3

Given a = 7 and common difference = 1.
General Term a_n = a +(n-1)d => a_n = 7 + 1(n-1) = 1n + 6
a₁₁ = 11 + 6 = 17
a₈ = 8 + 6 = 14
a₁₁ - a₈ = 3

45. In an arithmetic progression the first term is 6 and its common difference is 2. If the general term is a_n , find a₁₀ - a₆.

     A. 9

     B. 11

     C. 10

     D. 8

     E. 7

Correct Ans:8

Given a = 6 and common difference = 2.
General Term a_n = a +(n-1)d => a_n = 6 + 2(n-1) = 2n + 4
a₁₀ = 20 + 4 = 24
a₆ = 12 + 4 = 16
a₁₀ - a₆ = 8

46. In an arithmetic progression the first term is 5 and its common difference is 4. If the general term is a_n , find a₆ - a_3.

       A. 12

       B. 15

       C. 14

       D. 9

       E. 11

Correct Ans:12

Given a = 5 and common difference = 4.
General Term a_n = a +(n-1)d => a_n = 5 + 4(n-1) = 4n + 1
a₆ = 24 + 1= 25
a₃ = 12 + 1 = 13
a₆ - a₃ = 12

47.In an arithmetic progression the first term is 7 and its common difference is 6. If the general term is a_n , find a₁₂ - a₇.

     A. 31

      B. 30

      C. 32

      D. 27

      E. 29

Correct Ans:30

Given a = 7 and common difference = 6.
General Term a_n = a +(n-1)d => a_n = 7 + 6(n-1) = 6n + 1
a₁₂ = 72 + 1 = 73
a₇ = 42 + 1 = 43
a₁₂ - a₇ = 73 - 43 = 30

48. In an arithmetic progression the first term is 11 and its common difference is 2. If the general term is a_n , find a₁₂ - a₅.

      A. 15

      B. 17

      C. 14

      D. 11

      E. 13

Correct Ans:14

Given a = 11 and common difference = 2.
General Term a_n = a +(n-1)d => a_n = 11 + 2(n-1) = 2n + 9
a₁₂ = 24 + 9 = 33
a₅ = 10 + 9 = 19
a₁₂ - a₅ = 33 - 19 = 14

49. In an arithmetic progression the first term is 7 and its common difference is 2. If the general term is a_n ,find a₇ - a₃.

      A. 9

      B. 8

       C. 10

      D. 5

       E. 7

Correct Ans:8

Given a = 7 and common difference = 2.
General Term a_n = a +(n-1)d => a_n = 7 + 2(n-1) = 2n + 5
a₇ = 14 + 5 = 19
a₃ = 6 + 5 = 11
a₇ - a₃ = 8

50. In an arithmetic progression the first term is 6 and its common difference is 5. If the general term is a_n , find a₆ - a₄.

      A. 10

      B. 13

      C. 12

      D. 7

      E. 9

Correct Ans:10

Given a = 6 and common difference = 5.
General Term a_n = a +(n-1)d => a_n = 6 + 5(n-1) = 5n + 1
a₆ = 30 + 1= 31
a₄ = 20 + 1 = 21
a₆ - a₄ = 10