BMM10233 Chapter 2 Progressions

Chapter 2 | Progressions | BMM10233 | 12 of 204

Every progression is a sequence of number. We shall discuss three most important progressions here.

Arithmetic Progression (A.P.)

A sequence of numbers in which the difference between any two consecutive terms is constant is an Arithmetic progression.

The first term and the constant difference (called the common difference), denoted by ‘a’ and ‘d’ respectively are the

components of any A.P.

The general A.P. is given by a, a + d, a + 2d , a + 3d ...

Clearly the nth term of an A.P., Tn is given by T

n = a + (n – 1)d

Sum of n terms of an A.P.

Let a, a + d, a + 2d ... be the given A.P. Then, the sum of n terms of the A.P., Sn is given by,

[ ] [ ]n

n nS 2a (n 1)d or a 1

2 2= + − + where l is the last term of the A.P.

SOLVED EXAMPLES

Example: 1.

Find the 12th term of the following A.P.

1, 4, 7, 10 …

Sol:

Here a = 1, d = 3, n = 12

nT a (n 1)d= + −

12T 1 (12 1)3 1 33 34= + − = + =

Example: 2.

Find the 15th term of the following A.P.

1, – 4, – 9, – 14 … n

Sol:

Here a = 1, d = -5, n = 15

nT a (n 1)d= + −

15T 1 (15 1)( 5) 69= + − − = −

Example: 3.

Which terms of the A.P. 7, 10, 13, … is

(1) 283 (2) 148.

Preface

Progressions


Sol:

(1) Here a = 7, d = 3, Tn = 283.

nT a (n 1)d= + −

283 = 7 + (n –1) 3

283 = 7 + 3n – 3 = 4 + 3n

279n 93

3= =

Therefore 283 is the 93rd term.

(2) a = 7, d = 3, Tn = 148

nT a (n 1)d= + −

148 = 7 + (n –1)3

148 = 7 + 3n − 3

144 th3n 144 or n 48. Therefore 148 is the 48 term.3

= = =

Example: 4.

Find the sum of the arithmetic series 2 + 5 + 8 + … to 10 terms

Sol:

[ ]n

nS 2a (n 1)d

2= + + −

Here a = 2, n = 10, d = 3

( ) [ ]n

10S 2 2 10 1 3 5 4 27 5 31 155

2 = × + − = + = × =

Example: 5.

The first term of an A.P. is – 6, the last term is 244 and the sum of the series is 12019. Find the number of terms and the

common difference.

Sol:

Let n be the number of terms.

nn

S (a l)2

= +

Here a = – 6, 1 = 244, Sn = 12019

n12019 ( 6 244) 119n.

2= − + =

Hence, n = 101

Let d be the common difference

Then 244 = the 101st term = – 6 + 100d

5100d 250 or d 2.5

2= = =

Example: 6.

Find the sum of the first ten terms of the series

25 125 625log5 log log log ...

2 4 8+ + + +


Sol:

(log 5 + log 52 + ... log 5 10 ) – (log 2 + log 22 + … log 29)

=(log 5 + 2 log 5 + … 10 log 5 ) – ( log 2 + 2 log 2 + … + 9 log 2 )

= log 5 (1 + 2 + … 10 ) – log 2 ( 1 + 2 +… + 9)

10 11 9 10log5 log 2

2 2

55log5 45log 2.

× ×= −

= −

Example: 7.

What is the 21st term of the A.P. whose 4th term is 17 and 9th term is 42.

Sol:

Let ‘a’ be the first term and ‘d’ be the common difference.

4th term = 17 = a + (n – 1 )d = a + 3d

Therefore a + 3d = 17 .....(1)

9th term = 42 = a + (n – 1)d = a + (9 – 1) d = a + 8d

Therefore a + 8d = 42 .….(2)

On subtracting eqn. (1) from eqn. (2)

we get –5d = – 25

∴ d = 5

substituting d = 5 in (1),

a + 3 × 5 = 17 or a = 17 – 15 , or a = 2

Therefore, the 21st term = a + (21 – 1 ) 5 = 2 + 100 = 102.

Example: 8.

A man saved Rs. 16,500 in ten years. In each year after the first, he saved Rs. 100 more than he did in the preceding year. How

much did he save in the first year?

Sol:

Here Sn = 16,500, n = 10 and d =100

( )[ ]nn

S 2a n 1 d2

= + −

[ ]1016500 2a (10 1)100 5(2a 900)

2

10a 4500 10a 12000or a 1200.

= + − = +

= + ⇒ = =

Example: 9.

How many terms of the series 1 1 1

p q r− = must be taken that sum may be 66?

Sol:

Here [ ]n18 (n 1)3 66

2− + − =

2n 7n 44 0− − =

( ) ( )n 11 n 4 0, n 11 or 4− + = = − series is so 11 terms.

9, 6, 3, 0, 3, 6, 9, 12,15, 18, 21− − −


Example: 10.

How many terms of the series 1, 5, 9 … must be taken so that their sum is equal to 190.

Sol:

Here a = 1, d = 4, Sn = 190

( )[ ]nn

Applying the formula S 2a n 1 d2

= + −

[ ]n

190 2 1 (n 1)42

= × + −

or 380 = n[2 + 4n – 4] = n (4n – 2)

or 4n2 – 2n – 380 = 0

2n2 – n – 190 = 0

2n2 – 20n + 19n –190 = 0

2n(n – 10) + 19(n – 10) = 0

Therefore n = 10. Hence 10 terms must be taken.

Example: 11.

A man borrows Rs. 1000 and agrees to repay it with a total interest of Rs. 140 in 12 installments, each installment being less

than the one preceding it by Rs. 10. What should be his first installment?

Sol:

Here n = 12, d = –10 Sn = 1140.

Applying the formula ( )[ ]nn

S 2a n 1 d2

= + −

We have, [ ]12

1140 2a (12 1) ( 10)2

= + − −

1140 = 6(2a – 110) = 12a – 660 or 12a = 1800 or a = 150

Therefore his first installment is Rs. 150.

Example: 12.

Find the sum of the first P terms of the series whose thn term is 3n 1−

Slo:

By putting n 1= , and n P= respectively, we obtain

first term = 2, last term 3p 1= −

sum ( ) ( )p p2 3p 1 3p 1

2 2= + − = +

Example: 13.

If 11 A.M.S are inserted between 28 and 10, then the middle term is?

Let 1 2 3 11A , A , A , ... A be 11 A.M.S between 28 and 10

∴ 1 2 3 1128, A , A , A ... A , 10 are in A.P. also the number of terms = 13

13 110 a a 12d 28 12d= = + = +

10 28 3d

12 2

−= = −

total number of terms = 131 (odd)


middle term

th1h

7

13 17 a

2

+ = = =

1

3a 6d 28 6 19

2

= + = + − =

Geometric Progression (G.P.)

It is a sequence in which the ratio between any two consecutive terms is a constant. The common ratio (CR) is obtained by

dividing any term by the preceding one.

For example (1) 1, 3, 9, 27 … (2) 1 1

3, 1, , ...3 9

− + − (3) a, ar, ar2 ... are all in .G.P.

In (1) CR = 3 (2) 1

CR is3

− (3) CR = r

If a, b, c, d … are in GP then b c d

a b c= =

We shall denote the first term of the GP by ‘a’ and the common ratio by ‘r’

If a = 2, and r = 2, then the GP is 2, 4, 8, 16 …

The nth term of a G. P.

Let a be the st1 term and r the CR of a GP

nd 2 1

rd 2 3 1

th 3 4 1 th n 1

Then by definition,

2 term ar ar ,

3 term ar ar ,

4 term ar ar ... n term ar

−

−

− −

= =

= =

= = =

If l is the last term of the GP with n terms then n 1l ar −=

Corollary: when any two terms are given, the GP is completely determined.

To find the sum of ‘n’ terms of a Geometric series

2 n-1nLet S = a +ar +ar + ... + ar

Multiply by r we get, r. 2 n-1 nnS =ar +ar +...+ar ar+

By subtraction, n nn n nS r. S a ar orS (1 r) a(1 r )− = − − = −

n n

n

1 r a(r 1)S a where r 1

1 r r 1

− −∴ = = ≠ − −

If r > 1, we write

n n

n n

a(r 1) a(1 r )S and if r 1,wewriteS

r 1 1 r

− −= < =

− −


To find the sum to infinite terms of a G.P.

Consider the infinite G.P.

2 n-1a + ar +ar +...+ar ...+ ∞

n

n

a(1 r )S If r 1(i.e. 1 r 1)

1 r

−= < − < <

−becomes very small whenever ‘n’ assumes a very large value. Hence

nS becomes

almost equal to

a(1 0) a

1 r 1 r

−=

− −

This is the sum of infinite terms of the series and is denoted by nS

Thus a

S whenever r 11 r

= <∞ −

Example: 14.

Find the common ratio (CR) of the following GP.

(1) 4, 2, 1, … (2) 1, 0.1, 0.01, 0.001, … (3) 1 1

1, ,2 4

− − (4) 27, 9, 3, 1 …

Sol: Sol: Sol: Sol:

2 1CR

4 2= =

0.1CR 0.1

1= =

1/ 2 1CR

1 2

−= =−

9 1CR

27 3= =

Example: 15.

(1) Find the 10th, 15th and 20th term of the GP whose first term is 4 and CR is 2.

Sol:

Here a = 4, CR =2

n 1 10 1 910

n 1 15 1 1415

n 1 20 1 1920

T ar 4 2 4 2

T ar 4 2 4 2

T ar 4 2 4 2

− −

− −

− −

= = × = ×

= = × = ×

= = × = ×

(2) The first & last terms of a GP are 1 & 81 respectively. If the CR is 3, how many terms are there in the GP?

Sol:

a = 1, Tn = 81, r = 3

Given that ar n-1 = 81. 1× 3 n-1 = 81, 3n-1 = 34

∴ n –1 = 4 or n = 5. There are 5 terms in the GP

Example: 16.

(1) Find the sum of the series 1 + 3 + 9 + 27 + … to 10 terms.

Sol:

Here a = 1, r = 3, n = 10, r > 1

We know that

n 10

n 10

a(r 1) 1(3 1) 59048S ,S 29524.

r 1 3 1 2

− −= = = =

− −(2) Find the sum to n terms of 1, 4, 16 …


Sol:

Here a = 1, r = 4, nth term = n, r > 1

n n n

n

a(r 1) 1(4 1) 4 1S

r 1 4 1 3

− − −= = =

− −

(3) Find the sum of n terms of 7 + 77 + 777 + …

Sol:

7 + 77 + 777 + … to n terms = 7(1 + 11 + 111 + ... to n terms) = 7

9(9 + 99 + 999 + … to n terms)

( ) ( ) ( )

( ) ( )

( ) ( )

2 3

1 2 3

n

n

710 1 10 1 (10 1 ... to n terms

9

710 10 10 ... to n terms 1 1 1 ... to n terms

9

10 10 17 7 10n 10 1 n

9 10 1 9 9

= − + − + − +

= + + + − + + +

− = − = − − −

(4) Find the sum of the series 6 + 12 + 24 + … + 1536.

Sol:

Here a = 6, r = 2, Tn = 1536

To find n we know that l = arn –1 ∴1536 = 6.2n –1

2n-1 = 256 or 2n-1 = 28 or n –1 =8 or n = 9.

n 9

n

a(r 1) 6(2 1)S 6(512 1) 6(511) 3066.

r 1 2 1

− −= = = − = =

− −

Example: 17.

Find the sum of the series 2 n 1

1 1 11 ... ...

3 3 3 −+ + + + ∞

Sol:

n n

a 1 1 3S ,a 1, r S

11 r 3 21

3

= = = ∴ = =− −

Example: 18.

Find the sum to 6 terms of the series 1, 4, 16, ...

Sol:

n 6

n

a(r 1) 4 1 4095S 1365.

r 1 4 1 3

− −= = = =

− −

Example: 19.

The 6th terms of a G. P. is 32 and its eight terms if 128, then find the value of common ratio

Sol:

6a 32= 5ar 32= 8a 128= 7ar 128=


Divided equation (ii) to (i), and get

2r 4, r 2,= = ± neglect ( 2)−

Example: 20.

consider an infinite Geometric series with first term a and a common ratio r. if its sum is 4 and the second term is 3

4 then find

the value of a and r?

Sol:

a4

1 r=

−

a 4 4r= − ...... (i)

3ar

4= ...... (ii)

solving (i) and (ii) 1

a 3, r4

= =

Harmonic Progression (H.P.)

A set of number 1 2 3a , a , a ... , are said to be in H.P. if their reciprocal, i.e., 1 2 3

1 1 1, , ...

a a a are in A.P.

For example: 1/ 22 are in H.P. because 1, 2, 3, are in A.P.

The general H.P. is 1 1 1 1

, , , ... ...a a d a 2d a (n 1)d+ + + −

Mean

1. If x 1 2 na, a , a , ... a , y are in A.P. then the numbers 1 2 na, a , a , ... a , are called the arithmetic means between x and y.

2. If x 1 2 na, a , a , ... a , y are in G.P, then 1 2 na, a , a , ... a , are called the geometric mean between 1 2 na, a , a , ... a , x and y.

3. If x 1 2 na, a , a , ... a , y are in H.P., then 1 2 na, a , a , ... a , are the harmonic mean between x and y.

Arithmetic Mean (AM), Geometric Mean (GM) and Harmonic Mean (HM) between two numbers

1. If a, b, c are in A.P., b is the AM between a and c

a cb c c b or 2b a c i.e. b

2

+∴ − = − = + =

Thus the AM between a and c is a c

2

+

2. If a, b, c are in G.P, then b is the GM between a and c.

2b cor b ac i.e. b ac

a b∴ = = =

Hence the GM between a and c is ac


3. If a, b, c are in HP then b is the HM between a and c.

Hence 1 1 1 1 2 1 1

orb a c b b a b− = − = +

2 a c 2aci.e. b

b ac a c

+= ⇒ =

+

Thus the HM, Between a and c is2ac

a c=

+Note: If AG, H are the AM, GM & HM between two positive number a and b (a > b) then (i) G2 = AH (ii) a > A > G > H >b

Example: 21.

Find the 3rd term of the H.P. given that its 6th term is 1

61and the 10th term is

1

105.

Sol:

Let a and d be the first term and the common difference respectively of the corresponding A.P.

Then T6 = a + 5d = 61 ….. (1)

T10

= a + 9d = 105 ….. (2)

Solving (1) and (2) a = 6, d = 11

∴T3 = a + 2d = 6 + 2 (11) = 6 + 22 = 28

Hence the 3rd term of the H. P. is 1

28

Example: 22.

The 3rd and 7th terms of an H.P. are 6

19and

2

9respectively. Find the H.P.

Sol:

Let ‘a’ and ‘d’ be the first term and common difference of the corresponding A.P

3

19T a 2d

6∴ = + = ….. (1)

2

9T a 6d

2= + = ….. (2)

5 1Solving (1) and (2) a = and d =

2 3

5 17 19 2 6 6 the A.P. is , , ... and hence the required H.P. is , , ...

2 6 6 5 17 19∴

Note: If a, b, c, are any three distinct positive real numbers then

3a b cA , G = abc

3

+ += then 3a + b + c

A > G > i.e. (a + b + c) > 27 abc3

⇒


Example: 23.

Insert 5 AMs between 5 and 15.

Sol:

1 2 3 4 5

1 2 3 4 5

7

let a , a , a , a and a , are in A.P.

Therefore 5, a , a , a , a , a , 15 are in A.P.

a = 5 and T = a + 6d = 15

1 2

3 4

5

5 20 20 5 25Hence a = 5 + ; a = ;

3 3 3 3 3

25 5 5 35 a = 10 a = 10

3 3 3 3

35 5 40 a =

3 3 3

= + =

+ = + =

+ =

20 25 35 40 The required means are , , 10, and

3 3 3 3∴

Example: 24.

The H.M. between two numbers is 3

117

and their A.M. is 14. Find them.

Sol.

Let the numbers be a and b.

2ab 7 40 ab 40 11 40(a+b)=7aba+b 3 3 a+b 7

∴ = = ⇒ = ⇒ ..... (1)

a+balso = 14 a + b = 28

2⇒ ..... (2)

Solving (1) and (2) a = 8, b = 20.

Hence the required numbers are 8 and 20.

Example: 25.

If pth term of a H.P. is q and the qth term is p, what is the (pq)th term?

Sol:

Let a and d be the first term and common difference of the corresponding A.P. d, then

( )p

1T a p 1 d

q= + − = ..... (1)

q

1T a (q 1)d

p= + − = ..... (2)

Subtracting (2) from (1)

( ) ( )1 1 p q 1p 1 q 1 d or p q d d

q p pq pq

−− − + = − − = ⇒ =

Substituting in (1)


p 1 1 1 p 1 p p 1 1a , a = a

pq q q pq pq pq

− − − ++ = ⇒ − = ⇒ =

Hence the (pq)th term is

( )

pq

th

T a (pq 1)d

1 pq 1 pq = 1

pq pq pq

the pq term of the H.P. = 1.

= + −

−+ = =

∴


1. The product (32) (32)1/6 (32)1/36 ... ∞ is equal to (Access Code - 01302001)

(1) 16 (2) 64 (3) 32 (4) 0 (5) None of these

2. Find the 9th term of the H.P. 6, 4, 3, ... (Access Code - 01302002)

(1) 7

5(2)

6

5(3)

2

3(4)

1

3(5)

5

6

3. Sum the series to infinity 2 3 4 5 6

3 5 3 5 3 5...

4 4 4 4 4 4− + − + − + (Access Code - 01302003)

(1) 8

7(2)

7

17(3)

7

15(4)

8

17(5)

8

15

4. The third and sixth terms of an H.P. are 5

16 and

1

5 respectively. Find the 10th term. (Access Code - 01302004)

(1) 3

7(2)

2

7(3)

5

7(4)

4

7(5)

6

7

5. Find the nth term of the H.P. Whose first two terms are 6 and 3, respectively. (Access Code - 01302005)

(1) 6

n(2)

7

n(3)

5

n(4)

8

n(5)

4

n

6.2 3 4

2 3 2 3...

5 5 5 5+ + + + ∞ (Access Code - 01302006)

(1) 17

24(2)

15

24(3)

13

24(4)

11

24(5)

11

10

7. If a be the first term of a G. P., l the nth term and P the product of first n terms then P = (Access Code - 01302007)

(1) (al)n/2 (2) (a - l)n/2 (3) (a + l)n/2 (4) 0 (5) None of these

8. The sum of 3rd and 15th elements of an arithmetic progression is equal to the sum of 6th, 11th and 13th elements of the same

progression. Then which element of the series should necessarily be equal to zero? (Access Code - 01302008)

(1) 1st (2) 9th (3) 12th (4) 5th (5) None of these

9. Three numbers are in G. P. Their sum is 28 and their product is 512. The numbers are (Access Code - 01302009)

(1) 6, 9 and 13 (2) 2, 4 and 32 (3) 2, 8 and 18 (4) 2, 6 and 18 (5) 4, 8 and 16

10. Let Sn denote the sum of the first ‘n’ terms of an A. P. S

2n = 3S

n. Then the ratio S

3n/S

n is equal to (Access Code - 01302010)

(1) 4 (2) 6 (3) 8 (4) 10 (5) 12

Practice Exercise - 1


11. If log32, log

3 (2x - 5), log

3(2x - 7/2) are in arithmetic progression, then the value of x is equal to (Access Code - 01302011)

(1) 5 (2) 4 (3) 2 (4) 3 (5) 1

12. There are 8436 steel balls, each with a radius of 1 centimeter, stacked in a pile, with 1 ball on top, 3 balls in the second layer,

6 in the third layer, 10 in the fourth, and so on. The number of horizontal layers in the pile is (Access Code - 01302012)

(1) 34 (2) 38 (3) 36 (4) 32 (5) 30

13. Let T be the set of integers {3, 11, 19, 27, ... 451, 459, 467} and S be a subset of T such that the sum of no two elements of

S is 470. The maximum possible number of elements in S is (Access Code - 01302013)

(1) 32 (2) 28 (3) 29 (4) 30 (5) 35

14. Find the sum of n terms of the series 11 + 103 + 1005 + ... (Access Code - 01302014)

(1) n10(10 1)

9

−(2)

n10(10 1)n

9

−+ (3)

n210(10 1)

n9

−+ (4)

n210(10 1)

n9

+− (5) None of these

15. If the positive real number x, y, z satisfy y - x = z - y and xyz = 4, then what is the minimum possible value of y? (Access Code

- 01302015)

(1) 21/3 (2) 22/3 (3) 21/4 (4) 23/4 (5) 21/2

16. An intelligence agency forms a code of two distinct digits selected from 0, 1, 2, ..., 9 such that the first digit of the code is

non zero. The code, handwritten on a slip, can however potentially create confusion when read upside down - for example,

the code 91 may appear as 16. How many codes are there for which no such confusion can arise? (Access Code - 01302016)

(1) 80 (2) 78 (3) 71 (4) 69 (5) 85

17. Find the sum of the integers between 1 and 200 that are multiples of 7. (Access Code - 01302017)

(1) 2742 (2) 2842 (3) 2646 (4) 2546 (5) None of these

18. If in any decreasing arithmetic progression, sum of all its terms, except for the first term is equal to - 36, the sum of all its terms,

except for the last term, is zero, and the difference of the tenth and the sixth term is equal to - 16, then what will be first term

of this series? (Access Code - 01302018)

(1) 16 (2) 20 (3) -16 (4) -20 (5) None of these

19. Mohan ate half a pizza on Monday. He ate half of what was left on Tuesday and so on. He followed this pattern for one

week. How much of the pizza would he have eaten during the week? (Access Code - 01302019)

(1) 99.22% (2) 95% (3) 98.22% (4) 100% (5) 80%

20. If logxa, ax/2 and log

bx are in GP, then x is (Access Code - 01302020)

(1) loga(log

ba)

(2) loga(log

ba) + log

a(log

ab)

(3) − loga(log

ab)

(4) loga(log

ab) − log

a(log

ba)

(5) None of these


21. The sum of the first term and the fifth term of an AP is 26 and the product of the second term by the fourth term is 160. Find

the sum of the first seven terms of this AP. (Access Code - 01302021)

(1) 110 (2) 114 (3) 112 (4) 116 (5) 120

22. 30 trees are planted in a straight line at intervals of 5 meters. To water them, the gardener needs to bring water for each tree,

separately from a well, which is 10 meters from the first tree in line with the trees. How far will he have to walk in order to water

all the trees beginning with the first tree? Assume that he starts from the well. (Access Code - 01302022)

(1) 4785 metres (2) 4795 metres (3) 4800 metres (4) 4250 metres (5) None of these

23. If a clock strikes once at one o’clock, twice at two o’clock and twelve times at 12 o’clock and again once at one o’clock and

so on, how many times will, the bell be struck in the course of 2 days? (Access Code - 01302023)

(1) 156 (2) 312 (3) 78 (4) 288 (5) 300

24. How many three digit numbers are divisible by 6? (Access Code - 01302024)

(1) 102 (2) 130 (3) 151 (4) 966 (5) 150

25. There are 12 towns grouped into four zones with three towns per zone. It is intended to connect the towns with telephone

lines such that every two towns are connected with three direct lines if they belong to the same zone, and with only one

direct line otherwise. How many direct telephone lines are required? (Access Code - 01302025)

(1) 72 (2) 60 (3) 96 (4) 144 (5) 90

Use HB pencil only. Abide by the time-limit

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10


1. In a certain examination paper, there are n questions. For j = 1, 2, ... n, there are 2n - j students who answered j or more

questions wrongly. If the total number of wrong is 4095, then the value of n is(Access Code - 01302026)

(1) 8 (2) 11 (3) 10 (4) 9 (5) 12

2. The measures of the angles of a convex pentagon are in arithmetic series. The smallest angle of the pentagon(Access Code

- 01302027)

(1) Must be > 36° (2) Cannot be < 36° (3) Must be = 36°(4) May be of any measure (5) None of these

3. The number of ordered triplets of positive integers which are solutions of the equation x + y + z = 100 is(Access Code -

01302028)

(1) 3582 (2) 5081 (3) 6871 (4) 7081 (5) 4851

4. Three distinct numbers x, y, z form a geometric progression in that order, and x + y, y + z, z + x form an arithmetic progression

in that order. Find the common ratio of geometric progression.(Access Code - 01302029)

(1) – 2 (2) 2 (3) 0.5 (4) – 0.5 (5) None of these

5. A man receives a pension starting with Rs.100 for the first year. Each year he receives 90% of what he received the previous

year. Find the maximum total amount can he receive even if he lives forever.(Access Code - 01302030)

(1) Rs.1100 (2) Rs.1000 (3) Rs.1200 (4) Rs.900 (5) Rs.1000

6. If x be the first term, y be the nth term and p be the product of n terms of a GP, then the value of p2 will be(Access Code -

01302031)

(1) n 1(xy) − (2) (xy)n (3) 1 n(xy) − (4) n / 2(xy) (5) None of these

7. The first and the third terms of an arithmetic progression are equal, respectively, to the first and the third term of a geometric

progression, and the second term of the arithmetic progression exceeds the second term of the geometric progression by

0.25. Calculate the sum of the first five terms of the arithmetic progression if its first term is equal to 2.(Access Code -

01302032)

(1) 2.25 or 25 (2) 2.5 or 27.5 (3) 1.5 (4) 3.25 (5) 4.5

8. Product of the fourth term and the fifth term of an arithmetic progressoin is 456. Division of the ninth term by the fourth term

of the progression gives quotient as 11 and the remainder as 10. Find the first term of the progression.(Access Code -

01302033)

(1) − 52 (2) −42 (3) − 56 (4) − 66 (5) None of these

9. Four numbers are inserted between the numbers 4 and 39 such that an AP results. Find the biggest of these four numbers.

(Access Code - 01302034)

(1) 31.5 (2) 31 (3) 33 (4) 30 (5) 32



10. A number of saplings are laying at a place by the side of a straight road. These are to be planted in a straight line at a distance

interval of 10 meters between two consecutive saplings. Mithiles, the country’s greatest forester, can carry only one sapling

at a time and has to move back to the original point to get the next sapling. In this mannner he covers a total distance of 1.32

kms. How many saplings does he plant in the process.(Access Code - 01302035)

(1) 15 (2) 14 (3) 13 (4) 12 (5) 11

11. What will be the sum to n terms of the series 8 + 88 + 888 +...?(Access Code - 01302036)

(1) n8(10 9n)

81

−(2)

n 18(10 10 9n)

81

+ − −(3) n 18(10 10)− − (4) Cannot be said (5) None of these

12. Obtain the sum of all natural numbers in the first 1000 positive integers which are neither divisible by 5 nor by 2.(Access

Code - 01302037)

(1) 200000 (2) 300000 (3) 100000 (4) 400000 (5) None of these

13. If p, q, r, s are in harmonic progression and p > s, then(Access Code - 01302038)

(1) 1 1

pq qr< (2) q + r = p + s (3)

1 1 1 1

q r p s+ = + (4)

1 1 1

p q r− = (5) None of these

14. f(a, b) is a series of which the first three terms are (a + b)2, (a2 + b2) and (a - b)2. We add the first n terms of the series f(a, b)

and call it S(a, b). If a = 7, b = 3, then find S(7, 3) for n = 20.(Access Code - 01302039)

(a) − 5980 (2) 6000 (3) 6960 (4) 6760 (5) None of these

15. f(x) = 2x; where x is an integer. If we arrange the value of f(x), for x = 25, 24, 23 ... (continuously decreasing value of x), we get

an Arithmetic Progression (A.P.) whose first term is is 50. Find the maximum value of the sum of all the terms of the

A.P.(Access Code - 01302040)

(1) 600 (2) 625 (3) 650 (4) 700 (5) 750

SCORE SHEET

Use HB pencil only. Abide by the time-limit

1

2

3

1 2 3 4 5

1 2 3 4 5

1 2 3 4 5

4

5

6

1 2 3 4 5

1 2 3 4 5

1 2 3 4 5

7

8

9

1 2 3 4 5

1 2 3 4 5

1 2 3 4 5

10

11

12

1 2 3 4 5

1 2 3 4 5

1 2 3 4 5

13

14

15

1 2 3 4 5

1 2 3 4 5

1 2 3 4 5


1. (5)

2. (4)

3. (5)

7. (2)

8. (4)

9. (5)

13. (3)

14. (5)

15. (3)

10. (3)

11. (2)

12. (1)

4. (1)

5. (5)

6. (2)

1. (2)

2. (5)

3. (5)

4. (3)

5. (1)

Answer Key



6. (3)

7. (1)

8. (3)

9. (5)

10. (2)

11. (4)

12. (3)

13. (4)

14. (2)

15. (2)

16. (3)

17. (2)

18. (1)

19. (1)

20. (1)

21. (3)

22. (2)

23. (2)

24. (5)

25. (5)

BMM10233 Chapter 2 Progressions

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Transcript of BMM10233 Chapter 2 Progressions