Chapter 6. Progressions

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1 Chapter 6 Progressios Evidece is foud that Babyloias some 400 years ago, kew of arithmetic ad geometric progressios. Amog the Idia mathematicias, Aryabhata (470 AD) was the first to give formula for the sum of squares ad cubes of atural umbers i his famous work Aryabhata Idia mathematicia Brahmagupta (598 AD), Mahavira (850 AD) ad Bhaskara ( AD) also cosidered the sums of squares ad cubes. Arithmetic progressio(a.p) A arithmetic progressio (AP) is a list of umbers i which each term is obtaied by term addig a fixed umber d to precedig term, except the first term. The fixed umber d is called the commo differece Ex: 1,, 7, 10, 13. are i AP Here d = 3 Let a 1, a, a 3, a k, a k+ 1.. a be a AP. Let its commo differece be d, the d = a a 1 = a 3 a = = a k+1 a k = If the first term is a ad the commo differece is d the a, a + d, a + d, a + 3d,.. is a A.P. Geeral term of a A.P. Let a be the first term ad d be the commo differece of a A.P., The, its th term or geeral term is give by a = a + ( 1) d Ex: The 10 th term of the A.P. give by 5, 1 3, 7, is a 10 = 5 + (10 1) ( 4 ) = 31

2 If the umber of terms of a A.P. is fiite, the it is a fiite A.P. Ex: 13, 11, 9, 7, 5 If the umber of terms of a A.P. is ifiite, the it is a ifiite A.P. Ex: 4, 7, 10, 13, 16, 19, Three umbers i AP should be take as a d, a, a + d. Four umbers i AP should be take as a 3d, a d, a + d, a + 3d. Five umbers i AP should be take as a d, a d, a, a + d, a + d Six umbers i AP should be take as a 5d, a 3d, a d, a + d, a + 3d, a + 5d. a c If a, b, c are i AP, theb is called the arithmetic mea if a ad c. s a d. The sum of the first terms of a AP is give by 1 If the first ad last terms of a AP are a ad l, the commo differece is ot give. the s a l a = s s 1 The sum of first positive itegers s ( 1). Ex: sum of first 10 positive itegers Geometric progressio (G.P.) A Geometric Progressio is a list of umbers i which each term is obtaied by multiplyig precedig term with a fixed umber r except first term. This fixed umber is called commo ratio r. Ex: 3, 9, 7, 81,. are i G.P.

3 Here commo ratio r = 3 A list of umbers a 1, a, a 3,.a are i G. P. The the commo ratio r a a a1 a a 1 a The first term of a G.P. by a ad commo ratio r the the G.P is a, ar, ar,.. If the first term ad commo ratio of a G.P. are a, r respectively the th term a = ar 1. 1 Mark Questios 1. Do the irratioal umbers, 8, 18, 3... form a A.P? If so fid commo differece? Sol: Give irratioal umbers are, 8, 18, 3... d = a a Here commo differece is same. i.e The umbers are i A.P.

4 . Write first four terms of the A.P, whe the first term a ad commo differece d are give as follow. a = 1.5, d = 0.5 Sol: a 1 = a = 1.5, d = 0.5 a = a + d = = a 3 = a + d = (-0.5) = 1.75 a 4 = a + 3d = (-0.5) =.00 AP = -1.5, - 1.5, , Is the followig forms AP? If it, form a AP, fid the commo differece d ad write three more terms. Sol:, 8, 18, 3... Here a = d a a1 8 4 d a3 a d a4 a d is equal for all. So it forms a AP Next three terms a5 a4 d a6 a5 d a7 a6 d , 7, 98.

5 4. If a AP a = 6 + fid the commo differece Sol: Let a = 6 + a 1 = 6 (1) + = = 6 + = 8 a = 6 () + = 1 + = 14 a 3 = 6 (3) + = 18 + = 0 d = a a 1 = 14 8 = 6. Commo differece = I G.P., -6, 18, -54 fid a Sol: a = a 6 r 3 a 1 a = a.r -1 =.(-3) The 17 th term of a A.P exceeds its 10 th term by 7. Fid the commo differece. Sol: Give a A.P i which a 17 = a a 17 a 10 = 7 (a +16d) (a + 9d) = 7 7d = 7

6 7 d A ma helps three persos. He ask each or them to give their help aother three persos. If the chai cotiued like this way. What are the umbers obtaied this series Sol: First perso = 1 No.of perso take help from 1 st perso = 3 No.of perso take help from the persos take help from first perso = 3 = 9 Similarly, o.of persos take help 7, 81, 43, progressio 1, 3, 9, 7, 81, 43.. I the above progressio a 1 = 1, a = 3, a 3 = Commo ratio (r) a a 3 a a So, above progressio is i G.P. 8. Fid the sum of 8 terms of a G.P., whose th term is 3. Sol: I a G.P. th term (a ) = 3 a 1 = 3 1 = 3 a = 3 = 9 a 3 = 3 3 = 7. Geometric progressio = 3, 9, 7 First term (a) = 3 a 9 Commo ratio (r) 3 1 a 3 1

7 No.of terms () = 8 Sum of terms (s ) ar ( 1) r 1 Sum of 8 terms (s 8 ) I A.P th ter a = a + ( - 1) d explai each term i it. Sol: a = a + ( - 1) d a = First term = No.of terms d = Commo differece a = th term , 18, 54.. is it i G.P. What is the commo ratio? Sol: Give that 6, 18, 54. a 18 r 3 a 6 1 a3 54 r 3 a 18 a a a3 3 a 1 So, 6, 18, 54, i.. is i G.P. Commo ratio = 3.

8 11. Fid sum of series 7, 13, 19 upto 35 terms Sol: Give that 7, 13, 19.. a 1 = 7 d = a a, = 13 7 = 6 No.of terms () = 35 S a d Sum of terms 1 S = = What is 10 th term i the series 3, 8, 13.. Sol: Give series 3, 8, 13,. it is i A.P. First term (a) = 3 d = 8 3 = 5 = 10 a = a + ( - 1)d

9 = 3 + (10-1) (5) = = th term i give series a 10 = ca x +, x + 4 ad x + 9 be i A.P. Justify your aswer Sol: Give terms are: x +, x + 4, x + 9 a a 1 = (x + 4) (x + ) = a 3 a = (x + 9) (x + 4) = x + 9 x 4 = 5 a a 1 a 3 a. Give terms are ot i A.P. 14. I a G.P., first term is 9, 7 th term is 1 81 fid the commo ratio Sol: G.P. first term a 1 = 9 = 3 7 th term a a ar 81

10 1 th 6 7 term a7 ar st 1 term a 9 1 a r r Commo ratio r = Write the geeral terms of a AP ad GP. Sol: The geeral terms of AP are a, a + d, a + d, a + 3d. The geeral terms of GP are a, ar, ar, ar

11 Mark Questios 1. Determie the A.P. whose 3 rd term is 5 ad the 7 th term is 9. Sol: we have a 3 = a + (3 1) d = a + d = (1) a 7 = a + (7 1) d = a + 6d = () solvig the pair of liear equatios (1) ad (), we get a + d = (1) a + 6d = () d = -4 4 d 4 d = 1 substitute d = 1 i equ (1) a + d = 5 a + (1) = 5 a = 5 = 3 a = 3 ad d = 1 Hece, the required AP is 3, 4, 5, 6, 7,... How may two-digit umbers are divisible by 3? Sol: The list of two digit umbers divisible by 3 are 1, 15, These terms are i A.P. ( t t 1 = t 3 t = 3) Here, a = 1, d = 3, a = 99 a = a + ( -1)d 99 = 1 + ( - 1) 3

12 99 1 = ( -1) 3 87 = ( - 1) 3 1 = = 9 = = 30 so, there are 30 two- digit umbers divisible by Fid the respective term of a 1 = 5, a 4 = 1 9 fid a, a 3 i APs Sol: Give a 1 = a = (1) a 4 = a + 3d = () Solvig the equ (1) ad equ (), we get equ (1) equ () (a + 3d) a = a + 3d a = 1 4 3d = 9 d = d = a a d 5

13 a3 a d 8 13 a, a a = 38 ; a 6 = - fid a 1, a 3, a 4, a 5 Sol: Give a = a + d = (1) a 6 = a + 5d = () Equatio () equatio (1) (a + 5d) (a + d) = - 38 a + 5d a d = d = - 60 d 60 4 d = - 15 a = a + d = 38 a + (- 15) = 38 a 15= 38 a = a = 53. a 1 = a = 53 a 3 = a + d = 53 + (-15) = = 3 a 4 = a + 3d = (-15) = = 8 a 5 = a + 4d

14 = (-15) = = -7 a 1 = 53, a 3 = 3, a 4 = 8, a 5 = Which term of the A.P: 3, 8, 13, 18 is 78? Sol: a = 78 a = 3 d = a a 1 = 8 3 = 5 a = a + ( 1)d 78 = 3 ( - 1) 5 78 = = 5 5 = = = th term of the A.P is Fid the 31 st term of a AP whose 11 th term is 38 ad 16 th term is 73. Sol: Give a 11 = 38, a 16 = 73 ad a 31 =? a = a + ( -1) d a 11 = a + 10 d = (1)

15 a 16 = a + 15d = () () (1) a + 15d = 73 a + 10d = Substitute d = 7 i equ (1) a + 10 d = 38 a + 10 (7) = 38 a + 70 = 38 a = a = st term a 31 = a + 30d = (7) = = is the 31 st term. 7. Fid the sum of 7 + 5d = d Sol: Give terms are i A.P Here a = 7, d = a a 1 = , a 84

16 a = a + ( - 1) d = ( - 1) = = 3 1 = = + 1 = 3 s a l 3 s S I a AP give a = 5, d = 3, a = 50, fid ad s Sol: a = a + ( 1) d 50 = 5 + ( 1) 3 ( a = 5, d = 3, a = 50) 50 =

17 50 = = 3 3 = 48 = S a 1 d S = 8 [10 + (15) (3)] = 8 [ ] = 8 55 S 16 = I a AP give a 3 = 15, S 10 = 15, fid d ad a 10 Sol: a 3 = 15, S 10 = 15 a 3 = a + d = (1) S a 1 d 10 S10 a 10 1 d 15 5[a + 9d] = a9d 5 a + 9d = ()

18 Equ (1) ad equ () a + d = (1) a + 9d = () 1 a + 4d = 30 a + 9d = d = +5 5 d 5 d = -1 Substitute d = -1 i equ (1) a + d = 15 a + (-1) = 15 a = 15 a = 15 + = 17 a 10 = a + 9d = (-1) 17 9 = 8 d = -1 ad a 10 = The first ad the last terms of a AP are 17 ad 350 respectively. If the commo differece is 9, how may terms are there ad what is their sum? Sol: Give A.P i which a = 17 Last term = l = 350 Commo, differece, d = 9 We kow that, a = a + ( 1) d 350 = 17 + ( 1) (9)

19 350 = = = = = 38 Now S a l 38 S = = 6973 = 38; S = Which term of the G.P:,, 4 is 18? Sol: Here a =, r Let 18 be the th term of the GP The a = ar -1 = 18 () -1 =

20 1 6 1 m a a m = 6 1 = 1 = = 13 Hece 18 is the 13 th term of the G.P. 1. Which term of the G.P. is, 8, 3,.. is 51? Sol: Give G.P. is, 8, 3,.. is 51 a 8 a =, r 4 a a = 51 1 a = ar -1 = (4) -1 = 51 ( ) -1 = 9-1 = 9 a m a a m

21 -1 = 9 m a a m = = 5 51 is the 5 th term of the give G.P ,3, is 79? Sol: Give G.P. is 3,3, is 79 a = 3 r a = 79 a = a.r -1 = 79 3.(3) -1 = a m a a m

22 6 m a a m = 6 = 1 79 is the 1 th term of the give G.P. 14. I a ursery, there are 17 rose plats i the first row, 14 i the secod row, 11 i the third row ad so o. If there are rose plats i the last row, fid how may rows of rose plats are there i the ursery. Sol: Number of plats i first row = 17 Number of plats i secod row = 14 Number of plats i third row = 11 The series formed as 17, 14, 11, 8, 5, ; the term are i A.P. Here a = 17, d = = - 3 a = a = a + ( 1) d = 17 + ( 1) ( 3) = = 0 3 = 3 = 0 3 = 18

23 18 3 = 6 There are 6 rows i the ursery. 15. Which term of the sequece -1, 3, 7, 11. is 95? Sol: Let the A.P. -1, 3, 7, a = -1; d = 3 (- 1) = = 4; a = 95 a + ( - 1) d = ( -1) (4) = = = = 96 4 = = th term = A sum of Rs. 80 is to be used to award four prizes. If each prize after the first is Rs. 0 less tha its precedig prize. Fid the value of each Sol: The value of prizes form a A.P I A.P. d = -0 S = 80 of the prizes. = 4

24 a 1 d 80 4 a [a 60] = 80 a 60 = 80 a 60 = 140 a = a = 00 a 00 a = 100 The value of each of the prizes = Rs 100, Rs 80, Rs 60, Rs If the 8 th term of a A.P. is 31 ad the 15 th term is 16 more tha the 11 th term, fid the A.P. Sol: I a A.P. a 8 = 31 a 8 = a + 7d = 31 a 15 = 16 + a 11 a + 14d = 16 + a + 10d 14d 10d = 16 + a a 4 d = d 4 4 a + 7d = 31 ad d = 4 a + 7(4) = 31

25 a + 8 = 31 a = 31 8 a = 3 A.P. is 3, 7, 11, 15, Defie Arithmetic progressio ad Geometric progressio. Sol: Arithmetic Progressio: A arithmetic progressio (AP) is a list of umbers i which each term is obtaied by addig a fixed umber d to the precedig term, except the first term. The fixed umber d is called the commo differece. Geometric Progressio: A geometric progressio (G.P) is a list of umbers i which each term is obtaied by multiplyig precedig term with a fixed umber r except first term. This fixed umber is called commo ratio (r).

26 4 Mark Questios 1. If the 3 rd & the 9 th terms of a A.P. are 4 ad -8 respectively. Which term of this AP is zero? Sol: a 3 = 4, a 9 = - 8 a 3 = a + d = (1) a 9 = a + 8d = () () (1) we get a + 8d = -8 a + d = d = -1 1 d 6 d = - Substitute d = - i the followig equatios a 4 = a 3 + d = 4 + (-) = 4 = a 5 = a 4 + d = + (-) = = 0 5 th term of the A.P becomes zero.. Fid the 0 th term from the ed of A.P: 3, 8, Sol: a = 3, d = a a 1 = 8 3 = 5, a = 53 a = a + ( - 1) d 53 = 3 + ( - 1) (5) 53 3 = ( - 1) 5

27 = 50 = 51 The 0 th term from the other ed would be r + 1 = = 3. a 3 = a + 31d = (5) = = 158 The 0 th term is The sum of the 4 th ad 8 th terms of a A.P. is 4 ad the sum of the 6 th ad 10 th term is 44. Fid the first three terms of the A.P. Sol: 4 th + 8 th = 4 (a + 3d) + (a + 7d) = 4 a + 3d + a + 7d = 4 a + 10d = 4 (a + 5d) = 4 4 a 5d a + 5d = 1. (1) 6 th + 10 th = 44

28 (a + 5d) + (a + 9d) = 44 a + 5d + a + 9d = 44 a + 14d = 44 (a + 7d) = 44 a + 7d 44 a + 7d = () () (1) = a + 7d = a + 5d = d d = 10 d = 5. Substitute d = 5 i eq (1) We get, a + 5(5) = 1 a + 5 = 1 a = 1 5 a = - 13 The first three terms of A.P are a 1 = a = - 13 a = a + d = = - 8 a 3 = a + d = (5) = = -3.

29 4. Subba rao started work i 1995 at a aual salary of Rs 5000 ad received a icremet of Rs 00 each year. I which year did his icome reach Rs 7000? Sol: Year Subba rao salary , 500, 5400, 5600, 5800 is i A.P. a = a + ( - 1)d = ( 1) (00) = 7000 = = 7000 = = 7000 = 00 = = 00 = = 11. The 11 th is I the year 005 his icome reaches to Rs Give a =, d = 8, S = 90. Fid ad a. Sol: a = a + ( 1) d = + ( 1) 8

30 = = 8 6 a =, d = 8, S = 90 S a 1 d [ ] = = = = = 0 45 = = 0 ( -5) + 9 ( - 5) = 0 ( - 5) ( + 9) = 0 5 = = 0 = 5 = But we caot take egative values so, = 5

31 a 5 = a + 4d = + 4 (8) = + 3 = 34. = 5 ad a 5 = If the sum of first 7 terms of a A.P is 49 ad that of 17 terms is 89, fid the sum of first terms. Sol: The sum of first 7 terms of a A.P = 7. S a 1 d, where S = 49 7 = 7, the 49 a 7 1 d 49 a 6d 7 14 = a + 6d 14 = (a + 3d) 14 a 3d 7 a + 3d = (1) Ad the sum of 17 terms is 89, S a 1 d,s = 89, = The 89 a a 16d = a + 16d d

32 34 34 a 8d a 8d 17 a + 8d = () (1) () by solvig a + 8d = 17 a + 3d = d 5 d = 5d = 10 Substitute d = i eq(1), we get a + 3d = 7 a + 3 = 7 a + 6 = 7 a = 7 6 a = 1 S a d The first terms sum 1 a = 1, d =, the o substitutig, we get S S.1 1 S = The sum of first terms (S ) =.

33 7. If the sum of the first terms of a AP is 4, what is the first term (remember the first term is S 1 )? What is the sum of first two terms? What is the secod term? Similarly, fid the 3 rd, the 10 th ad the th terms. Sol: The sum of the first terms of a A.P is 4 First term a 1 = S 1 = = 4 1 = 3 ( = 1) First sum of the two terms = 4 = 8 4 = 4 S 3 = = 1 9 = 3 a = S S 1 = 4 3 = 1 Third term (a 3 ) = S 3 S = 3 4 = -1 S 10 = = = -60 S 9 = = = -45 Teth term (a 10 ) = S 10 S 9 = - 60 (-45) = = 15 S = 4 S 1 = 4( - 1) ( - 1) = 4 4 ( + 1) = The th term a = S S -1 a = 4 ( ) = = 5

34 S 1 = 3, S = 4, a = 1, a 3 = -1, a 10 = -15, a = A sum of Rs 700 is to be used to give seve cash prizes to studets of a school for their overall academic performace. If each prize is Rs 0 less tha it s precedig prize, fid the value of each of the prizes. Sol: First term = Rs a Each price is Rs 0 less tha it s precedig prize, the the remaiig prize of gift (a - 0), (a - 40) (a - 10), the a, (a - 0), (a - 40) (a - 10) forms a A.P. Here S = 700, = 7, a = a, a = a 10, o substitutig these so, S a a values we get a a 700 a = a = a a 30 a = 160 Each value of the prize Rs 160, Rs 140, Rs 10, Rs 100, Rs 80, Rs 60, Rs 40.

35 9. The umber of bacteria i a certai culture triples every hour if there were 50 bacteria preset i the culture origially. The, what would be umber of bacteria i fifth, teth hour. Sol: The o of bacteria i a culture triples every hour. No of bacteria i first hour = 50 No of bacteria i secod hour = 3 50 = 150 No of bacteria i third hour = = , 150, would forms a G.P. First term (a) = 50 t Commo ratio(r) = t th term a = ar -1 No of bacteria i 5 th hour = = = 4050 No of bacteria i 10 th hour = = = rd, 5 th, 10 th hours of bacteria umber = 450, 4050, The 4 th term of a G.P is 3 Sol: The 4 th term of G.P, ad the seveth term is the seveth term is. Fid the Geometric series. 81 i.e. 3 ar (1) ar () 81

36 ,the we get 1 ar ar r 7 r 3 3 r 3 Now substitute 3 r i eq(1), we get a. a a, r 4 3 The A.P. a, ar, ar, ar 3,. 3 9, 9, 9, ,,1,, If the geometric progressios 16, 54, 18 ad,,,... have their th term equal. Fid its value of? Sol: 16, 54, 18 a 54 1 Here a = 16, r a

37 The th term =ar 1 = (1) 3,,, a 81 Here a, r a th 1 1 term a. r. 3...() 81 Give that th terms are equal From (1)& () m m a. a a = 8 [if the bases are equal, expoets are also equal] = = 5 The 5 th terms of the two G.P. s are equal.

38 1. Fid the 1 th term of a G.P. whose 8 th term is 19 ad the commo ratio is. Sol: Give G.P a 8 = 19 & r = a = a.r 1 a 8 = a () 8-1 = a. 19 a a1 a. r 11 = 3 10 = = I a A.P d, 3 rd terms are 14 & 18 ad fid sum of first 51 terms? Sol: d term: 3 rd term: a + d = 14.. (1) a + d = 18.. () a + d = 14 a + d = d = -4 d = 4 Substitute d = 4 i eq (1) a + 4 = 14 a = 14 4 a = 10 S a 1 d 51 S

39 = =5610 The sum of first 51 term = I a A.p, the sum of the ratio of the m ad terms i m :, the show that m th term ad th terms ratio is (m- 1) : ( 1). Sol: AP, first term = a Commo differece = d m Sm a m 1 d S a 1 d Give S S m m m a m 1 d m a 1 d [a + (m - 1) d] = [a + ( -1)d] m a ( - m) = d [( -1)m (m -1)] a( -m) = d ( -m) d = a Tm a ( m 1)a a am a T a ( 1)a a a a am a a m1 a a a 1 T T m m 1 1

40 15. The sum of,, 3 terms of a A.P are S 1, S, S 3 respectively prove that S 3 = 3 (S S 1 ) Sol: I a A.P. first term is a ad the commo differece is d. S1 a 1 d 1 S a 1 d 3 S3 a 3 1 d 3 S S1 a 1 d a 1 d S S1 a 3 1 d 3 3S S a 3 1 d S 1 3 S 3 = 3 (S S 1 )

41 Multiple Choice Questios 1. The th term of G.P is a = ar -1 where r represets [ ] a) First term b) Commo differece c) Commo ratio d) Radius. The th term of a G.P is ( 0. 5 ) -1 the r = [ ] a) 5 b) 1 7 c) 1 3 d) I the A.P 10, 7, 4. -6, the 11 th term from the last is. [ ] a) 40 b) 3 c) 3 d) Which term of the G.P 1, 1, 1... is 1? [ ] a) 1 b) 8 c) 7 d) Noe 5. 1,, 3, a =. [ ] a) b) 0 c) 1 d) 6. I a A.P a = 7, d = 5 the a 18 =. [ ] a) 71 b) 78 c) 87 d) =. [ ] a) 5050 b) 5049 c) 5115 d) ,,, s81 [ ] a) 3418 b) 891 c) 3963 d) I G.P, 1 st term is, Q commo ratio is 3 the 7 th term is [ ] a) 1458 b) 1458 c) 79 d) 79

42 10. 1,, 4, 8,. is a. Progressio [ ] a) A.P b) G.P c) Both d) Noe of these 11. Commo differece i 1 3,1,... [ ] a) 1 b) 1 c) d) ,3,3 3...is a [ ] a) A.P b) G.P c) Harmoic progressio d) Ifiite progressio a, d,the 8 th term of a A.P is [ ] 3 3 a) 7 3 b) 9 3 c) 9 9 d) Arithmetic progressio i which the commo differece is 3. If is added to every term of the progressio, the the commo differece of ew A.P. [ ] a) 5 b) 6 c) 3 d) 15. I a A.P. first term is 8 commo differece is, the which term becomes zero [ ] a) 6 th term b) 7 th term c) 4 th term d) 5 th term 16. 4, 8, 1, 16,. is series [ ] a) Arithmetic b) Geometric c) Middle d) Harmoic 17. Next 3 terms i series 3, 1, -1, -3 [ ] a) -5, -7, -9 b) 5, 7, 9 c) 4, 5, 6 d) -9, -11, -13

43 18. If x, x + & x + 6 are the terms of G.P. the x [ ] a) b) -4 c) 3 d) I G.P. a p + q = m, a p q =. The a p = [ ] a) m b) m c) m d) m Progressio, the th term is [ ] a) b) c) +1 d) a 1 = 37, d = 3, the S 1 = [ ] a) 64 b) 46 c) 4 d) 60. I the garde, there are 3 roses i the first row, i the d row there are 19. At the last row there are 7 trees, how may rows of rose trees are there? [ ] a) 10 b) 9 c) 11 d) 7 3. From 10 to 50, how may multiples of 4 are [ ] a) 40 b) 60 c) 45 d) The taxi takes Rs. 30 for 1 hour. After for each hour Rs. 10, for how much moey ca be paid & how it forms progressio [ ] a) Geometric progressios b) Harmoic progressio c) Series Progressios d) Arithmetic progressio Key 1) C ) D 3) C 4) C 5) B 6) B 7) B 8) D 9) A 10) B 11) B 1) B 13) B 14) C 15) D 16) A 17) A 18) A 19) C 0) A 1) B ) B 3) B 4) D

44 1. The sum of first 0 odd umbers. 10, 7, 4,.. a 30 = = 4. I the G.P 5, 5, 1, Bit Blaks 1. r = 5 5. The reciprocals of terms of G.P will form 6. If, x, 7 7 are i G.P. The x = = 8. If a, b, c are i G.P, the b a = 4x 5x 9. x,,,... a I a G.P a 4 = ,,, 1 are i The 10 th term from the ed of the A.P; 4, 9, is 13. I a G.P a 1 = 14. I a A.P s s 1 = terms = 16. I a series a 3, a ,,, 17.. A.P, the th term

45 18. a 3 = 5 & a 7 = 9, the fid the A.P 19. The th term of the G.P (0.5) -1, the the commo ratio 0. 4, 8, 16, 3 the fid the commo ratio is 1. The th term t the t 4 1. I a A.P l = 8, s = 144 & total terms are 9, the the first term is 3. I a A.P 11 th term is 38 ad 16 th term is 73, the commo differece of A.P is 4. I a garde there are 3 rose flowers i first row ad 9 flowers i d row, ad 6 flowers i 3 rd row, the how may rose trees are there i the 6 th row is 5. 5, 1, 3, 7. Progressio, the 6 th term is 6. I Arithmetic progressio, the sum of th terms is 4, the first term is Key 1) 400; ) 77; 3) 5050; 4) 1 ; 5) Geometric Progressio; 5 6) 1; 7) 55; 8) c b ; 9) 8 x ; 10) ar 3 ; 3 11) G.P.; 1) 09; 13) ar - ; 14) a ; 15) 70; 16) ; 17) ; 18) 3, 4, 5, 6, 7; 19) 0.5; 0) ; 1) 4 ; ) 4; 3) 7; 4) 17; 5) 15; 6) 3. 5

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