2. The ratio of the fifth term to the twelfth term of a sequence in an arithmetic progression is

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1 Chapte 7 Seqeces ad seies No calclato:. The secod tem of a aithmetic seqece is 7. The sm of the fist fo tems of the aithmetic seqece is. Fid the fist tem, a, ad the commo diffeece, d, of the seqece. (Total 4 maks). The atio of the fifth tem to the twelfth tem of a seqece i a aithmetic pogessio is 6. If each tem of this seqece is positive, ad the podct of the fist tem ad the thid tem is, fid the sm of the fist 00 tems of this seqece. (Total 7 maks). A aithmetic seqece has 5 ad as its fist two tems espectivel. Wite dow, i tems of, a expessio fo the th tem, a. Fid the mbe of tems of the seqece which ae less tha The sm of the fist tems of a aithmetic seqece is S =. Fid the th tem. 5. Fid the sm of the positive tems of the aithmetic seqece 85, 78, 7, Fid a expessio fo the sm of the fist 5 tems of the seies (Total 4 maks) (Total maks) (Total maks) l x x + l x l x l x givig o aswe i the fom l, whee m,. m (Total 5 maks) 7. Fid the sm to ifiit of the geometic seies (Total maks) 8. The th tem,, of a geometic seqece is give b = (4) +, +. Fid the commo atio. Hece, o othewise, fid S, the sm of the fist tems of this seqece. (Total maks) 9. Coside the ifiite geometic seies

2 Chapte 7 Seqeces ad seies x x x... Fo what vales of x does the seies covege? Fid the sm of the seies if x =.. (Total maks) 0. A seqece { } is defied b 0 =, =, + = whee +. Fid,, 4. () (i) Expess i tems of. (ii) Veif that o aswe to pat (i) satisfies the eqatio + =. (). The sm of the fist tems of a aithmetic seqece { } is give b the fomla S = 4. Thee tems of this seqece,, m ad, ae cosective tems i a geometic seqece. Fid m.. I a aithmetic seqece the secod tem is 7 ad the sm of the fist five tems is 50. Fid the commo diffeece of this aithmetic seqece.. The sm of the fist tems of a seies is give b S =, whee +. Fid the fist thee tems of the seies. Fid a expessio fo the th tem of the seies, givig o aswe i tems of. 4. The fist ad foth tems of a geometic seies ae 8 ad espectivel. Fid the sm of the fist tems of the seies; the sm to ifiit of the seies. (4) 5. The commo atio of the tems i a geometic seies is x. State the set of vales of x fo which the sm to ifiit of the seies exists. ()

3 Chapte 7 Seqeces ad seies If the fist tem of the seies is 5, fid the vale of x fo which the sm to ifiit is 40. (4) 6. Fid the sm of the ifiite geometic seqece 7, 9,,,.... () (Total maks) 7. A geometic seqece has all positive tems. The sm of the fist two tems is 5 ad the sm to ifiit is 7. Fid the vale of the commo atio; the fist tem. Calclato eqied: 8. The sm of a ifiite geometic seqece is, ad the sm of the fist thee tems is. Fid the fist tem. 9. Coside the aithmetic seies (Total maks) Fid a expessio fo S, the sm of the fist tems. Fid the vale of fo which S = The fist fo tems of a aithmetic seqece ae, a b, a +b + 7, ad a b, whee a ad b ae costats. Fid a ad b.. The thee tems a,, b ae i aithmetic pogessio. The thee tems, a, b ae i geometic pogessio. Fid the vale of a ad of b give that a b.. A sm of $5 000 is ivested at a compod iteest ate of 6.% pe am. (c) Wite dow a expessio fo the vale of the ivestmet afte fll eas. What will be the vale of the ivestmet at the ed of five eas? The vale of the ivestmet will exceed $0 000 afte fll eas. (i) Wite a ieqalit to epeset this ifomatio. (ii) Calclate the miimm vale of.. The sm to ifiit of a geometic seies is. The sm of the fist fo tems is 0 ad all the tems ae positive.

4 Chapte 7 Seqeces ad seies Fid the diffeece betwee the sm to ifiit ad the sm of the fist eight tems. 4. A sm of $00 is ivested. If the iteest is compoded aall at a ate of 5 pe ea, fid the total vale V of the ivestmet afte 0 eas. 5 If the iteest is compoded mothl at a ate of pe moth, fid the miimm mbe of moths fo the vale of the ivestmet to exceed V. 5. A ifiite geometic seies is give b k 4x k. Fid the vales of x fo which the seies has a fiite sm. Whe x =., fid the miimm mbe of tems eeded to give a sm which is geate tha Coside the aithmetic seies Fid the least mbe of tems so that the sm of the seies is geate tha Let a be the fist tem ad d be the commo diffeece, the 4 a + d = 7 ad S 4 = (a + d) = a d 7 4a 6d a = 5, d = 8 (A) (C)(C [4]. Let the aithmetic seqece be witte as a, a + d, a + d,... a 4d 6 The a d So a + 5d = 6a + 66d 7a = 4d a = d. Sice each tem is positive, both a ad d ae positive. We ae give a(a + d) =, settig a = d, we get d(d + d) = 8d =. d =. Hece, d = ad a = 4 ad sm to 00 tems of this seqece is 00 {()(4) + (00 l)}. = 0 00 [7]. a = 5 ad d = 8 a = a + ( )d a = 8 (C) 4

5 Chapte 7 Seqeces ad seies 8 < < 40 < o < 50 o < 5 8 Theefoe, thee ae 50 tems less tha 400. (C) [4] 4. = S S = [ ] [( ) ( )] = 6 5 (C) OR + = 6 4 = = 6 5 (C) [] 5. Aithmetic pogessio: 85, 78, 7,... = 85, d = 7 = + ( ) d = 85 7( ) = 9 7 Ths, > 0 povided. The eqied sm = S = ( + ) = (85 + ). = 559 (C) [] 6. METHOD l x + l x x l x l + = l x + (l x l ) +(l x l ) + (l x l ) + 5 S 5 = ( +( )d) = ( l x 4 l ) = 5 l x 595 l = l x 70 l x = l 595 (Accept m = 70, = 595) (N) METHOD l x x x x x x... x + l l l... l I the deomiato, the sm of the powes of is (0 + 4) = 595 The eqied expessio is l 70 x 595 (Accept m = 70, = 595) (N) [5] 5

6 Chapte 7 Seqeces ad seies 7. The sm to ifiit of a geometic seies is S =, = < (fom fomlae ad statistical tables) I this case, = ad =, theefoe S = 6 = 5 o 7. [] 8. = 9 = 4 (C) 48 OR = (4) = 4 (C) (4) ( ) 48(4 ) S = ( ) = 6(4 ) (C) [] 9. x The seies coveges povided < <. This gives.5 < x <.5 o x< (C) Whe x =., the commo atio is = 0.8 ad the sm is 0.8 = 5 (C) [] 0. = () () = (4) () 4 = (8) (4) = 4 = 8 4 = 6 (i) Cojecte is = (C) (ii) ( ) ( ) = ( ) ( ) = ( ) = ( ) = + 6

7 Chapte 7 Seqeces ad seies = + (AG). ( S ) S 0 d 8 50, ad m i geometic pogessio m m m 0 50 m 50 ( 50 ot possible sice d is positive) m 50 8( m) m 7 m (C6). Usig = + d + d = 7 A Usig S 5 4d d = 0 Attemptig to solve d = Note: Special case: = 7, S 5 = 50 = 0 d = gais fll maks. A AN. S = S S 8 6 S (C) S S ( ) ( ) 4 4 (C) 7

8 Chapte 7 Seqeces ad seies 4. 8 = = 7 A S 8 M 7 = A S 8 = 7 = A 5. 0 < x < x < 0 AN = 5 40 = 5 M = x = 8 A x = log (= ) A 8 Note: The sbstittio = x ma be see at a stage i the soltio. 8

9 Chapte 7 Seqeces ad seies 6. = S S M AN Attemptig to show that the eslt is te fo = M LHS = a ad RHS = a a A Hece the eslt is te fo = Assme it is te fo = k a aa... a k a k M Coside = k + : a a a... a k a k k k a a M a k k a Note: Note: k a a a a = A Awad A fo a eqivalet coect itemediate step. k a a k a = A Illogical attempted poofs that se the eslt to be poved wold gai MA0A0 fo the last thee above maks. k k The eslt is te fo = k it is te fo = k + ad as it is te fo =, the eslt is poved b mathematical idctio. Note: To obtai the fial R mak a easoable attempt mst have bee made to pove the k + step. RN0 [0] 9

10 Chapte 7 Seqeces ad seies a 7. a( + ) = 5; = 7 5 = 7 o a 54a = 0 = (C) a = 7 a = 9 (C) 8. 7 ad + + = 7 ( )( + + ) = 6 = givig = 7 Theefoe, = 9. (C) [] 9. S = ( +( )) = ( +) (C) ( + ) = = 0 9 => = 0 o = = 0 (C4) 0. (a + b + 7) (a b) = (a b) (a b) (a + b + 7) = (a + b + 7) (a b) 0 = a + 6b + 4 b = (C) a = (C). a = b ad b = a a + a = 0 a =, b = 4 (C6) Note: Do ot awad the fial fo a =, b = ol. Notes: Awad (C5) if both soltios a =, b = 4 ad a =, b = appea. If o wokig show, awad (C) fo oe coect vale (.06) 0

11 Chapte 7 Seqeces ad seies A N Vale = $5000(.06) 5 (= $ ) = $6790 to sf (Accept $6786, o $6786.5) AN (c) (i) 5000(.06) > 0000 (o (.06) > ) AN (ii) Attemptig to solve the above ieqalit log (.06) > log > eas A N Note: Cadidates ae likel to se TABLE o LIST o a GDC to fid. A good wa of commicatig this is sggested below. Let =.06 x Whe x =, =.958, whe x =, =.086 x = ie eas A 4 a a. S ad S4 0 a M 4 0 = 0.5 ad a = S S Note: Allow FT o icoect a / AA M AN 4. V = 00( ) 0 V = $65 (accept $65.) AN l l 00 = 5.65 MA AN

12 Chapte 7 Seqeces ad seies 5. = 4 x 4x M < 4 x < 5 x AN x =. a = 0.8 = 0.4 S So Solvig gives > A 7 tems ae eeded AN4 Note: Geeatig tems of the seies to fid that 7 tems ae eeded is a alteative method. 6. Usig S d with = 6 ad d = 7 S 7 9 o eqivalet A S Solvig > 0000 o eqivalet fo fo eg ( > 0) M > The least mbe of tems is 55. AN4