H2 Mathematics Arithmetic & Geometric Series ( )

Similar documents
07 - SEQUENCES AND SERIES Page 1 ( Answers at he end of all questions ) b, z = n

Chapter 2 Infinite Series Page 1 of 11. Chapter 2 : Infinite Series

Option 3. b) xe dx = and therefore the series is convergent. 12 a) Divergent b) Convergent Proof 15 For. p = 1 1so the series diverges.

PURE MATHEMATICS A-LEVEL PAPER 1

Worksheet: Taylor Series, Lagrange Error Bound ilearnmath.net

1985 AP Calculus BC: Section I

A Simple Proof that e is Irrational

Time : 1 hr. Test Paper 08 Date 04/01/15 Batch - R Marks : 120

z 1+ 3 z = Π n =1 z f() z = n e - z = ( 1-z) e z e n z z 1- n = ( 1-z/2) 1+ 2n z e 2n e n -1 ( 1-z )/2 e 2n-1 1-2n -1 1 () z

Probability & Statistics,

National Quali cations

MONTGOMERY COLLEGE Department of Mathematics Rockville Campus. 6x dx a. b. cos 2x dx ( ) 7. arctan x dx e. cos 2x dx. 2 cos3x dx

Review Exercises. 1. Evaluate using the definition of the definite integral as a Riemann Sum. Does the answer represent an area? 2

Chapter Five. More Dimensions. is simply the set of all ordered n-tuples of real numbers x = ( x 1

SOLVED EXAMPLES. Ex.1 If f(x) = , then. is equal to- Ex.5. f(x) equals - (A) 2 (B) 1/2 (C) 0 (D) 1 (A) 1 (B) 2. (D) Does not exist = [2(1 h)+1]= 3

+ x. x 2x. 12. dx. 24. dx + 1)

Discrete Fourier Transform (DFT)

STIRLING'S 1 FORMULA AND ITS APPLICATION

ln x = n e = 20 (nearest integer)

Digital Signal Processing, Fall 2006

Solution to 1223 The Evil Warden.

National Quali cations

Today is. first we went to the park, and then we went to the library

Restricted Factorial And A Remark On The Reduced Residue Classes

Calculus & analytic geometry

Chapter (8) Estimation and Confedence Intervals Examples

Problem Set 6 Solutions

DTFT Properties. Example - Determine the DTFT Y ( e ) of n. Let. We can therefore write. From Table 3.1, the DTFT of x[n] is given by 1

Additional Math (4047) Paper 2 (100 marks) y x. 2 d. d d

Chapter 1. Chapter 10. Chapter 2. Chapter 11. Chapter 3. Chapter 12. Chapter 4. Chapter 13. Chapter 5. Chapter 14. Chapter 6. Chapter 7.

They must have different numbers of electrons orbiting their nuclei. They must have the same number of neutrons in their nuclei.

Chapter 9 Infinite Series

Statistics 3858 : Likelihood Ratio for Exponential Distribution

( ) ( ) ( ) 2011 HSC Mathematics Solutions ( 6) ( ) ( ) ( ) π π. αβ = = 2. α β αβ. Question 1. (iii) 1 1 β + (a) (4 sig. fig.

On a problem of J. de Graaf connected with algebras of unbounded operators de Bruijn, N.G.

امتحانات الشهادة الثانوية العامة فرع: العلوم العامة

Triple Play: From De Morgan to Stirling To Euler to Maclaurin to Stirling

MILLIKAN OIL DROP EXPERIMENT

Narayana IIT Academy

DFT: Discrete Fourier Transform

Session : Plasmas in Equilibrium

WBJEEM MATHEMATICS. Q.No. μ β γ δ 56 C A C B

COLLECTION OF SUPPLEMENTARY PROBLEMS CALCULUS II

NEW VERSION OF SZEGED INDEX AND ITS COMPUTATION FOR SOME NANOTUBES

Chapter Taylor Theorem Revisited

NET/JRF, GATE, IIT JAM, JEST, TIFR

4. (5a + b) 7 & x 1 = (3x 1)log 10 4 = log (M1) [4] d = 3 [4] T 2 = 5 + = 16 or or 16.

Further Results on Pair Sum Graphs

Fooling Newton s Method a) Find a formula for the Newton sequence, and verify that it converges to a nonzero of f. A Stirling-like Inequality

Ordinary Differential Equations

Student s Printed Name:

LECTURE 13 Filling the bands. Occupancy of Available Energy Levels

Problem Value Score Earned No/Wrong Rec -3 Total

1973 AP Calculus BC: Section I

Introduction to Arithmetic Geometry Fall 2013 Lecture #20 11/14/2013

ENGG 1203 Tutorial. Difference Equations. Find the Pole(s) Finding Equations and Poles

Homework #3. 1 x. dx. It therefore follows that a sum of the

Law of large numbers

CHAPTER CHAPTER. Discrete Dynamical Systems. 9.1 Iterative Equations. First-Order Iterative Equations. (b)

Lectures 9 IIR Systems: First Order System

5.1 The Nuclear Atom

10. Joint Moments and Joint Characteristic Functions

Abstract Interpretation: concrete and abstract semantics

Chapter 4 - The Fourier Series

On the approximation of the constant of Napier

Multiple Short Term Infusion Homework # 5 PHA 5127

Mock Exam 2 Section A

How many neutrino species?

Chapter 3 Fourier Series Representation of Periodic Signals

Introduction to Quantum Information Processing. Overview. A classical randomised algorithm. q 3,3 00 0,0. p 0,0. Lecture 10.

MAT 182: Calculus II Test on Chapter 9: Sequences and Infinite Series Take-Home Portion Solutions

Washington State University

Reliability of time dependent stress-strength system for various distributions

Bipolar Junction Transistors

The pn junction: 2 Current vs Voltage (IV) characteristics

Linear Algebra Existence of the determinant. Expansion according to a row.

7' The growth of yeast, a microscopic fungus used to make bread, in a test tube can be

ASSERTION AND REASON

ENGI 3424 Appendix Formulæ Page A-01

Part B: Transform Methods. Professor E. Ambikairajah UNSW, Australia

ARITHMETIC PROGRESSION

An Introduction to Asymptotic Expansions

Cramér-Rao Inequality: Let f(x; θ) be a probability density function with continuous parameter

Derivation of a Predictor of Combination #1 and the MSE for a Predictor of a Position in Two Stage Sampling with Response Error.

2008 AP Calculus BC Multiple Choice Exam

Some Common Fixed Point Theorems for a Pair of Non expansive Mappings in Generalized Exponential Convex Metric Space

Thomas J. Osler. 1. INTRODUCTION. This paper gives another proof for the remarkable simple

(Reference: sections in Silberberg 5 th ed.)

CLASS XI CHAPTER 3. Theorem 1 (sine formula) In any triangle, sides are proportional to the sines of the opposite angles. That is, in a triangle ABC

Part B: Transform Methods. Professor E. Ambikairajah UNSW, Australia

Higher order derivatives

Gaps in samples of geometric random variables

A Review of Complex Arithmetic

EXERCISE - 01 CHECK YOUR GRASP

Physics 302 Exam Find the curve that passes through endpoints (0,0) and (1,1) and minimizes 1

International Journal of Advanced and Applied Sciences

A Propagating Wave Packet Group Velocity Dispersion

Discrete Mathematics and Probability Theory Fall 2014 Anant Sahai Homework 11. This homework is due November 17, 2014, at 12:00 noon.

APPENDIX: STATISTICAL TOOLS

Total Prime Graph. Abstract: We introduce a new type of labeling known as Total Prime Labeling. Graphs which admit a Total Prime labeling are

Transcription:

H Mathmatics Arithmtic & Gomtric Sris (08 09) Basic Mastry Qustios Arithmtic Progrssio ad Sris. Th rth trm of a squc is 4r 7. (i) Stat th first four trms ad th 0th trm. (ii) Show that th squc is a arithmtic progrssio ad stat th commo di rc. (iii) Fid th sum of th first 0 trms ad th sum of th xt 0 trms. (i),, 5, 9,..., (ii) u r 4r 7 u u (4 7) [4( ) 7] 4 (costat) Thrfor th squc is a arithmtic progrssio with d 4. (iii) S 0 0 S 50 50 S 50 S 0 4050 [( ) + (0 )(4)] 700 [( ) + (50 )(4)] 4750. Th first trm of a fiit arithmtic progrssio is. Th sum of th first 8 trms is 58 ad th sum of th whol sris is 5. Calculat (i) th commo di rc, (ii) th umbr of trms i th sris, (iii) th last trm. (i) S 8 8 [() + (8 )d] 58 ) d.5 (ii) S [() + ( )(.5)] 5 ) +5 00 0 ) 0 or 65 (rjctd as is a positiv itgr) (iii) u 0 + (0 )(.5) 0.5

Gomtric Progrssio ad Sris. A gomtric progrssio has first trm 5 ad commo ratio. (i) Stat th first four trms ad th 0th trm. (ii) Fid th sum of th first 0 trms. (iii) Fid th sum of th first trms. (iv) Fid th last valu of for which th sum of th first trms xcds.8. (v) Fid th sum to ifiity. (i) 5, 0, 0 9, 40 7,...,5 (ii) S 0 h 5 (iii) S 5 0 i 9 h 5 5 0 i (iv) 5 >.8 < 5 lg < lg 5 > lg 5 lg >6. ) last valu of is 7 (v) S 5 5 4. Th scod trm of a gomtric progrssio is 8 ad its sum to ifiity is 8. Fid th valu of th commo ratio ad th first trm of th gomtric progrssio. Dtrmi th sum of th first 0 trms, corrct to sigificat figurs. u ar 8 S a r 8 u ar a S r 8 8 ) r( r) 4 9 ) r r 4 9 0 ) r or 4 (rjctd as r < ) u ar 8 ) a 54 S 0 h 54 0 i 40.5

Practic Qustios. Th sum of th first trms of a squc is giv by S p+q, ad that S 6 ad S 5. (i) Fid th valus of p ad q. (ii) Show that th squc follows a arithmtic progrssio. Hc, or othrwis, stat th valu of th commo di rc. (i) S p + q S p +9q 6 S 5 5p + 5q Solvig simultaously, p.7 ad q 0. (ii) u S S (.7 +0. ) h.7( ) + 0.( ) i 0. +.6 u u (0. +.6) [0.( ) +.6] 0. (costat) Thrfor th squc is a arithmtic progrssio with d 0... Th sum of th first 0 trms of a arithmtic progrssio is 50 ad th sum of th xt 0 trms is 50. Fid th sum of th first 00 trms of this progrssio. S 0 50 S 40 S 0 50 ) S 40 0 S 0 0 [a + (0 )d] 50 ) a + 9d 5 () S 40 40 [a + (40 )d] 0 ) a + 9d 0 () Solvig () ad () simultaously, a 9 8 ad d 4 S 00 00 9 8 + (00 ) 4 750

. Fid th sum to trms of th sris x + x x +... + ( )r x r r +... Giv that th sris is covrgt, stat th st of possibl valus of x, ad th sum to ifiity. Th sris is gomtric with first trm a ad commo ratio r x. S x x 9 x +x Giv that th sris is covrgt, r < ) x < ) <x< S x 9 x + 4. Th sum of th first trms of a sris is giv by 5 5. (i) Show that th sris is a gomtric sris. (ii) Explai why th sris covrgs ad fid th sum to ifiity. (i) S 5 5 u S S 5 5 5 5 5 5 (5 ) 5 5 5 5 u u 5 5 5 (costat) Thrfor th sris is gomtric with r 5. (ii) Sic r 5 <, th sris covrgs. S 5 5 4

5. Th first trm of a gomtric progrssio is 8. Th sum of its first t trms is o-ighth of th sum of th rciprocals of ths trms. Show that th sum of th first sv trms of th origial gomtric progrssio is th sam as th sum of th rciprocals of th sv trms. Giv (S 0 ) origial 8 (S 0) rciprocals 8 h i 9 8 r 0 < 0 8 r r 8 : ; r h i 5 r 0 0 r r r r 5 r 0 r r 9 5r 9 r 0 r 0 ) 5r 9 ) r Usig r, (S 7 ) origial 8 r7 r 7 8 (S 7 ) rciprocals h 8 r r 7 i 7 8 6. Th sum of th first 00 trms of a arithmtic progrssio is 0 000. Th first, scod ad fifth trms of this progrssio ar thr coscutiv trms of a gomtric progrssio. Fid th first trm, a, ad th o-zro commo di rc, d, of th arithmtic progrssio. S 00 00 [a + (00 )d] 0 000 ) a + 99d 00 () u u u 5 u ) a + d a a +4d a + d ) (a + d) a (a +4d) ) d (d a) 0 () Substitut () ito () : d [d (00 99d)) 0 d (00d 00) 0 ) d 0 (rjctd) or d Hc a. 5

7. A boy rus from a startig poit O to ad from a sris of poits, A, A, A,..., icrasigly far away i a straight li. Th boy starts at O ad rus stag from O to A ad back to O, th stag from O to A ad back to O, ad so o. Th distacs btw th poits ar such that OA 4m,A A 4m,A A 8m,A A 4 6 m ad A A + A A for,, 4,... (i) Fid th distac ru by th boy wh h complts th first 0 stags. (ii) Hc fid th distac from O, ad th dirctio of travl, aftr h has ru xactly 4 km. (04/P/ modifid) (i) Th cotxt dscribs a gomtric progrssio with first trm a 4 8 ad commo ratio r. S 0 8 0 884 (ii) S 8( ) 8( ) S 4 000 ) 8( ) 4 000 75 lg lg 75 0.7 Th athlt has compltd 0 stags ad is ruig th th stag. 4 000 884 586 u 8 0 89 89 4096 Th athlt is 586 m ito th th stag. Sic h has ru mor tha half of th th stag, h is at a distac 76 m from O, ruig towards O. 6

8. (i) A compay is drillig for oil. Usig machi A, th dpth drilld o th first day is 56 mtrs. O ach subsqut day, th dpth drilld is 7 mtrs lss tha o th prvious day. Drillig cotius daily up to ad icludig th day wh a dpth of lss tha 0 mtrs is drilld. What dpth is drilld o th 0th day, ad what is th total dpth wh drillig is compltd? [6] (ii) Usig machi B, th dpth drilld o th first day is also 56 mtrs. O ach subsqut day, th dpth drilld is 8 9 of th dpth drilld o th prvious day. How may days dos it tak for th dpth drilld to xcd 99% of th thortical maximum total dpth? [4] (i) Th cotxt dscribs a arithmtic progrssio with first trm a 56 ad commo di rc d 7. Dpth drilld o th 0th day, u 0 a +( )d 56 + (0 )( 7) 9 m Dpth drilld o th th day, u 56 + ( )( 7) 6 7 <0 ) >6. Thrfor th last day is th 7th day. 6 7 Total dpth drilld, S 7 a +( )d 7 [(56) + (7 )( 7)] 480 m (ii) Th cotxt dscribs a gomtric progrssio with first trm a 56 ad commo di rc d 8 9. (0/P/9) Total dpth drilld, S > 99% thortical maximum total dpth, S ) a( r ) > 0.99 a r r r > 0.99 8 9 r < 0.0 < 0.0 lg 8 9 < lg 0.0 lg 0.0 > lg 8 9 >9. Thrfor it taks 40 days. 7

9. Two musical istrumts, A ad B, cosist of mtal bars of dcrasig lgths. (i) Th first bar of istrumt A has lgth 0 cm ad th lgths of th bars form a gomtric progrssio. Th 5th bar has lgth 5 cm. Show that th total lgth of all th bars must b lss tha 57 cm, o mattr how may bars thr ar. [4] Istrumt B cosists of oly 5 bars which ar idtical to th first 5 bars of istrumt A. (ii) Fid th total lgth, L cm, of all th bars of istrumt B ad th lgth of th th bar. [] (iii) Ufortuatly, th maufacturr misudrstads th istructios ad costructs istrumt B wrogly, so that th lgths of th bars ar i arithmtic progrssio with commo di rc d cm. If th total lgth of th 5 bars is still L cm ad th lgth of th 5th bar is still 5 cm, fid th valu of d ad th lgth of th logst bar. [4] (i) Lt th lgth of th first bar b a ad th commo ratio b r. a 0, ar 4 5 ) r 4 5 0 4 ) r 4 S a r 0 4 4 56. Thrfor, th total lgth of all th bars must b lss tha 57 cm. 4 (009/P/8) (ii) L S 5 a( r5 ) r h 0 4 4 4 5 i 4 7.6 7 Lgth of th th bar ar 0 4 4 0 (iii) Lt th lgth of th first bar b b. b + (5 )d 5 ) b 5 4d L 5 b + (5 )d ) 5 (5 4d) + 4d 7.6 ) d 0.49086 Sic d<0, th first bar is th logst bar. Lgth of th logst bar b 5 4( 0.49086) 6.8 cm 8

0. O Jauary 00 Mrs A put $00 ito a bak accout, ad o th first day of ach subsqut moth sh put i $0 mor tha i th prvious moth. Thus o Fbruary sh put $0 ito th accout ad o March sh put i $0 ito th accout, ad so o. Th accout pays o itrst. (i) O what dat did th valu of Mrs A s accout first bcom gratr tha $5000? [5] O Jauary 00 Mr B put $00 ito a savigs accout, ad o th first day of ach subsqut moth h put aothr $00 ito th accout. Th itrst rat was 0.5% pr moth, so that o th last day of ach moth th amout i th accout o that day was icrasd by 0.5%. (ii) Us th formula for th sum of a gomtric progrssio to fid a xprssio for th valu of Mr B s accout o th last day of th th moth (whr Jauary 00 was th st moth, Fbruary 00 was th d moth, ad so o). Hc fid i which moth th valu of Mr B s accout first bcam gratr tha $5000. [5] (iii) Mr B watd th valu of his accout to b $5000 o Dcmbr 00. What itrst rat pr moth, applid from Jauary 00, would achiv this? [] (0/P/4) 9

(i) Th cotxt dscribs a arithmtic progrssio with first trm a 00 ad commo di rc d 0. S (a +( )d) > 5000 [(00) + 0( )] > 5000 (90 + 0) > 5000 + 9 000 > 0 ) < 4.5 or >.5 ) It taks 4 moths for th valu to xcd $5000, i.. Dcmbr 00. (ii) O th last day of th th moth: th st $00 has compoudd tims ito 00 (.005) th d $00 has compoudd ( ) tims ito 00 (.005) th rd $00 has compoudd ( ) tims ito 00 (.005) th th $00 has compoudd oc ito 00 (.005) Total valu of th accout 00 (.005) +... + 00 (.005) + 00 (.005) + 00 (.005) Usig th formula for th sum of a GP, S 00(.005) (.005 ).005 0 00 (.005 ) S 0 00 (.005 ) > 5000.005 >.487 lg.005 > lg.487 lg.487 > lg.005 >44.5 ) It taks 45 moths for th valu to xcd $5000, i.. Sptmbr 004. (iii) O th last day of Novmbr 00, 5. O Dcmbr 00, S 5 + 00 5000 00x x 5 + 00 5000 x x x 5 49 x From GC, x.08. Thrfor th rquird itrst rat is.8%. 0

Furthr Practic Qustios. A arithmtic progrssio has first trm a ad commo di rc 0. Th sum of th first trms 0 000 is 0 000. Exprss a i trms of, ad show that th th trm is + 5( ). Giv that th th trm is lss tha 500, show that 0 + 000 < 0, ad hc fid th largst possibl valu of. S [a +( )(0)] 0 000 ) a 0 000 5( ) u a +( )(0) 0 000 5( ) + 0 ( ) 0 000 +5( ) 0 000 +5( ) < 500 000 + ( ) < 00 0 + 000 < 0 ) 7.04 <<7.96 Hc th largst possibl valu of is 7.. A squc u,u,u,...is such that u ad u + u. Show that th sum of th first trms of th progrssio is ( 6). Th squc is a arithmtic progrssio with a ad d. S [a +( )d] [( ) + ( )( )] ( 6)

. Th first thr trms, all positiv, of a crtai gomtric progrssio ar x 4, x ad 5x. Fid (i) th valu of x, (ii) th valu of th fourth trm. (i) u u ) x 5x u u x 4 x ) x (5x )(x 4) ) 4x x + 48 0 ) x 8x + 0 ) x 6 or x (rjctd as x 4 > 0) (ii) u 5x 8 r x x 4 u 4 u r 54 4. A arithmtic progrssio has trms ad a commo di rc d. Prov that th di rc btw th sum of th last k trms ad th sum of th first k trms is ( k) kd. Sum of th last k trms S S k Sum of th first k trms S k Di rc (S S k ) S k [a +( )d] k [a +( k )d]... ( k) kd k [a +(k )d]

5. A arithmtic squc u,u,u,...is dfid by u r r for r Z +. (i) Fid th last valu of such that u + u 6 + u 9 +... + u xcds 700, whr is a multipl of. a ur (ii) Aothr squc of ral umbrs v,v,v,...is dfid by v r for r Z +,whra is a costat. Show that v,v,v,...forms a gomtric squc. Stat th st of valus of P a for which th sum to ifiity xists ad show that k v r r,whrk is a costat a to b dtrmid. (i) u,u,u,...is a arithmtic progrssio with first trm ad commo di rc. u,u 6,u 9,...,u is a arithmtic progrssio with first trm 8 ad commo di rc 9. Lt S k k [(8) + (k )(9)] > 700 ) 9k +7k 400 > 0 ) k>9.05 or k< 9.8 Hc, at last 0 trms ar rquird, i.. th last valu of is 60. (ii) v r v r a u r a u r a ur u r a (costat) a. Thrfor th squc is a gomtric progrssio with r Th sum to ifiity xists wh r a < ) a < X r v r v r a u a a a a a

6. Pics of dcrasig lgths ar cut from a ribbo. Th lgths of th pics cut follow a arithmtic progrssio with th 6th pic ad th 6th pic cut big of lgths 9 cm ad 5 cm rspctivly. (i) Fid th lgth of th first pic cut ad th commo di rc of th arithmtic progrssio. (ii) Assumig that th ribbo is su citly log, fid th gratst umbr of such pics that ca b cut from th ribbo ad also th last possibl lgth of th ribbo i this cas. (i) u 6 a +5d 9 u 6 a + 5d 5 Solvig simultaously, a 0 ad d 0. (ii) Lt u 0 + ( ) 5 > 0 ) <0 Thrfor th gratst umbr of pics that ca b cut is 00. Last possibl lgth, S 00 00 [(0) + (00 )( 0.)] 00 cm 4