Chapter 8 - Sequences and Series

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1. Introduction - Sequence and Series

Sequence is ordered list of numbers following a rule. Example 2,5,8,11,14,... each term increases by 3. Series is sum of terms in a sequence Example 2+5+8+11+14.

Notation: a1 first term, a2 second term, an nth term term number n. Example sequence 3,6,12,24,... a1=3 a2=6 a3=12 pattern multiply by 2.

Types: Arithmetic Progression AP constant difference d add same number each time Example 2,5,8,11 d=3. Geometric Progression GP constant ratio r multiply same number each time Example 3,6,12,24 r=2 multiply by 2.

Finite sequence has limited terms Example 1,2,3,4,5 five terms, Infinite sequence never ends ... goes to infinity Example 2,4,6,8,... even numbers infinite.

Sequence ordered list numbers following rule Example 2,5,8,11,14 each increases by 3 Series sum terms sequence Example 2+5+8+11+14 | Notation a1 first term a2 second an nth term term number n Example 3,6,12,24 a1=3 a2=6 a3=12 pattern multiply 2 | Types AP constant difference d add same number Example 2,5,8,11 d=3 GP constant ratio r multiply same number Example 3,6,12,24 r=2 | Finite limited terms Example 1,2,3,4,5 five terms Infinite never ends ... infinity Example 2,4,6,8 even infinite

Introduction - Sequence and Series

2. Arithmetic Progression AP - nth term and Sum

AP each term increases by constant difference d. Definition: a_{n+1} - a_n = d constant. Example 2,5,8,11,14 d=5-2=3 constant.

nth term: a_n = a + (n-1)d where a first term d common difference n term number. Example AP 2,5,8,11 a=2 d=3 find 10th term a10 =2+(10-1)×3=2+9×3=2+27=29.

Sum of first n terms: S_n = n/2 [2a+(n-1)d] or S_n = n/2 (a + l) where l last term a_n =a+(n-1)d. Example sum first 5 terms of 2,5,8,11,14 n=5 a=2 d=3 S5=5/2[2×2+(5-1)×3]=5/2[4+12]=5/2×16=5×8=40. Also S_n = n/2(a+l) =5/2(2+14)=5/2×16=40.

Example theatre seats Row1 10 seats each row 4 more than before Row20? AP a=10 d=4 n=20 a20=10+(19)×4=10+76=86 seats Row20 86 seats. Total first 20 rows S20=20/2[2×10+19×4]=10[20+76]=10×96=960 total seats 960.

AP each term increases constant difference d a_{n+1}-a_n=d constant Example 2,5,8,11,14 d=3 | nth term a_n=a+(n-1)d a first term d common difference n term number Example AP 2,5,8,11 a=2 d=3 find 10th a10=2+9×3=29 | Sum first n terms S_n=n/2[2a+(n-1)d] or S_n=n/2(a+l) where l last term a_n Example sum first 5 terms 2,5,8,11,14 n=5 a=2 d=3 S5=5/2[4+12]=40 Also n/2(a+l)=5/2(2+14)=40 | Example theatre Row1 10 each row 4 more Row20 AP a=10 d=4 n=20 a20=10+19×4=86 Row20 86 Total first 20 rows S20=20/2[20+76]=10×96=960 total 960

Arithmetic Progression AP

3. Geometric Progression GP - nth term and Sum

GP each term multiplied by constant ratio r. Definition: a_{n+1}/a_n = r constant ratio. Example 3,6,12,24,48 r=6/3=2 multiply by 2.

nth term: a_n = a × r^{n-1} where a first term r common ratio n term number. Example GP 3,6,12,24 a=3 r=2 find 6th term a6=3×2^{6-1}=3×2^5=3×32=96.

Sum first n terms when r≠1: S_n = a(1-r^n)/(1-r) or S_n = a(r^n-1)/(r-1) same formula multiply numerator denominator by -1. Use first when r<1 second when r>1 for positive. Example GP 3,6,12,24,48,96 n=6 a=3 r=2 S6=3(2^6-1)/(2-1)=3(64-1)/1=3×63=189 total over 6 terms.

Infinite GP sum when |r|<1 less than 1 magnitude: S_∞ = a/(1-r) infinite sum converges to finite value. Example infinite GP 1,1/2,1/4,1/8,... a=1 r=1/2 |r|<1 S_∞=1/(1-1/2)=1/(1/2)=2 sum infinite terms 2.

Example bacteria culture starts 3 bacteria doubles each day r=2 a=3 after 6 days a6=96 bacteria total over 6 days S6=189 bacteria total counting all days.

GP each term multiplied constant ratio r Definition a_{n+1}/a_n=r constant Example 3,6,12,24,48 r=2 multiply 2 | nth term a_n=a×r^{n-1} a first term r common ratio n term number Example GP 3,6,12,24 a=3 r=2 find 6th a6=3×2^5=96 | Sum first n terms r≠1 S_n=a(1-r^n)/(1-r) or a(r^n-1)/(r-1) same multiply -1 Use first r<1 second r>1 Example GP 3,6,12,24,48,96 n=6 a=3 r=2 S6=3(64-1)/1=189 total over 6 terms | Infinite GP sum when |r|<1 S_∞=a/(1-r) infinite sum finite Example 1,1/2,1/4,1/8 a=1 r=1/2 S_∞=1/(1-1/2)=2 sum infinite 2 | Example bacteria starts 3 doubles r=2 a=3 after 6 days a6=96 total over 6 days S6=189 total counting all days

Geometric Progression GP

$GP$

4. AM GM HM - Means and Relation

Arithmetic Mean AM of two numbers a and b: AM = (a+b)/2 average.

Geometric Mean GM of a and b: GM = √(ab) square root of product, defined for positive numbers.

Example a=4 b=9 AM=(4+9)/2=13/2=6.5 GM=√(4×9)=√36=6. Observe AM≥GM 6.5≥6.

General AM≥GM≥HM inequality: For positive numbers, Arithmetic Mean ≥ Geometric Mean ≥ Harmonic Mean. Equality when a=b all equal.

Harmonic Mean HM = 2ab/(a+b) reciprocal average.

Example 2,8 AM=(2+8)/2=5 GM=√(16)=4 HM=2×2×8/(10)=32/10=3.2 So 5≥4≥3.2 AM≥GM≥HM true.

Use: GM inserted between a and b gives GP a,GM,b is GP if GM²=ab. AM inserted gives AP a,AM,b is AP if AM=(a+b)/2.

Example Insert 3 AMs between 2 and 14? AP with 5 terms total a=2 l=14 n=5 d=(l-a)/(n-1)=(14-2)/4=12/4=3 AP 2,5,8,11,14 three AMs 5,8,11.

AM two numbers a and b AM=(a+b)/2 average | GM a and b GM=√(ab) square root product positive numbers | Example a=4 b=9 AM=(4+9)/2=6.5 GM=√36=6 AM≥GM 6.5≥6 | General AM≥GM≥HM positive numbers Arithmetic ≥ Geometric ≥ Harmonic Equality when a=b all equal | HM=2ab/(a+b) reciprocal average | Example 2,8 AM=5 GM=√16=4 HM=32/10=3.2 So 5≥4≥3.2 AM≥GM≥HM true | Use GM inserted between a and b gives GP a,GM,b is GP if GM²=ab AM inserted gives AP a,AM,b is AP if AM=(a+b)/2 | Example Insert 3 AMs between 2 and 14 AP 5 terms a=2 l=14 n=5 d=(14-2)/4=3 AP 2,5,8,11,14 three AMs 5,8,11

AM GM HM Means Relation

5. Sum of Special Series - n, n², n³

Special sums formulas useful for many problems:

Sum of first n natural numbers 1+2+3+...+n = n(n+1)/2. Example n=5 sum=5×6/2=15 1+2+3+4+5=15.

Sum of squares of first n natural numbers 1²+2²+3²+...+n² = n(n+1)(2n+1)/6. Example n=3 sum=3×4×7/6=84/6=14 1+4+9=14.

Sum of cubes of first n natural numbers 1³+2³+...+n³ = [n(n+1)/2]^2 = (sum n)^2 square of sum natural numbers. Example n=3 sum= [3×4/2]^2=6^2=36 1+8+27=36.

These formulas used to find sum of AP of squares cubes etc.

Example sum 1×2+2×3+3×4+...+n(n+1) = sum n²+n = sum n² + sum n = n(n+1)(2n+1)/6 + n(n+1)/2 = n(n+1)(n+2)/3.

Special sums formulas useful | Sum first n natural 1+2+...+n = n(n+1)/2 Example n=5 sum=5×6/2=15 1+2+3+4+5=15 | Sum squares 1²+2²+...+n² = n(n+1)(2n+1)/6 Example n=3 sum=3×4×7/6=84/6=14 1+4+9=14 | Sum cubes 1³+2³+...+n³ = [n(n+1)/2]^2 = (sum n)^2 square sum natural Example n=3 sum=6^2=36 1+8+27=36 | These used find sum AP squares cubes etc | Example sum 1×2+2×3+...+n(n+1) = sum n²+n = sum n² + sum n = n(n+1)(2n+1)/6 + n(n+1)/2 = n(n+1)(n+2)/3

Sum of Special Series n, n², n³

6. Practical Applications - Compound Interest EMI Growth

Sequences and series used in finance and growth:

Compound interest: Amount A = P(1+r/n)^{nt} where P principal r annual rate n times compounded per year t years. This is GP with ratio (1+r/n). Example Invest $1000 at 5% annual compounded annually for 3 years A=1000(1+0.05)^3=1000×1.157625=1157.62 interest earned 157.62.

EMI Equal Monthly Instalment: EMI = [P×r×(1+r)^m]/[(1+r)^m-1] where P loan amount r monthly rate annual rate/12 m tenure months. Used for home loan car loan personal loan fixed monthly payment repay loan over time. Example Loan 10000 annual 12% monthly r=1% =0.01 tenure 24 months EMI≈470.73 per month.

Population growth: Population grows exponentially GP growth curve. Example bacteria doubles each day exponential growth model. Small steady growth rate → rapid increase over time.

Finance and investing: Savings and investments grow via compound interest exponential GP. Loans and mortgages EMI helps plan affordable monthly payments.

Science and biology: Sequences model patterns in nature and data.

Population planning: Growth models help plan housing resources services.

Sequences series used finance growth | Compound interest Amount A=P(1+r/n)^{nt} P principal r annual rate n times compounded per year t years This is GP ratio (1+r/n) Example Invest $1000 at 5% compounded annually 3 years A=1000×1.157625=1157.62 interest 157.62 | EMI Equal Monthly Instalment EMI=[P×r×(1+r)^m]/[(1+r)^m-1] P loan amount r monthly rate annual/12 m tenure months Used home loan car loan personal loan fixed monthly payment repay over time Example Loan 10000 annual 12% monthly r=1% 0.01 tenure 24 months EMI≈470.73 per month | Population growth Population grows exponentially GP growth curve Example bacteria doubles each day exponential growth Small steady growth rate → rapid increase over time | Finance investing Savings investments grow via compound interest exponential GP Loans mortgages EMI helps plan affordable monthly payments | Science biology Sequences model patterns nature data | Population planning Growth models help plan housing resources services

Practical Applications - Interest EMI Growth

7. Worked Examples - AP GP Word Problems

Example 1 Theatre: Theatre has 10 seats in Row1 each row has 4 more than before Find seats in Row20 and total seats first 20 rows. AP a=10 d=4 n=20 a20=10+(20-1)×4=10+76=86 Row20 86 seats Total S20=20/2[2×10+19×4]=10[20+76]=10×96=960 total 960 seats. Graph straight line linear seats vs row number.

Example 2 Bacteria: Bacteria culture starts 3 bacteria doubles each day How many bacteria after 6 days? Find total over first 6 days. GP a=3 r=2 n=6 a6=a×r^{n-1}=3×2^5=96 bacteria after 6 days total S_n=a(1-r^n)/(1-r)=3(1-2^6)/(1-2)=3(1-64)/(-1)=3×63=189 total over 6 days exponential growth curve Day1=3 Day2=6 Day3=12 Day4=24 Day5=48 Day6=96.

Quick comparison AP vs GP: Feature Pattern AP Add same number each time (+d) GP Multiply same number each time (×r) Common Value AP common difference d GP common ratio r Example AP 5,8,11,14,17 Example GP 2,6,18,54,162 Graph AP straight line linear GP exponential curve growing faster Typical Use Cases AP evenly spaced events seats steps GP population growth interest doubling.

Key reminders: Identify if pattern adding (+d) or multiplying (×r) first Always write down a,d or r,n before using formulas Check units answer in full sentence word problems.

Example 1 Theatre 10 seats Row1 each row 4 more Find seats Row20 total first 20 rows AP a=10 d=4 n=20 a20=10+19×4=86 Row20 86 Total S20=20/2[20+76]=10×96=960 total 960 Graph straight line linear seats vs row | Example 2 Bacteria starts 3 doubles each day How many after 6 days total over first 6 days GP a=3 r=2 n=6 a6=3×2^5=96 after 6 days total S_n=3(1-2^6)/(1-2)=3×63=189 total over 6 days exponential Day1=3 Day2=6 Day3=12 Day4=24 Day5=48 Day6=96 | Quick comparison AP vs GP Feature Pattern AP Add same number (+d) GP Multiply same number (×r) Common Value AP common difference d GP common ratio r Example AP 5,8,11,14,17 Example GP 2,6,18,54,162 Graph AP straight line linear GP exponential curve growing faster Typical Use AP evenly spaced events seats steps GP population growth interest doubling | Key reminders Identify pattern adding (+d) or multiplying (×r) first Always write down a,d or r,n before formulas Check units answer full sentence word problems

Worked Examples - AP GP Word Problems

8. Formulas Short Tricks - Quick Revision

Must remember:

Sequence ordered list numbers rule Series sum terms sequence a1 first term an nth term.

AP a_{n+1}-a_n=d constant difference d. nth term a_n=a+(n-1)d. Sum S_n=n/2[2a+(n-1)d]=n/2(a+l) where l=a_n last term.

GP a_{n+1}/a_n=r constant ratio r. nth term a_n=a×r^{n-1}. Sum S_n=a(1-r^n)/(1-r)=a(r^n-1)/(r-1) r≠1. Infinite sum |r|<1 S_∞=a/(1-r).

AM=(a+b)/2 GM=√(ab) HM=2ab/(a+b). AM≥GM≥HM equality when a=b.

Insert n AMs between a and b: d=(b-a)/(n+1) AP with n+2 terms. Insert n GMs: r=(b/a)^{1/(n+1)} GP with n+2 terms.

Special sums: Σn = n(n+1)/2 Σn² = n(n+1)(2n+1)/6 Σn³ = [n(n+1)/2]^2 = (Σn)^2.

Compound interest A=P(1+r/n)^{nt} GP ratio (1+r/n). EMI=[P×r×(1+r)^m]/[(1+r)^m-1] P loan r monthly rate annual/12 m tenure months.

Quick comparison AP add same number (+d) straight line linear evenly spaced events seats steps GP multiply same number (×r) exponential curve population growth interest doubling.

Trick: Identify pattern adding (+d) or multiplying (×r) first write down a,d or r,n before using formulas.

Must remember Sequence ordered list numbers rule Series sum terms sequence a1 first term an nth term | AP a_{n+1}-a_n=d constant difference d nth term a_n=a+(n-1)d Sum S_n=n/2[2a+(n-1)d]=n/2(a+l) where l=a_n last term | GP a_{n+1}/a_n=r constant ratio r nth term a_n=a×r^{n-1} Sum S_n=a(1-r^n)/(1-r)=a(r^n-1)/(r-1) r≠1 Infinite sum |r|<1 S_∞=a/(1-r) | AM=(a+b)/2 GM=√(ab) HM=2ab/(a+b) AM≥GM≥HM equality when a=b | Insert n AMs between a and b d=(b-a)/(n+1) AP n+2 terms Insert n GMs r=(b/a)^{1/(n+1)} GP n+2 terms | Special sums Σn=n(n+1)/2 Σn²=n(n+1)(2n+1)/6 Σn³=[n(n+1)/2]^2=(Σn)^2 | Compound interest A=P(1+r/n)^{nt} GP ratio (1+r/n) EMI=[P×r×(1+r)^m]/[(1+r)^m-1] P loan r monthly rate annual/12 m tenure months | Quick comparison AP add same number (+d) straight line linear evenly spaced events seats steps GP multiply same number (×r) exponential curve population growth interest doubling | Trick Identify pattern adding (+d) or multiplying (×r) first write down a,d or r,n before using formulas

Formulas Short Tricks