Arithmetic Sequences and Series  

What is a Sequence?

A Sequence is a set of things (usually numbers) that are in order.

Infinite or Finite

If the sequence goes on forever it is called an infinite sequence,otherwise it is a finite sequence

Examples:

{1, 2, 3, 4 ,...} is a very simple sequence (and it is an infinite sequence)

{20, 25, 30, 35, ...} is also an infinite sequence

{1, 3, 5, 7} is the sequence of the first 4 odd numbers (and is a finite sequence)

{4, 3, 2, 1} is 4 to 1 backwards

{1, 2, 4, 8, 16, 32, ...} is an infinite sequence where every term doubles

{a, b, c, d, e} is the sequence of the first 5 letters alphabetically

{f, r, e, d} is the sequence of letters in the name "fred"

{0, 1, 0, 1, 0, 1, ...} is the sequence of alternating 0s and 1s (yes they are in order, it is an alternating order in this case)

In Order

When we say the terms are "in order", we are free to define what order that is! They could go forwards, backwards ... or

they could alternate ... or any type of order you want!

Like a Set

A Sequence is like a Set, except:

the terms are in order (with Sets the order does not matter) the same value can appear many times (only once in Sets)

Example: {0, 1, 0, 1, 0, 1, ...} is the sequence of alternating 0s and 1s.

The set would be just {0,1}

Notation

Sequences also use the same notation as sets:

list each element, separated by a comma,

{3, 5, 7, ..}

and then put curly brackets around the whole thing.

The curly brackets { } are sometimes called "set brackets" or "braces".

A Rule

A Sequence usually has a Rule, which is a way to find the value of each term.

Example: the sequence {3, 5, 7, 9, ...} starts at 3 and jumps 2 every time:

As a Formula

Saying "starts at 3 and jumps 2 every time" is fine, but it doesn't help us calculate the:

10th term, 100th term, or

nth term, where n could be any term number we want.

So, we want a formula with "n" in it (where n is any term number).

So, What Would A Rule For {3, 5, 7, 9, ...} Be?

Firstly, we can see the sequence goes up 2 every time, so we can guess that a Rule will be something like "2 times n" (where "n"

is the term number). Let's test it out:

Test Rule: 2n

n Term Test Rule1 3 2n = 2×1 = 22 5 2n = 2×2 = 43 7 2n = 2×3 = 6

That nearly worked ... but it is too low by 1 every time, so let us try changing it to:

Test Rule: 2n+1

n Term Test Rule1 3 2n+1 = 2×1 + 1 = 32 5 2n+1 = 2×2 + 1 = 53 7 2n+1 = 2×3 + 1 = 7

That Works!

So instead of saying "starts at 3 and jumps 2 every time" we write this:

2n+1

Now we can calculate, for example, the 100th term:

2 × 100 + 1 = 201

Many Rules

But mathematics is so powerful we can find more than one Rule that works for any sequence.

Example: the sequence {3, 5, 7, 9, ...}

We have just shown a Rule for {3, 5, 7, 9, ...} is: 2n+1

And so we get: {3, 5, 7, 9, 11, 13, ...}

But can we find another rule?

How about "odd numbers without a 1 in them":

And we would get: {3, 5, 7, 9, 23, 25, ...}

A completely different sequence!

And we could find more rules that match {3, 5, 7, 9, ...}. Really we could.

So it is best to say "A Rule" rather than "The Rule" (unless you know it is the right Rule).

Notation

To make it easier to use rules, we often use this special style:

xn is the term n is the term number

Example: to mention the "5th term" you just write: x5

So a rule for {3, 5, 7, 9, ...} can be written as an equation like this:

xn = 2n+1

And to calculate the 10th term we can write:

x10 = 2n+1 = 2×10+1 = 21

Can you calculate x50 (the 50th term) doing this?

Here is another example:

Example: Calculate the first 4 terms of this sequence:

{an} = { (-1/n)n }

Calculations:

a1 = (-1/1)1 = -1 a2 = (-1/2)2 = 1/4

a3 = (-1/3)3 = -1/27 a4 = (-1/4)4 = 1/256

Answer:

{an} = { -1, 1/4, -1/27, 1/256, ... }

  Fibonacci Sequence

This is the Fibonacci Sequence

0, 1, 1, 2, 3, 5, 8, 13, 21, 34, ...

The next number is found by adding the two numbers before it together:

The 2 is found by adding the two numbers before it (1+1) The 21 is found by adding the two numbers before it

(8+13) etc...

Rule is xn = xn-1 + xn-2

That rule is interesting because it depends on the values of the previous two terms.

Rules like that are called recursive formulas.

The Fibonacci Sequence is numbered from 0 onwards like this:

n = 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 ...

xn = 0 1 1 2 3 5 8 13 21 34 55 89 144233

377 ...

 

Example: term "6" would be calculated like this:

x6 = x6-1 + x6-2 = x5 + x4 = 5 + 3 = 8

Arithmetic Sequences

In an Arithmetic Sequence the difference between one term and the next is a constant.

In other words, you just add some value each time ... on to infinity.

Example:

1, 4, 7, 10, 13, 16, 19, 22, 25, ...

This sequence has a difference of 3 between each number. Its Rule is xn = 3n-2

In General you could write an arithmetic sequence like this:

{a, a+d, a+2d, a+3d, ... } where:

a is the first term, and d is the difference between the terms (called the "common difference") And you can make the rule by: xn = a + d(n-1)

(We use "n-1" because d is not used in the 1st term

Exercise:  Find the next term and the general formula for the following:

A.{2, 5, 8, 11, 14, ...}

B.{0, 4, 8, 12, 16, ...}

C.{2, -1, -4, -7, -10, ...}

For each of these three sequences there is a common difference. In the first sequence the common difference is d = 3, in the second sequence the common difference is d = 4, and on the third sequence the common difference is d = -3.  We will call a sequence an arithmetic sequence if there is a common difference. The general formula for an arithmetic sequence is

an = a1  + (n - 1)d

Example

What is the difference between the fourth and the tenth terms of {2,6,10,14,...) ,We have a10 - a4  =  (10 - 4)d  =  6(4)  =  24

  Arithmetic Series

First we see that

1+ 2 + 3 + ... + 100  =  101 + 101 + ... + 101 (50 times)  =  101(50) In general               

        1 + 2 + 3 + ... + n  =   n(n+1)2

Example : What is S  =  1 + 4 + 7 + 10 + 13 +... +  46 Solution :S  =  1 + (1 + 1(3)) + (1 + 2(3))  + (1 + 3(3)) + ... + (1 + 15(3)) =  (1 + 1 + ... + 1) + 3(1 + 2 + 3 + ... + 15)=  16 + 3(15)(16)/2 Or Alternatively

Sn = n/2(a1 + an)

Example

How much will I receive over my 35 year career if my starting salary is $40,000, and I receive a 1,000 salary raise for each year I work here?

Solution

We have the series:

        40,000 + 41,000 + 42,000 + ... + 74,000

        =  35/2 (40,000 + 74,000)  =  $1,995,500

An important result : if a,b,c are in A.P. then b – a = c

– b ⇨ 2b = a+c , d = b−an+1 and nth term = Tn = Sn – Sn-1

Arithmetic meam between two numbers a & b is (a+b)/2 is denoted by A.M.

A.P.

Question:3 In an A.P., the first term is 2 and the sum of the first five terms is one-fourth of the next five terms.Show that 20th term is -112.

Answer: S5 = ¼ (S10 - S5) ⇨ 5 S5 = S10 ⇨ 5.5/2 [2a+(n-1)d] = 10/2[2a+(n-1)d] ⇨ d = -6 ∵a=2 a20 = a+19d = 2+19×(-6) = -112.Question: 5 In an A.P., if pth term is 1/q and qth term is 1/p,

prove that the sum of first pq terms is

½ (pq+1), where p≠q.

Answer: pth term = a+(p-1)d = 1/q………..(1)

qth term= a+(q-1)d = 1/p……….(2) By subtracting (1)&(2) , we get d(p-1-q+1) =1q - 1

p ⇨ d = 1pq by putting the value of d

In (1) a = 1pq ⇨ Spq = pq/2 [ 2×

1pq +(pq-1) 1

pq ] = ½ (pq+1).

Question: 9 The sum of first n terms of two arithmetic progressions are in the ratio

5 n+49 n+6 . Find the ratio of their 18th

term.

Answer: ∑ of n terms of first A . P .

∑ of n terms of second A . P . =

5 n+49 n+6

n2[2 a1+(n−1 )d 1]

n2[2 a2+(n−1 )d 2]

= 5 n+49 n+6

[2 a1+ (n−1 )d 1][2 a2+ (n−1 )d 2]

= 5 n+49 n+6

……(1) ⇨ 18 th termof first A . P .18th termof second A . P .

= a 1+17 d 1a 2+17 d 2

By putting n=35 in (1) 2 a1+ (35−1 ) d 1

2 a2+¿¿ = 5× 35+49 ×35+6 =

179321 .

Question:11 Sum of first p, q, r terms of an A.P. are a,b,c respectively. Prove that

ap (q - r) +

bq (r - p) +

cr (p – q) = 0.

Answer: Sp = p/2 [2a1 + (p – 1)d] = a

Sq = q/2[2a1 + (q – 1)d] = b

Sr = r/2[2a1 + (r – 1)d ] = c

1/2 [2a1 + (p – 1)d]×(q – r) = ap

× (q - r)………………(i)

1/2[2a1 + (q – 1)d]× (r – p) = bq ×

(r – p)………………(ii)

1/2[2a1 + (r – 1)d ] × (p – q) = cr ×

(p – q)………………(iii)

(i) + (ii) + (iii)

ap (q - r) +

bq (r - p) +

cr (p – q) = ½{2a1 (q – r + r – p + p -

q) + d [ (p-1)(q-r) + (q-1)(r-p) + (r-1)(p-q) ]}

= ½ { 0 + d [ pq – q –pr +r + qr – r – pq +p + rp –p – qr +q]} = 0

Question:15 If aⁿ+bⁿ

an−1+bn−1 is the A.M. between a and b, then find

the value of n.

[Hint : a ⁿ+b ⁿ

an−1+bn−1 = a+b

2 ⇨ (a ⁿ+b ⁿ)2 = (a+b)(an−1+bn−1) ⇨ 2an +2

bn = an+abn-1+ban-1+bn

an-1(a-b) - bn-1(a-b) = 0 ⇨(a-b)(an-1 - bn-1) = 0 ⇨ (a-b)=0 or (an-1 - bn-1) =0 ⇨ n=1 ∵ a≠b.

Question: 16 Between 1 and 31 , m numbers have been inserted in such a way that the resulting sequence. In an A.P. and ratio of 7th and (m – 1)th numbers is 5: 9.

Answer: Let a2, a3, a4,…………am+1 as 1, a2, a3, a4,…………am+1, 31

am+2 = a1+ (m+1) d = 31 and a1 = 1

31 = 1 + (m+1) d ⇨ d = 30/(m+1) or d =(b – a)/ (n+1)

According to statement a 8am =

59 ⇨

a 1+7 da 1+(m−1)d =

1+7×30

(m+1)

1+(m−1)×30

(m+1)

=59 ⇨ m = 14.

Geometric Sequences

In a Geometric Sequence each term is found by

multiplying the previous term by a constant.

Example:

2, 4, 8, 16, 32, 64, 128, 256, ... This sequence has a factor of 2 between each number.

Its Rule is xn = 2n

In General you could write a geometric sequence like this:

{a, ar, ar2, ar3, ... }

where:

a is the first term, and

r is the factor between the terms (called the "common ratio")

r can be calculated by dividing any two consecutive terms

in a geometric sequence. The formula for calculating r is...

...where n is any positive integer greater than 1.

Note: r should not be 0.

When r=0, you get the sequence {a,0,0,...} which is not geometric

And the rule is:

xn = ar(n-1)

(We use "n-1" because ar0 is the 1st term)

Finding the Sum of a Series

Given our generic arithmetic sequence a1, a2, a3, a4, ..., an, we can look at it as a series: a1 + a2 + a3 + a4 + ... + an. There exists a formula that can add a finite list of these numbers and a formula for an infinite list of these numbers. Here are the formulas...

and

...where Sn is the sum of the first n numbers, a1 is the first number in the sequence, r is the common ratio of the sequence, and -1 < r < 1 for infinite series.

***Introduction The process of adding infinitely many numbers is at the heart of the mathematical concept of a numerical series.

Let ‘s denote the sum of the series just considered:

Let's multiply both sides by 1/2

and subtract the second line from the first. All terms on the right

side except for the 1 will cancel out!

We have shown that

One also says that this series converges to 2.

Let's play the same game for a general q instead of 1/2:

multiply both sides by q

then, subtract the second line from the first:

The series

is called the geometric series. It is the most important

series you will encounter!

S∞ = a1−r , a is first term , r is the common ratio.

As n→∞ rn → 0 for |r|<1.

Geometric mean between two numbers a & b is √ab i.e.,

G2 = ab or G = √ab , Harmonic mean (H.M.)= 2aba+b

If a , b, c are in G.P. then b/a = c/b ⇨ b2 = ac. A – G = a+b2

- √ab ≥ 0

⇨ A ≥ G≥H(Harmonic mean)

R = ¿

Special sequences: (i) 1+2+3+…………..+n = n(n+1)2

(ii) 12+22+32+……..+ n2 = n (n+1 )(2n+1)6

(iii) 13+23+33+………+n3 = ( n(n+1)2 )2

Example:

Find the sum of the series

First, factor out the 5 from upstairs and a 2 from downstairs:

.

The series in the parentheses is the geometric series with ,

but the first term, the "1" at the beginning is omitted.

Thus, the series sums up to

There is a slightly slicker way to do this. Do you see how?

Try it yourself!Find the sum of the series

G.P.

Question: 21 Find four numbers forming a G.P. in which

the third term is greater than the first term by 9, and the

second term is greater than the 4th term by 18.

Answer: In G.P. a, ar, ar2, ar3 are first four terms.

ar2 – a = 9 and ar3 - ar = 18 ,

dividing by above results , we get r =-2 and a=3, then terms

are 3, -6, 12 and -24.

Question:23 If the first and the nth term of a G.P. are a and b resp.

and if P is the product of n terms , prove that P2 = (ab)n .

Answer: P = a.ar. ar2. ar3……..arn-1 = an. r(1+2+3+……+n-1) =

an . rn (n−1 )

2 [+2+3=……+n=n(n+1)/2]

P2 = ( an . rn (n−1 )

2 )2 = a2n . r2n (n−1)

2

= [a.(a rn-1]n =(ab)n ∵ b = nth term = a rn-1

Question:24 Show that the ratio of the sum of first n

terms of a G.P. to the sum of terms from (n+1)th to (2n)th Term

is 1

rn .

Answer: Sn = a(1−rn)1−r

……….(i)

The sum of terms from (n+1)th to (2n)th = arn(1−rn)1−r

……(ii)

[ a,ar,ar2,…….arn-1, arn,arn+1, ……ar2n-1]

Question:28 The sum of two numbers is 6 times their G.P.,

show that numbers are in the ratio (3+2√2¿: ¿3-2√2).

Answer: a+b

2√ab = 31 by c & d

a+b+2√aba+b−2√ab =

3+13−1 ⇨ √a+√b

√a−√b = √ 21 , again by c& d,

we get

√a√b = √2+1

√ 2−1 , then squaring on the both sides ,we get

the answer.

(MISC.)

Question:4 Find the sum of all numbers between 200 and 400 which are divisible by 7.

Answer: 203, 210, 217, 224, ……. ,339.

399 = 203 + (n-1)7 ⇨ 196 = (n-1)7 ⇨ n = 29, so Sn = n/2 (a+l) = 29/2(203=399) =29 ×301 = 8729.

Question: 5 Find the sum of integers from 1 to 100 that are divisible by 2 or 5.

Answer: Divisible by 2 are 2,4,6,……..,94,96,98,100 or divisible by 5 are 5,10,15…….95,100.

nth term = 100 = 2+(n-1) × 2 ⇨ n = 50

10 terms as 10,20,30,……100 of 5,10,15…….95 are already used in 2,4,6,…….100 (100-40[50-10]=60)

The sum of remaining 60 terms is 60/2[2+100] = 3060.

Question:11 A G.P. consists of an even number of terms. If the sum of all the terms is 5 times the sum

Of terms occupying odd places, then find the common ratio.

Answer: a, ar, ar2, ar3 ,…………arn-1 in G.P.

S2n = 5[ a+ ar2 +ar4+……n terms]

a(1−r2 n)1−r

= 5. a¿¿ ⇨ 1= 5

1+r ⇨ r = 4.

Question: 18 If a and b are the roots of x2 – 3x +p = 0 and c, d are the roots of x2 – 12x + q = 0,

where a,b,c,d form an G.P. Prove that (q+p) : (q-p) = 17:15.

Answer: a+b = 3, ab = p and c+d = 12, cd = q

a, b, c, d are in G.P. i.e., b=ar, c = ar2 d= ar3

a+b= a+ar=a(1+r) = 3 …..(i) and c+d=12 ⇨ ar2 + ar3 = ar2(1+r) = 12….(ii) , by dividing (i)&(ii)

1/r2 = ¼ ⇨ r = ±2 , then q+ pq−p =

cd+abcd−ab =

a ² r (r ⁴+1)a ² r (r ⁴−1) =17:15.

Question:20 If a, b, c are in A.P. ; b, c, d are in G.P. and 1/c, 1/d, 1/e are in A.P. Prove that a, c, e are in G.P.

Answer: 2b =a+c…..(i)

c2 =bd…..(ii)

2/d = 1/c+1/e = e+cce ⇨ d =

2 cec+e ........(iii)

We shall prove c2 = ae

From (i),(ii) &(iii) we get,

c2 = ¿)(2 cec+e

¿ = ace+c ² e

c+e ⇨ c2 (c+e) = ace+c2e ⇨ c3 + c2e =

ace+c2e ⇨ c2 = ae.

Question:23 Find the sum of 3+7+13+21+31+.........

Answer: an = n(n+1)+1 ,n=1,2,3….

= n2+n+1

Sn = ∑1

n

k ² + ∑1

n

k +n = n (n+1 )(2n+1)

6 + n(n+1)2 +n.

Question:27 A farmer buys a used tractor for Rs. 12000. He pays Rs. 6000 cash and agrees to pay the balance in annual instalments of Rs. 500 plus 12% interest on the unpaid amount. How much will the tractor cost him?

Answer: Total cost =Rs. 12000 , paid cash = 6000 then balance = Rs. 6000

No. of instalments of Rs. 500 = 12

Interest on 1st instalment = Rs. 6000× 12× 1

100 =Rs.720

1st instalment = Rs.(500+720)=1220

Interest on 2nd instalment = Rs. 5500× 12× 1

100 =Rs.660

2nd instalment = Rs.(550+660)=1160 and so on…

Total amount in instalments = (1220+1160+1100+……to 12 terms)

S= 12/2[2× 1220 + (12-1)(-60)] = 10680 , therefore total paid by farmer = Rs. 6000+10680=16680.

Question: 32 150 workers were engaged to finish a job in a certain number of days. 4 workers dropped

Out on second day, 4 more workers dropped out on third day and so on.It took 8 more days to finish the work. Find the number of days in which the work was completed.

Answer: First term is 150 common difference is -4 (4 workers dropped every day)

n/2[2×150+(n-1)×-4] = n(152-2n)

Had the workers not dropped out , then the work would have finished in (n-8) days with 150 workers working on each day . The total no. of workers who would have worked all then days are 150(n-8)

n(152-2n) = 150(n-8) ⇨ n=25 ∵ n≠-24.

ASSIGNMENT OF SEQUENCE & SERIES

Question:1 Find k so that 2/3, k, 5k/8 are in A.P.

Question:2 If the roots of (b-c)x2+(c-a)x+(a-b) = 0 are equal, then a,b,c are in A.P.

[Hint: take Discriminant D=0]Question:3 There are n A.M.’s between 7 and 85 such that (n-3)th mean : nth mean is 11 : 24.Find n.[Hint: 7, a2, a3, a4,…………..an+1, 85 ⇨ an+2 = 85 ⇨ d= 78n+1 same as in your ncert book problem, find n=5]Question:4 Prove that the sum of n terms of the series

11+103+1005+ ……..is 109 (10n - 1) + n2

[HINT: Sn = (10+1)+(100+3)+ (1000+5)+……….n terms

= (10+102+103+……n term)+(1+3+5+……n terms)

G.P. A.P.

Question:5 How many terms of the series 2+2√ 2 +4+…..will amount to 30+14 √ 2 .

[Hint: use Sn = a(1−rn)1−r

= 30+14 √ 2 ,a=2 & r =√ 2, then n=7]

Question:6 Find the sum of (1+1

22) + (12 +

1

24) + (1

22+ 1

26) +…….to ∞[Hint: (1 +

12 +

1

22+…..) + (1

22+ 1

24+… ), answer is 7/3 ,use S∞ =

a1−r

, a is first term , r is the common ratio.]Question:7 Show that (x2+xy+y2) , (z2+zx+x2), (y2+zy+z2) are consecutive terms of A.P. if x, y, z are in A.P.[Hint: let a = (x2+xy+y2) , b = (z2+zx+x2) ,c = (y2+zy+z2) use b-a = c-b , then you will get if (x+z)2 – y2 = y(x+y+z) ⇨ x+z = 2y]Question:8 If a, b, c, d are in G.P. prove that a2-b2, b2-c2, c2-d2 are also in G.P. [HINT: take a, b= ar, c=ar2, d = ar3 , show (b2-c2)/ (a2-b2)= (c2-d2)/ (b2-c2) by putting values of a, b, c, d]

Question:9 If one geometric mean G and two arithmetic mean p and q be inserted between two quantities Show that G2 = (2p-q)(2q-p).[Hint: G2 = ab, a, p, q, b are in A.P. d(common difference)= (b-a)/3 . p = a+d= 2 a+b

3 , q = a+2d = a+2 b3 put these values in R.H.S.]Question:10 If the value of 1+2+3+…….+n = 28, then find the value of 12+22+32+………..+n2 .**Question:11 If g1, g2 be two G.M.’s between a and b and A is the A.M. between a & b, then prove that

g 1²g 2 + g 2²

g 1 = 2A.[Hint: 2A = a+b, b/g2 = g2/g1 = g1/a ⇨ a = g12 / g2 , b = g22 /g1.]**Question:12 If a is the A.M. of b, c and two geometric means between b , c and G1,G2, then prove that G13 = g23 . Answer: 2a = b+c , c = ar3 ⇨ r = ¿ , put in G1= br & G2= br2 .

**Question:13 If x = 1+a+a2+a3+……….∞ and y = 1+b+b2+b3+………. ∞ Then prove that 1+ab+a2b2+….. ∞ = xy

x+ y−1Answer: x = 1+a+a2+a3+……….∞ = 11−a y = 1+b+b2+b3+………. ∞ = 1

1−b take R.H.S. xyx+ y−1 = 1

1−ab = (1-ab)-1 = 1+ab+a2b2+

….. ∞ [ by binomial expansion (1-x)-1 = (1+x+x2+x3+…..)]**Question:14 An A.P. consists of n(odd)terms and it’s middle term is m. Prove that Sn = mn. Answer: m = mid term = T(n+1)/2 = a+(n+1

2 - 1)d ⇨ 2m= 2a+(n-1)d , Sn = n/2[2a+(n-1)d]=mn.**Question:15 Sum of infinity the series 1

2 + 12+4 + 1

2+4+6 +………..[Hint: 1

2+4+6+……….n terms = 12(1+2+3+………n terms) = 1

2n(n+1)

2 =

1n(n+1)

⇨ S∞=1 [ 11.2

+ 12.3

+......∞ =1 as given short-cut method on blog)].

**Question:16 Prove that 23rd term of sequence 17, 1615 , 15

25 ,

1435 ,………is the first negative term.

[Hint: a=17, d=-4/5 and let nth term be first negative term ⇨ 17+(n-1)(-4/5) < 0 ⇨ n > 89/4 ⇨ n=23]

**Question:18 Prove that 1+23 +

63² +

10

33 + 14

34 + ………..∞ = 3.

Answer: we can write above series as 1 + 23 [1+

33 +

5

32+

…….],where a=1,d=2 and r= 1/3, then use formula of combined

A.P.& G.P(Arithmetic –geometric series) [a

1−r + dr

(1−r ) ² ] or you

can do by another method S - 13 S = {1+

23 +

63² +

10

33 + 14

34 +

………..∞} – {13 +

2

32 + 6

33 +10

34 + 14

35 + ………..∞} ⇨ 2 S

3 = 1+ (23 - 1

3) + 4

32 {1+ 13+ 1

32+…..∞} =2 ⇨ S=3 a/(1-r)**Question: 19 If there are distinct real numbers a,b,c are in G.P. and a+b+c = bx , show that x ≤ -1 or x ≥ 3.[Hint: take D ≥ 0 , a+ar+ar2 = (ar)x ⇨ r2+(1-x)r+1=0]**Question:20 (i) If first term of H.P. is 1/7 and 2nd term is 1/9, prove that 12th term is 1/29.

[Hint: as 1/a,1/b,1/c are in H.P. ⇨ a,b,c are in A.P. therefore first and second terms are in A.P. will be 7,9 a=7, d= 2 , then find a12] (ii) In an increasing G.P., the sum of first and last term is 66, and product of the second and last but one term is 128. If the sum of the series is 126, find the number of terms in the series.[Hint: a+arn-1 = 66 , (ar)( arn-2) = 128 and Sn = 126 ⇨ r=2 and n = 6] 

Terms

Arithmetic Sequence  -  A sequence in which each term is a constant amount greater or less than the previous term. In this type of sequence, a n+1 = a n + d , where d is a constant. Common Ratio  -  In a geometric sequence, the ratio r between each term and the previous term. Convergent Series  -  A series whose limit as n→∞ is a real number. Divergent Series  -  A series whose limit as n→∞ is either ∞ or - ∞ . Explicit Formula  -  A formula for the n th term of a sequence of the form a n = some function of n .

Finite Sequence  -  A sequence which is defined only for positive integers less than or equal to a certain given integer. Finite Series  -  A series which is defined only for positive integers less than or equal to a certain given integer. Geometric Sequence  -  A sequence in which the ratio between each term and the previous term is a constant ratio. Index of Summation  -  The variable in the subscript of Σ . For a n , i is the index of summation. Infinite Sequence  -  A sequence which is defined for all positive integers. Infinite Series  -  A series which is defined for all positive integers. Recursive Sequence  -  A sequence in which a general term is defined as a function of one or more of the preceding terms. A sequence is typically defined recursively by giving the first term, and the formula for any term a n+1 after the first term. Sequence  -  A function which is defined for the positive integers. Series  -  A sequence in which the terms are summed, not just listed. Summation Notation  -  a n = a 1 + a 2 + a 3 + a 4 + ... + a n . The symbol Σ and its subscript and superscript are the components of summation notation. Term  -  An element in the range of a sequence. A sequence is rarely represented by ordered pairs, but instead by a list of its terms.