# Progression and Series

In this topic we will study Sequence which is a succession of numbers formed according to some definite rule.

Example : 1, 3, 5, 7,9 â€¦â€¦. is a sequence, here each term of the sequence can be obtained by adding 2 to the preceding term. (All the odd integers)

There are two types of sequence .i.e Finite sequence and Infinite sequence. A sequence is said to be a finite or infinite depending upon number of terms. For finite n we call it finite sequence and for infinite n we call it infinite sequence. If X = R then we call the sequence a real sequence and if X = C, we call it a complex sequence.

If {fn} be a sequence then an expression of the form f1 + f2 + â€¦â€¦ + fn is called Series. In other word a series is the sum of the terms of the sequence.

# Sequence and pattern

Not necessary, if the terms of a sequence follow certain pattern or rule then it is called progression. Here we are discussing following progression:

(i) Arithmetic Progression (A.P.)

(ii) Geometric Progression (G.P.)

(iii) Harmonic Progression (H.P.)

**Arithmetic Progression (A.P.) :- **A sequence is said to be in Arithmetic Progression when they increase or decrease by a constant number. This constant number is called the common difference (c.d.) of the arithmetic progression.

1, 3, 5, 7, â€¦â€¦â€¦...... c.d. = 2

â€“7, â€“3, 1, 5, 9 â€¦â€¦â€¦ c.d. = 4

8, 5, 2, â€“1, â€“4 â€¦â€¦â€¦ c.d. = â€“3

Since we are adding â€˜dâ€™ (common difference) each time (negative value of d accounts here for subtraction) to get next number in the sequence. So by close inspection we can easily say n^{th} term of an A.P. will be given by t_{n} = a + (nâ€“1)d.

S**um of the first n term of an A.P.**

Our next interest is to find the sum of first â€˜nâ€™ terms of an A.P. Let us denote it by S_{n}

S_{n} = {a} + {a + d} + {a + 2d} +â€¦+ {a + (nâ€“1)d} â€¦ (1)

we can write the above series in reverse way also.

S_{n} = {a+(nâ€“1)d} + {a+(nâ€“2)d} + {a+(nâ€“3)d} +â€¦â€¦+ {a} â€¦ (2)

Adding (1) and (2) 2S_{n} = {2a(nâ€“1)d} + {2a(nâ€“1)d} + (2a+(nâ€“1)d}+â€¦+ {2a+(nâ€“1)d}

or 2 S_{n} = n {2a+(nâ€“1)d}

S_{n} = n Ã— {2a + (nâ€“1)d} / 2 â€¦ (3)

Once can also remember the above formula, in this way

S_{n} = first term+last term/2 Ã— number of terms = ((a + (a + (n-1)d))/2) x n

Suppose, now you have to find out the sum from j-th term to k-th term of an A.P.

S_{j.k} = j^{th} term+k^{th} term/2 Ã— (k â€“ j + 1)

(We are also including here j-th term so number of terms = k â€“ j + 1)

or S_{j.k} = a+(jâ€“1)d+a+(kâ€“1)d/2 (k â€“ j + 1)

or S_{j.k} = (kâ€“j+1)/2 {2a + (j â€“ k + 2)d} â€¦ (4)

Let sum of n terms of a series be n (2nâ€“1). Find its mth terms.

Let S_{m} and S_{mâ€“1} denote the sum of first m and (m â€“ 1) terms respectively.

S_{m} = T_{1} + T_{2} + T_{3} + â€¦â€¦. + T_{mâ€“1} + T_{m}

S_{m} = T_{1} + T_{2} + T_{3} + â€¦â€¦. + T_{mâ€“1}

Subtracting

S_{m} â€“ S_{mâ€“1} = T_{m}

â‡’ T_{m} = (m(2mâ€“1))â€“(mâ€“1)(2(mâ€“1)â€“1))

= (2m^{2} â€“ m)â€“(2m^{2} â€“ 5m + 3)

= 4m â€“ 3

The sums of n terms of two A.P.â€™s are in the ratio 3n + 2: 2n+3. Find the ratio of their 10^{th} terms.

Let a, a + d, a + 2d, a + 3d, â€¦â€¦â€¦â€¦â€¦

A, A + D, A + 2D, A + 3D, â€¦â€¦â€¦â€¦â€¦â€¦

be two A.P.â€™s n/2[2a+(nâ€“1)d]/n/2[2A+(nâ€“1)d] = 3n+2/2n+3 (given)

â‡’ a+nâ€“1/2d/A+nâ€“1/2D = 3n+2/2n+3

â‡’ To get the ratio of 10^{th} terms put nâ€“1/2 = 9 or n = 19

â‡’ a+9d/A+9D = 3(19)+2/2(19)+3 = 59/41

* *

# Arithmetic mean between two numbers

When three numbers are in A.P., the middle term is called arithmetic mean (AM) of the two others.

Let a and b are two numbers.

And A be the arithmetic mean between two numbers.

So a A b are in A.P.

or, A â€“ a = b â€“ A

or, 2A = a + b

A =a+b/2

So, in general

A.M. = Sum of the numbers/2 â€¦â€¦ (5)

Similarly we can find the two arithmetic means between two number. A_{1} and A_{2} are two arithmetic mean between two member a and â€˜bâ€™ then a, A_{1}, A_{2} b are in A.P.

Now using the formula, t_{n} = a+(nâ€“1)d

or, d = bâ€“a/3 [ here n = 4]

so, A_{1} = a + d = a +bâ€“a/3 = 2a+b/3

and A_{2} = a + 2d = a + 2(bâ€“a)/3 = a+2b/3

**Basic Principle of AP :**

(1) Addition/Subtraction of constant number to each term of an A.P. also results an A.P.

e.g. suppose a_{1}, a_{2}, a_{3}, â€¦, a_{n} are in A.P.

then a_{1} + k, a_{2}, k, a_{3} + k â€¦â€¦, a_{n} + k and

a_{1} â€“k, a_{2} â€“ k, a_{3} â€“ k, â€¦â€¦, a_{nâ€“k} will also be in A.P.

where k R

(2) Multiplication/Division by a constant number to each term of an A.P. also results an A.P.

e.g. suppose a_{1}, a_{2}, a_{3}, â€¦, a_{n} are in A.P.

then ka_{1}, ka_{2}, ka_{3} â€¦â€¦, ka_{n} and

a_{1}/k, a_{2}/k, a_{3}/k will also be in A.P.

where k ? R and k â‰ 0

(3) Addition/Subtraction of two A.P.â€™s also results an A.P.

e.g. suppose a_{1}, a_{2}, a_{3}, â€¦, a_{n} and

b_{1}, b_{2}, b_{3}, â€¦, b_{n}, are in A.P.

then a_{1}+b_{1}, a_{2}+b_{2}, a_{3}+b_{3}, â€¦â€¦, a_{n}+b_{n}. (c.d.) = (a_{2-} a_{1}) + (b_{2-} b_{1})

a_{1}â€“b_{1}, a_{2}â€“b_{2}, a_{3}â€“b_{3}, â€¦â€¦, a_{n}â€“b_{n} will also be in A.P. and the common difference will be (a_{2-} a_{1}) â€“ (b_{2-} b_{1})

(4) Reversing the order of an A.P. also results an A.P.

e.g. Suppose a_{1}, a_{2}, a_{3}, â€¦â€¦, a_{n} are in A.P.

then a_{n}, a_{nâ€“1}, â€¦â€¦, a_{3}, a_{2}, a_{1} will also be in A.P.

**Note:**

For simplicity assume numbers in A.P. in the following format. If three numbers are in A.P assume them as: Î±â€“ÃŸ, Î±, Î±+ÃŸ (here first term is Î±â€“ÃŸ and c.d. is ÃŸ). For four numbers in A.P.: Î±â€“3ÃŸ, Î±â€“ÃŸ, Î± + ÃŸ, Î±+3ÃŸ (here first term is Î±-3ÃŸ and c.d. is 2ÃŸ). Five convenient numbers in A.P. are Î±â€“2 ÃŸ, Î±â€“ÃŸ, Î±â€“ÃŸ, Î±, Î± + ÃŸ, Î± + 2 ÃŸ.(c.d. = ÃŸ)

Basically sum of the above sets of variables eliminate one variable which make easy to solve the problems.

** **

1. If nth term of an series is t_{n} = A n + B, then

t_{n} - t_{nâ€“1} = A n + B â€“ A(nâ€“1)â€“B

= A = constant

i.e. the series is in A.P.

2. If t_{n} = A n^{2} + B n + c, then

t_{n} â€“ t_{nâ€“1} = A(n^{2}) + B_{n} + C â€“ (A(nâ€“1)^{2} + B(nâ€“1) + c)

= A (2n + 1) + B = âˆª(n) (say) U_{n} â€“ U_{nâ€“1} = 2A

which is constant, i.e. if the difference of difference of the terms of a series is constant, then the nth term is quadratic in n.

3. Similarly, if the 3rd degree difference between the terms of a series is constant then the nth term of the series is a cubic in n i.e.

t_{n} = An^{3} + Bn^{2} + Cn + D

and so on.

â€¢ If a fixed number is added to (or subtracted from) each term of a given A.P., then the resulting sequence is also an A.P. with the same common difference as that of the given A.P.

â€¢ If each term of an A.P. is multiplied by a fixed number (say k) (or divided by a non-zero fixed number), the resulting sequence is also an A.P. with the common difference multiplied by k.

â€¢ If a_{1}, a_{2}, a_{3}, â€¦â€¦ and b_{1}, b_{2}, b_{3}, â€¦â€¦ are two A.P.â€™s with common differences d and dâ€™ respectively then a_{1} + b_{1}, a_{2} + b_{2}, + a_{3} + b_{3}, â€¦ is also an A.P. with common difference d + dâ€™.

â€¢ If we have to take three terms in an A.P., it is convenient to take them as a â€“ d, a, a + d. In general, we take a â€“ rd, a â€“ (r â€“ 1)d, â€¦., a â€“ d, a, a + rd in case we have to take (2r + 1) terms in an A.P.

â€¢ If we have to take four terms, we take a â€“ 3d, a â€“ d, a + d, a + 3d. In general, we take a â€“(2râ€“1)d, a â€“(2râ€“3)d,â€¦, aâ€“d, a+d, â€¦, a + (2râ€“1)d, in case we have to take 2r terms in an A.P.

â€¢ If a_{1}, a_{2}, a_{3}, â€¦â€¦, a_{n} are in A.P., then a_{1} + a_{n} = a_{2} + a_{nâ€“1} = a_{3} + a_{nâ€“2} = â€¦â€¦ and so on.

# Arithmetic Mean(s)

â€¢ If three terms are in A.P., then the middle term is called the arithmetic mean (A.M.) between the other two i.e. if a, b, c are in A.P. then b = a+c/2 is the A.M. of a and c.

â€¢ If a_{1}, a_{2}, â€¦â€¦, a_{n} are n numbers, then the arithmetic mean (A) of these numbers is A = 1/n (a_{1} + a_{2} + a_{3} +â€¦+ a_{n}).

â€¢ The n numbers A_{1}, A_{2}, â€¦â€¦, A_{n} are said to be A.M.â€™s between the numbers a and b if a, A_{1}, A_{2}, â€¦â€¦, A_{n}, b are in A.P. If d is the common difference of this A.P., then b = a + (n + 2 â€“ 1)d â‡’ d = bâ€“a/n+1.

â‡’ A_{1} = a + bâ€“a/n+1 = na+b A_{2} = a + 2(bâ€“a)/n+1 ,â€¦, A_{n} = a + n(bâ€“a)/n+1 = a+nb/n+1.

If 1^{st} and 2^{nd} terms of an A.P. are 1 and â€“3 respectively, find the nth term and the sum of the first n terms.

1^{st} term = a, 2^{nd} term = a + d where a = 1, a + d = â€“3,

â‡’ d = â€“4(common difference of A.P.)

â‡’ a_{n} = a + (n â€“ 1)d = 1 + (nâ€“1)(â€“4) = 5 â€“ 4n.

and S_{n} = n/2{a + a_{n}} = n/2{1 + 5 â€“ 4n} = n(3 â€“ 2n).

If 6 arithmetic means are inserted between 1 and 9/2, find the 4^{th} arithmetic mean.

Let a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, a_{6} be six arithmetic means

â‡’ 1, a_{1}, a_{2}, â€¦â€¦, a_{6}, 9/2 will be in A.P.

Now, 9/2 = 1 + 7d â‡’ 7/2 = 7d â‡’ d = 1/2.

Hence a_{4} = 1 + 4 (1/2) = 3.