AP ( Arithmatic Progression)

SEQUENCE AND PROGRESSION

Syllabus in IIT JEE : Arithmetic, geometric and harmonic progressions, arithmetic, geometric  and harmonic means, sum of finite arithmetic and geometric progressions, infinite geometric series, sum of square and cubes of the first n natural numbers.


1.    Introduction :
    A sequence is a set of terms which may be algebraic, real or complex numbers, written according to  definite rule and the series thus formed is called a progression.
    e.g.    0, 1, 7, 26.............. (rule is n^3 – 1)
             1, 4, 7, 10 .............
             2, 4, 6, 8, ............... etc.

Short story :
    Leading to historical development and origin of the chapter - Fredric Karl Gauss.

Common Sequence :
1. AP
2. GP
3. HP
4. AGP
5. Miscellaneous

3(a)    Arithmetic Progression :
    It is a sequence whose terms increase or decrease by a fixed number. Fixed number is called the common difference. If 'a' is the first term and 'd' is the common difference, then the standard appearance of an A.P. is
        a + (a + d) + (a + 2d) + ......... + (a + {n-1}d )
    and   nth or last term is given by
        Tn = a + (n – 1)d
NOTE: If   d > 0,  then increasing AP
            If   d < 0, then decreasing AP
            If   d = 0, then all terms remain constant

Examples :
(i)    If 6th and 11th term of an A.P. are respectively 17 and 32. Find the 20th term.   [Ans. 59]
(ii)    In an A.P. if  tp = q   and   tq = p  then find the rth term.        [Ans.  p + q – r]
(iii)    In an A.P. if  a2 + a5 – a3 = 10   and  a2 + a9 = 17  then find the 1st term and the common  difference.                        [a1= 13 and d = – 1]
(iv)    If pth, qth and rth term of an A.P. are respectively a, b, and c then prove that
        a(q – r) + b(r – p) + (p – q) = 0

Sum of n terms of AP  = n[ a + (n-1) d ] / 2

Remember that :    (i)    sum of first n natural number is  n ( n + 1) / 2
                (ii)    sum of first n odd natural number is  n^2
                (iii)    sum of first n even natural number is  n(n + 1)

3(c)    Highlights about an A.P.
(i)    If each term of an A.P. is increased, decreased, multiplied or divided by the same non zero  number, then the resulting sequence is also an AP.

(ii)    Three numbers in AP can be taken as a – d ,  a ,  a + d ; four numbers in AP can be taken  as a – 3d,  a – d, a + d,  a + 3d ;  five numbers in AP are  a – 2d, a – d ,  a, a + d, a + 2d  &  six terms in  AP are  a – 5d, a – 3d, a – d, a + d, a + 3d, a + 5d  etc.

(iii)    The common difference can be zero, positive or negative.

(iv)    The sum of the two terms of an AP equidistant from the beginning  &  end  is constant and equal to the sum of first  &  last terms.

(v)    If the number of terms in an A.P. is even then take it as  2n  and if odd then take it as (2n+1)

(vi)    For any series, Tn = S(n) – S(n – 1). In a series if Sn is a quadratic function of n or Tn is a linear function of n, then the series is an A.P.

(vii)    If  a, b, c are in A.P.  Þ  2b = a + c.

Note :
(1)    If a, b, c are in A.P. then prove that
    (a)    b + c ; c + a ; a + b are also in A.P.
    [Hint:    b + c = a + d + a + 2d etc, c + a = a + 3d + a = 3a + 3d etc.]

    (b)    (b + c)^2 – a^2 ; (c + a)^2 – b^2  ; (a + b)^2 – c^2 are also in A.P.
    [Hint: b + c – a, c + a – b, a + b –c are in A.P.
        b + c – a = a + d + a + 2d – a = a + 3d etc.  ]

3(d)    Arithmetic mean :
    Definition : When three quantities are in A.P. then the middle one is called the Arithmetic Mean of the other two.
    e.g.  a, b, c are in A.P. then 'b' is the arithmetic mean between 'a' and 'c' and a + c = 2b.
    It is to be noted that between two given quantities it is always possible to insert any number of terms such that the whole series thus formed shall be in A.P. and the terms thus inserted are called the arithmetic means.

To insert 'n' AM's between a and b.
    Let A1, A2, A3 ........ An are the n means between a and b.
    Hence  a, A1A2, ........ Anb is an A.P. and b is the (n + 2)th terms.
    Hence  b = a + (n + 1)d   =>  d = (b-a)/(n+1)