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Progressions

Sum of n terms of an Arithmetic Progression

When Carl Friderich Gauss, the great German Mathematician, was in elementary school, his teacher gave a problem of finding the sum of first 100 natural numbers. He immediately replied that the sum is 5050.

He wrote the first 100 natural numbers as given below:
               S =   1 + 2 + . . . . + 99 + 100
Reversing  S =   100 + 99 + . . . . + 2 + 1
Adding these two, he got
            2S =   101 + 101 + . . . . + 101 + 101   (100 times)
              =   101 * 100.
    →S   =   101 * (100/ 2)
          =   101 * 50  
          =   5050
We will now use the same technique to find the sum of the first n terms of an A.P. Let the first term of an A.P be a and common difference d. Let Sn denote the sum of the n terms of A.P then

Sn   =   a + [a + d] + [a + 2d] + . . . . + [a + (n – 2)d] + [a + (n – 1)d]

Rewritting the terms in reverse order, we have
Sn   =   [a + (n – 1)d] + [a + (n – 2)d] + . . . . + [a + 2d] + [a + d] + a

On adding Sn + Sn  =  [2a + (n – 1)d] + [2a + (n – 1)d] + . . . . . . .+ [2a + (n – 1)d] + [2a + (n – 1)d] (n times).

                    2Sn   =   n[2a + (n – 1)d]

                  Sn  =   [2a + (n – 1)d].

            Simply, S  =   [2a + (n – 1)d].

We can also write Sn as [a + a + (n – 1)d]= [a + an] [ a + (n – 1)d = an].

If there are only 'n' terms in an A.P. and an = l then S = [a + l].
The above one is useful when the first and last terms of the A.P are known.

Properties of Arithmetic Progression

1. If a, b, c are in A.P then 2b = a + c.
2. If a, b, c are in A.P and k ≠ 0, then the following are also in A.P
(i). a + k, b + k, c + k
(ii). a – k, b – k, c – k.
(iii). ak, bk, ck
(iv). a/k, b/k, c/k.

Arithmetic Progressions

Sequence: A set of numbers arranged in a definite order according to some definite rule (or rules) is called a 'sequence'. Each number of the set is called a 'term' of the sequence.

Finite and Infinite sequence: A sequence is called finite or infinite according to the number of terms in it is finite or infinite.

General term of sequence: Let us consider the sequence of "cubes" of natural numbers: 1, 8, 27, 64, . . . . The different terms of a sequence are usually denoted by t1, t2, t3, . . . etc. Here, the subscript (always a natural number) denotes the position of the term in the sequence. Thus, in the above sequence t1 = 1, t2 = 8, t3 = 27 . . . . . etc. Hence, First term = t1 = 1, Second term = t2 = 8, Third term = t3 = 27, . . . etc. In general, nth term = tn, which is called "general term" of the sequence.

Often, it is possible to express the rule which generates the various terms of the sequence in terms of algebraic formula. In the above sequence 1, 8, 27, 64, 125 . . ., nth term = tn = n3. Thus, the rule for the above sequence is n3, where n is any natural number.

Series: If a1, a2, . . ., an is a sequence of numbers, then the expression a1 + a2 + . . . + an + . . . is called series associated with the given sequence.

A "series" may be finite or infinite. It is common to represent a series compactly using the Σ (sigma) symbol indicating a summation as:

  • ai = a1 + a2 + a3 + . . . + an for a finite series.
  • ai = a1 + a2 + a3 + . . . + an + . . . for an infinite series.

Note that 'i' takes the values from '1' to the number indicated at the top of the sigma symbol.

Arithmetic progression: An arithmetic progression is a sequence in which each term, except the first term is obtained by adding a fixed number (positive or negative or zero) to the term immediately preceeding it. This fixed number is the difference of two successive terms, for this reason it is called as the common difference usually denoted by d.

Quantities are said to be in A.P when they increase or decrease by a common difference. The common difference is formed by subtracting any term of the sequence from that which follows it.

Thus, If  t1, t2, t3, . . . . . , tn are the terms in an A.P and the common difference is d then
t2 = t1 + d d = t2 – t1    or    t3 = t2 + d d = t3 – t2    or    t4 = t3 + d d = t4 – t3
t5 = t4 + d d = t5 – t4    or    t6 = t5 + d d = t6 – t5    or    t7 = t6 + d d = t7 – t6
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - tn = tn - 1 + d d = tn – tn - 1
In above, If the first term  t1 = a, then t2   =   t1 + d   =   a + d   =   a + (2 – 1)d
t3   =   t2 + d   =   [a + d] + d   =   a + 2d   =   a + (3 – 1)d
t4   =   t3 + d   =   [a + 2d] + d  =   a + 3d   =   a + (4 – 1)d
t5   =   t4 + d   =   [a + 3d] + d   =   a + 4d   =   a + (5 – 1)d
- - - - - - - - - - tn   =   tn – 1 + d   =   a + (n – 1)d,   which is called "general term of A.P".
By substituting n = 1, 2, 3, . . . we get   a, a + d, a + 2d, a + 3d, a + 4d, . . . . . . . . represents an arithmetic progression where "a" is the first term and "d" is the common difference. This is called the "general form of an A.P" and the common difference is given by "tn – tn – 1".

Arithmetic mean: When three quantities are in A.P. the middle one is said to be the arithmetic mean of the other two. Thus, if a, b, c are three quantities in A.P then b is called the arithmetic mean of a, c and it is given by b =. Hence, arithmetic mean of two numbers is half of their sum.

Sum of n terms of Geometric Progressions

Consider the following Geometric Progression with n terms:
a, ar, ar2, ar3, ar4 , . . . . . . . . . . . . . , ar n – 1

Let Sn   =   a, ar, ar2, ar3, ar4 , . . . . . . . . . ., ar n – 1         - - - - - (i)

Now, multiply the equation (i) with r then,
r * Sn   =   ar, ar2, ar3, ar4, ar5 , . . . . . . . . . ., ar n         - - - - - (ii)

Now, subtract (ii) from (i) , then,
    Sn – r * Sn   =   a – arn
Sn[1 – r]   =   a[1 – rn]
        Sn   =   .
Sum of n terms in a G.P   =   .



Find the sum of first 10 terms of G.P. 3, 6, 12, 24, . . . . . ?

Sol : Given that G.P. is 3, 6, 12, 24, . . . . .
We have to find the sum of 10 terms i.e., n = 10
We know that the common ratio is given by tr = rtr – 1
r = tr / tr – 1
r = t2 / t1 [By substituting r = 2]
Here, t1 = 3 , t2 = 6

Common ratio(r) = 6/3 = 2
we know that sum of n terms in g.p is
sn = a[1 – r]n] / 1 – r

Sum of 10 terms = S10
                           = (3)[1 – (2)10] / [1 – 2]
                           = (3)[1 – 1024] / – 1 = 3069.

How many terms of a G.P. 4, 8, 16, 32, 64, . . . . are needed to give the sum 8188 ?

Sol : Given G.P. is 4, 8, 16, 32, 64 . . . . . . . . .
We have to find number of terms i.e., n = ?
We know that the common ratio is given by tr = rtr – 1
r = tr / tr – 1
r = t2 / t1 [By substituting r = 2]
Here, t1 = 4 , t2 = 8

Common ratio(r) = 8/4 = 2
we know that sum of n terms in g.p is
sn = a[1 – r]n] / 1 – r

8188 = (4)[1 – (2)n] / [1 – 2]
8188 = (4)[2n - 1] / [2 – 1]
8188 = (4)[2n] - 4
8192 = 4(2)n
2048 = 2n
211 = 2n
n = 11
Thus, the sum of 11 terms in the G.P.make 8188.

Geometric Progressions

Sequences, in which each term, except the first, is obtained by multiplying the term immediately preceding it by a fixed non zero number are called Geometric Progressions, briefly written as G.P.

Thus, G.P is a series in which the ratio of each term (except the first term) to the preceding term is a constant. This constant is called the "common ratio (r)".

If the first term is 'a' and the common ratio is 'r'(≠ 0), then the Geometric Progression is a, ar, ar2, ar3, . . . . . which is called the 'general form of G.P', and the common ratio is given by tr = r tr – 1, which is constant.

General term of a G.P: If a, ar, ar2, ar3, ar4, . . . . represents a G.P then
t1 = a = ar1 – 1 t2 = ar = ar2 – 1 t3 = ar2 = ar3 – 1 t4 = ar3 = ar4 – 1
t5 = ar4 = ar5 – 1 t6 = ar5 = ar6 – 1 t7 = ar6 = ar7 – 1 t8 = ar7 = ar8 – 1
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - tn = arn – 1
Hence, the general term in G.P. is arn – 1
  • Example: 2, 4, 8, 16, . . . . , 2n. Write a G.P. whose first term is 6 and common ratio is 2 ?
    Sol : Given that, first term = 6, common ratio = 2
    We know that if a is the first term and r is the common ratio, then the geometric progression is a, ar, ar2, ar3, ar4 , . . . .
    Here, a = 6, r = 2
    So, the required G.P. = 6, (6)(2), (6)(2)2, (6)(2)3, (6)(2)4 , . . . . . .
                               = 6, 12, 6(4), 6(8), 6(16), . . . .
                               = 6, 12, 24, 48, 96, . . . .
  • Example: Write a G.P whose first term is 5 and common ratio is – 5 ?
    Sol : Given that, first term = 5, common ratio = – 5
    We know that if a is the first term and r is the common ratio, then the geometric progression is a, ar, ar2, ar3, ar4 , . . . .
    Here, a = 5, r = – 5
    So, the required G.P = 5, (5)(– 5), (5)(– 5)2, (5)(– 5)3, (5)(– 5)4 , . . . .
                              = 5, – 25, 5(25), 5(– 125), 5(625), . . . .
                              = 5, – 25, 125, – 625, 3125, . . . .
  • Example: A particle moves 125 metres in the first minute, 250 metres in the second minute, 500 metres in the third minute and so on. How much time it takes to travel 63.875 kilometres ?

    Sol : Given that, particle is moving 125 metres in first minute, 250 metres in second minute, 500 metres in third minute, . . . .
    That is, 125, 250, 500, . . . .
    Clearly it is a G.P with first term = 125, common ratio = 250/125 = 2.
    t1 = a = 125, t2 = 250, t3 = 500, . . . .
    Now, we have to find how much time it takes to travel 63.875 km.
    Sn = 63.875 km
             = 63.875 × 1000 m [ 1 kilometre = 1000 metres]
             = 63875 metres.
    we know that sum of n terms in g.p is
    sn = a[1 – r]n] / 1 – r

    63875 = (125)[1 – (2)n] / [1 – 2]
    63875 = (125)[2n - 1] / [2 – 1]
    [2n - 1] = 63875 / 125
    [2n - 1] = 511
    2n = 512
    2n = 29
    n = 9
    Thus, the particle takes 9 minutes to travel 63.875 km.

Geometric mean: When three quantities are in G.P. the middle one is said to be the geometric mean of the other two. Thus, if a, b, c are in G.P. then b is the geometric mean of a, c and it is given by b = √(ac). We can also write this as b2 = ac.

Harmonic Progressions

A sequence is said to be in Harmonic Progression, when their reciprocals are in arithmetic progression. Thus, corresponding to every harmonic progression there is an arithmetic progression. The quantites a, b, c, d . . . . are in H.P if , , , , . . . . . are in A.P.

We know that general form of an A.P is a, a + d, a + 2d, a + 3d, . . . .

Then the "general form of H.P" is , , , , . . . . . . .

where t1 = , t2 = , t3 = , t4 = , . . . .

Hence, the General term of H.P (tn) = .

  • In a H.P. 3rd term is 1/7 and 13th term is 1/27. Write H.P ?

    Sol : Given, 3rd term = 1/7, 13th term = 1/27.
    t3 = 1/7, t13 = 1/27
    1 / a + (3 – 1)d = 1/7, 1 / a + (13 – 1)d = 1/27
    1/ (a + 2d) = 1/7, 1/ (a + 12d) = 1/27
    Thus, a + 2d = 7 - - - - - (i) , a + 12d = 27 - - - - - (ii)
    By solving equations (i) and (ii) we get a, d values.
    From equation (i) we can write that a = 7 – 2d
    Substitute this value in equation (ii)
    7 – 2d + 12d = 27
    10d = 20
    d = 2
    Substitute this d value in equation (ii)
    a + 12(2) = 27
    a + 24 = 27
    a = 3
    We know that if a is the first term and d is the common difference, then H.P is 1 / a , 1 / a + d , 1 / a + 2d , 1 / a + 3d , . . . . .
    So, the required H.P = 1 / 3 , 1 / 3 + 2 , 1 / 3 + 2(2) , 1 / 3 + 3(2) , . . . . .
    = 1/3, 1/5, 1/7, 1/9, . . . .
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