Concept: Progression

 

INTRODUCTION

Terms arranged in a definite order are called sequences. Terms in a sequence may be numbers, letters, symbols or words.

In this chapter we will discuss various types of sequences of numbers. The concepts covered here are also useful in solving problems related to numbers.

SEQUENCE

A sequence is a logically ordered list of elements related to each other. We can identify the patterns followed by terms in a sequence and use the patterns to find the other terms of the sequence, sums of the terms in the sequence or to identify properties of the sequence.

Terms of a sequence are generally denoted by T1, T2, T3, …, Tn.

Let us have a look at some basic patterns followed by sequences. This is not an exhaustive list. However, it gives a fair idea of how to analyze sequences.

Sum Or Difference

In this type of sequence, each term is determined by increasing or decreasing the previous term by some quantity. This quantity may be fixed or it may follow some logical rule by itself.

For example, consider the sequence 1, 3, 7, 13, 21, 31,...

You will observe that the differences between the successive terms are multiples of 2. Hence the sequence could be broken up as

T1 = 1 + 2 × 0 = 1
T2 = 1 + 2 × 1 = 3
T3 = 3 + 2 × 2 = 7
T4 = 7 + 2 × 3 = 13
T5 = 13 + 2 × 4 = 21
T6 = 21 + 2 × 5 = 31

Thus the nth term of the sequence can be written as

Tn = Tn-1 + 2(n – 1)

We can see that the nth term depends on the previous term as well as its position in the sequence.

Hence,
T7 = 31 + 2 × 6 = 43

 

Progression

Arithmetic Progression, Geometric Progression and Harmonic Progression are special types of sequences which have a fixed relationship amongst their terms. We will study the relationships and properties of these three types of progressions in detail.

 

ARITHMETIC PROGRESSION

Quantities are said to be in Arithmetic Progression (A.P.) when they increase or decrease by a constant value known as their common difference, denoted by d. In other words, the difference between any two consecutive terms in an A.P. is constant. The first term of an A.P. is generally denoted by a.

  • If every term of an A.P. is greater than the previous term, the A.P. is said to be an increasing A.P.

  • If every term of an A.P. is lesser than the previous term, the A.P. is said to be a decreasing A.P.

Firt term of an A.P. is denoted by 'a'
common difference is denoted by 'd'

For example,
AP1 ⇒ 1, 2, 3, 4,… where a = 1 and d = 1
AP2 ⇒ 3, 7, 11, 15,… where a = 3 and d = 4
AP3 ⇒ 8, 2, -4, -10,… where a = 8 and d = -6


nth term of an AP

The first, second, third,… nth terms of an A.P. are denoted by T1, T2, T3, …, Tn respectively

T1 = a = a + (1 - 1)d
T2 = T1 + d = a + d = a + (2 – 1)d
T3 = T2 + d = a + 2d = a + (3 – 1)d and so on.

Continuing thus, the nth term of an A.P. is,
Tn = a + (n – 1)d


Example: Find the fifteenth term of the A.P. -3, -9, -15, …

Solution:
Tn = a + (n – 1)d
Here, a = -3, d = -9 – (-3) = -6 and n = 15
∴ T15 = -3 + (15 - 1) × (-6)
∴ T15 = -87


Example: The ninth term exceeds the fifth term of an A.P. by 32. The sum of the ninth and fifth terms is 114. Find the eighth term of the A.P.

Solution:
T9 = a + (9 - 1)d = a + 8d
T5 = a + (5 – 1)d = a + 4d
T9 – T5 = 32
⇒ 4d = 32
⇒ d = 8
Also, T9 + T5 = 2a + 12d
⇒ 114 = 2(a + 6d)
⇒ a + 6d = 57
∴ T8 = a + 7d = a + 6d + d = 57 + 8 = 65


Example: The 54th and the 4th terms of an A.P. are -61 and 64 respectively. Find the 23rd term.

Solution:
a + 53d = -61 …(i)
a + 3d = 64 … (ii)
Subtracting, we get,
50d = -125
d = -5/2
a = 64 – 3d = 143/2
Hence, the 23rd term = a + 22d
= 143/2 + 22 × (-5/2) = 33/2


Sum Of n terms Of an A.P.

Let the first term and common difference of an A.P. containing n terms be a and d respectively. Let Tn be the n term of the A.P. Then, the sum of n terms of the A.P. is

Sn = n2(2a + (n - 1)d)

Explanation:
Tn = a + (n – 1)d
Let the sum of n terms of the A.P. be denoted by Sn. Then,
Sn = a + (a + d) + (a + 2d) + … + (Tn – 2d) + (Tn – d) + Tn …(1)
The same equation can be written as
Sn = Tn + (Tn – d) + (Tn – 2d) + … + (a + 2d) + (a + d) + a …(2)
Adding (1) and (2),
∴ 2Sn = (a + Tn) + (a + Tn) + (a + Tn) … n times
∴ 2Sn = n(a + Tn)
∴ Sn = n2 × [a + Tn] …(3)
∴ Sn = n2 × [a + a + (n – 1)d]
∴ Sn = n2 [2a + (n – 1)d] …(4)

Equation (3) gives us the sum of the n terms of an A.P. in terms of its first and last terms.
Equation (4) gives us the general formula for finding the sum of n terms of an A.P.


Example: The sixth and eighth terms of an A.P. are 38 and 52, respectively. Find the sum of the first twelve terms of the A.P.

Solution:
T6 = a + 5d = 38
T8 = a + 7d = 52
Solving the two equations, d = 7 and a = 3
∴ The sum of 12 terms of the A.P. is
S12 = 123 × [2 × 3 + (11)7] = 498


Example: How many terms of the series -12, -9, -6, … must be taken so that the sum may become 78?

Solution:
⇒ n/2 × [2a + (n – 1)d] = 78
⇒ n/2 × [-24 + (n – 1)3] = 78
⇒ n × (n – 9) = 52
⇒ n = 13


Average Of The Terms Of An A.P.

The average or the arithmetic mean of n terms of an A. P.
= Snn = n2 × (a+Tn)n
∴ The average of n terms of an A.P.
= (a+Tn)2 …(1)
∴ The average of n terms of an A.P.
= (a+d+Tn-d)2 …(2)
∴ The average of n terms of an A.P.
= (a+2d+Tn-2d)2 …(3)

Continuing thus, we see that the average of the terms of an A.P. is equal to the average of its first and nth terms, second and (n – 1)th terms, third and (n – 2)th terms and so on. In general, the average of the terms of the A.P. is equal to the average of the kth term from the beginning and the kth term from the end, or it is equal to the average of any two terms of the A.P. that are equidistant from the beginning and the end.
Also, if n is even the average of the terms of the A.P. is equal to the average of its
(n/2)th and (1 + n/2)th terms.
If n is odd, the average of the terms of the A.P. is equal to the
(n + 1)2th terms of the A.P.


Example: The sum of the first nine terms of an A.P. is 387. Find the fifth term.

Solution:
The average of the first 9 terms of the A.P. = (9 + 1)2th term
∴ The 5th term of the A.P. = 387/9 = 43

If any two consecutive terms of an arithmetic progression are known, the series can be completely determined.

Example: The fourteenth and fifteenth terms of an A.P. are 25 and 32 respectively. Find the 30th term, sum of the first 30 terms and the first term of the A.P.

Solution:
T14 = a + 13d = 25
T15 = a + 14d = 32
∴ d = 7 and a = -66
T30 = - 66 + 29 × 7 = 137
The sum of the first 30 terms = 30/2 × (-66 + 137) = 1065


Example: A teacher observes that the marks that the students in her class have scored are all different. She arranges her students in a line in increasing order of their marks such that difference in marks scored by any two students next to each other is 4. The lowest marks that any student has scored are 11. The sum of the marks that all her students have scored is 585. Find the marks scored by the student standing in the middle of the line.

Solution:
The marks of the students standing in a line form an A.P. with a = 11 and d = 4.
Let there be n students.
Sn = n/2 × [2a + (n – 1)d]
∴ 585 = n/2 × [22 + (n – 1)4]
∴ (n – 15)(2n + 39) = 0
∴ n = 15
∴ The 8th student stands in the middle of the line.
His marks = 585/15 = 39


Example: A child has some marbles which he arranges into rows such that every row has n marbles more than the row preceding it, the first row having only 1 marble. There are 23 marbles in the fourth and fifth rows put together. Which of 25, 35 and 45 could be the number of marbles in the last row?

Solution:
The number of marbles in the rth row = 1 + (r – 1)n
By conditions,
1 + 3n + 1 + 4n = 23
∴ n = 3
∴ The number of marbles in the rth row = 1 + 3(r – 1) = 3r – 2
∴ The number of marbles in the last row has to be of the form 3r – 2, where r is an integer.
This is true only for 25.
∴ The number of marbles in the last row could be 25.


Remember:
  • When three terms are in Arithmetic progression, the middle term is the arithmetic mean of the other two.
  • It is always convenient to take three terms in an A.P. as (a – d), a and (a + d). Similarly, four terms in an A.P. could be taken as a – 3d, a - d, a + d and a + 3d; five terms could be taken as a – 2d, a – d, a, a + d, a + 2d. The advantage of representing terms in this way is that the sum of terms is then obtained in only one unknown.
  • 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 A.P.
    • In case the terms are increased or decreased by some quantity, the common difference of the new A.P. remains equal to that of the original A.P.
    • In case the terms are multiplied or divided by a constant c (c ≠ 0), the common difference d accordingly changes to d  c or d/c .

If the terms of a sequence are given in terms of n, then whether or not the sequence is an A.P. will depend on the difference between two consecutive terms. If this difference is independent of n, the sequence is an A.P., otherwise not.

For example,
If the nth term of a sequence is Tn = 3n + 2,
Then, Tn + 1 = 3n + 5
Tn + 1 – Tn = 3
The sequence of numbers given by Tn = 3n + 2 is an A.P.
But, if Tn = 3n2 + 2n – 1, then the difference
Tn + 1 – Tn = 6n + 5, which is not independent of n.
∴ The sequence of numbers given by Tn = 3n2 + 2n – 1 is not an A.P.
    The number of elements in an arithmetic series from n1 to n2, with a step size (or common difference) of m is 1 + (n2 - n1)/m


Example: How many multiples of 13 lie between 1000 and 5000? What is the sum of all these multiples?

Solution:
The lowest multiple of 13 that is greater than 1000 is 1001. The greatest multiple of 13 that is lesser than 5000 is 4992.
∴ There are (4992 - 1001)/13 + 1
= 308 multiples of 13 between 1000 and 5000.
The sum of all these multiples is
S = 308/2 × (1001 + 4992)
∴ S = 922922


If the sum of the first p terms of an A.P. is equal to the sum of the first q terms of the A.P. such that p and q are different, then the sum of (p + q) terms of the A.P. is zero.

Explanation:
p/2 × [2a + (p – 1)d] = q/2 × [2a + (q – 1)d]
∴ 2ap + p(p – 1)d = 2aq + q(q – 1)d
∴ (q – p)[2a + (p + q – 1)d] = 0
⇒ p = q or [2a + (p + q – 1)d] = 0
But p ≠ q
∴ [2a + (p + q – 1)d] = 0. Hence, the sum of (p + q) terms of the A.P. is zero.


Example: The sum of the first 16 terms of an A.P. is equal to the sum of the first 24 terms of the A.P. Find the sum of the first 40 terms of the A.P.

Solution:
∵ S16 = S24
∴ S(16 + 24) = 0
∴ S40 = 0


Example: The sum of 3rd and 15th elements of an arithmetic progression is equal to the sum of 6th, 11th and 13th elements of the same progression. Then which element of the series should necessarily be equal to zero?

(a) 1st
(b) 9th
(c) 12th
(d) None of the above

Solution:
Assume that the first term of the progression is a and the common difference is d.
∵ T3 + T15 = T6 + T11 + T13
∴ (a + 2d) + (a + 14d) = (a + 5d) + (a + 10d) + (a + 12d)
∴ a + 11d = 0
But, T12 = a + 11d
∴ The 12th term of an arithmetic progression is 0.
Hence, option 3.


Example: Tarushi has 45 coins to be arranged in a line in heaps such that every heap has a fixed number of coins more than the preceding heap. There have to be at least three heaps and the first heap has to have only one coin. In how many ways can Tanishka arrange the coins?

Solution:
As every heap has a fixed number of coins more than that in the previous heap, the number of coins in the heaps form an A.P.
The sum of the terms in this A.P. = the total number of coins = 45
∴ If there are n heaps and the last heap has k coins,
n/2 (1+k) = 45
∴ n(1 + k) = 90
Also, k = 1 + (n – 1)d where d is the common difference of the A.P.
90 = 1 × 90 = 2 × 45 = 3 × 30 = 5 × 18 = 6 × 15 = 9 × 10
As the number of heaps ≥ 3, we do not consider the first two cases.
If n(1 + k) = 3 × 30, the first heap has 1 coin, the second has 15 and the third has 29.
If n(1 + k) = 5 × 18, the first heap has 1 coin and the last has 17. The middle three have 5, 9 and 13 coins respectively.
If n(1 + k) = 6 × 15, the number of coins in all heaps cannot be integers.
If n(1 + k) = 9 × 10, the first heap has 1 coin, the second has 2, the third has 3 and so on. The last heap has 9 coins.
∴ There are 3 ways in which she can arrange the coins.


Example: Thirty-one magazines are arranged from left to right in order of increasing prices. The price of each magazine differs by Rs. 2 from that of each adjacent magazine. For the price of the magazine at the extreme right a customer can buy the middle magazine and an adjacent one. Then:

(1) The adjacent magazine referred to is at the left of the middle magazine.
(2) The middle magazine sells for Rs. 36.
(3) The most expensive magazine sells for Rs. 64.
(4) None of these is correct.

Solution:
Let the price of the cheapest magazine i.e. the one at the extreme left be x.
Difference, between the prices of two adjacent magazines is Rs. 2
∴ Price of extreme right or the costliest magazine will be x + 30 × 2 = x + 60
Now, the price of the magazine in middle (the 16th position) = x + 15 × 2
= x + 30
The price of the magazines, adjacent to the one in the middle is x + 28 or x + 32 depending on whether it is on the left or right of the middle magazine respectively.
Suppose, x + 60 = x + 28 + x + 30
∴ x + 60 = 2x + 58
∴ x = 2
And if x + 60 = x + 30 + x + 32
∴ 60 = x + 62
∴ x = -2 (which is not possible)
So the adjacent magazine is the one whose price is x + 28 i.e. one to the left of the middle magazine.
Hence, option 1.

 

GEOMETRIC PROGRESSION

Quantities are said to be in Geometric Progression (G.P.) when they increase or decrease by a constant factor. The constant factor is called the common ratio, denoted by r, and it is found by dividing any term by the preceding term.

If the first term is positive and common ratio is greater than 1 (or if the first term is negative and the common ratio is less than 1 and positive), the G.P. is an increasing G.P.

If the first term is positive and the common ratio is less than 1 and positive (or if the first term is negative and the common ratio is greater than 1), the G.P. is a decreasing G.P.

In other words, if all terms are greater than the preceding terms, the G.P. is an increasing G.P. otherwise it is a decreasing G.P.

If the first term is a, the terms of the progression are a, ar, ar2, ar3


The nth Term Of A G.P.

If T1, T2, T3, …, Tn denote the terms of a G.P., then
T1 = a = ar1 – 1
T2 = ar = ar2 - 1
T = ar2> = ar3 -1

Continuing thus, the nth term of the geometric progression is given by,
Tn = arn – 1


Example: Find the fifth term of the G.P. whose first term is 3 and the common ratio is 1/3.

Solution:
a = 3 and r = (1/3)
The 5th term = ar5 – 1
= 3 × (1/3)(5-1)
= 1/27


Example: The product of the first five terms of a G.P. is 28. Find the third term.

Solution:
a × ar × ar2 × ar3 × ar4 = 28
∴ a5r10 = 28
∴ (ar2)5 = 28
∴ ar2 = 285
∴ The third term = 285


Example: The product of the first three consecutive terms of an increasing G.P. is 216 and their sum is 21. Find the fourth term of this G.P.

Solution:
Let the three terms of the G.P. be a/r, a and ar respectively.
Then,
a/r × a × ar = 216
∴ a3 = 216
∴ The second term = a = 6
Also, a/r + a + ar = 21
∴ 6r2 + 6r + 6 = 21r
∴ 6r2 – 15r + 6 = 0
∴ (2r – 1)(r – 2) = 0
∴ r = 2 or r = 1/2
∵ The G.P. is an increasing G.P.
∴ r = 2
The fourth term of this G.P. is ar2 = 6 × 22 = 24


Example: The first two terms of a geometric progression add up to 12. The sum of the third and the fourth terms is 48. If the terms of the geometric progression are alternately positive and negative, then the first term is

(a) -2
(b) -4
(c) -12
(d) -8

Solution:
Let the first term and the ratio of the Geometric Progression be a and r respectively.
∴ a + ar = 12
∴ a(1 + r) = 12 … (1)
Also, ar2 + ar3 = 48
∴ ar2(1 + r) = 48… (2)
Dividing (2) by (1),
ar2(1+r)a(1+r) = 4812 = 4
∴ r2 = 4
∴ r = ± √4 = ±2
Since the terms of the Geometric Progression are alternately positive and negative, r = - 2
∴ From (1), a(1 – (-2)) = 12
∴ a = -12
Hence, option (c).


Sum of n terms of a G.P.

The sum of n terms of a G.P. is a(rn-1)r-1

Explanation:
Let Sn be the sum of n terms of a G.P.
∴ Sn = a + ar + ar2 + ar3 + ….+ arn – 2 + arn – 1
∴ r × Sn = ar + ar2 + ar2 + … + arn – 1 + arn
∴ rSn – Sn = arn – a
∴ Sn(r – 1) = a(rn> – 1)
∴Sn = a(rn-1)r-1 …(1)


Example: Find the sum of 5 terms of the series 1/5, 1/2, 5/4,…

Solution:
The given series is a G.P. with a = 1/5 and r = 5/2
∴S5 = 1/5 × (5/25-1)5/2-1
∴S5 ≈ 12.88


Example: Find the sum of n terms of the series 9.01 + 99.001 + 999.0001 + …

Solution:
Sn = 9.01 + 99.001 + 999.0001 + … upto n terms
∴ Sn = (9 + 99 + 999 + … upto n terms) + (0.01 + 0.001 + 0.0001 + … upto n terms)
∴ Sn = [(10  1) + (100  1) + (1000  1) + … upto n terms] + (0.01 + 0.001 + 0.0001 + … upto n terms)
Let S1 = (10 - 1) + (100 – 1) + (1000 – 1) + … upto n terms
and S2 = 0.01 + 0.001 + 0.0001 + …upto n terms
∴ S1 = (10 + 100 + 1000 + …) – n
∴S1 = 10 × 10n-110-1 - n
∴S2 = 0.010.1n-10.1-1 = 0.01 × 1-0.1n0.9
∴Sn = 10 × 10n-110-1 + 0.01 × 1-0.1n0.9 - n


Example: A man decides to save 10% of his income every year. But from the second year onwards, he manages to save only 90% of his savings percentage in the previous year. How much money does he save at the end of five years if his income in the first year is Rs. 2,00,000 and his income increases every year by 10%?

Solution:
Savings in the first year = x = 200000 × 0.1 = 20000
Savings in the second year = (200000 ×1.1) × (0.1 × 0.9) = 0.99x
Savings in the third year = (0.99)2x
Similarly, savings in the fourth and fifth years are (0.99)3x and (0.99)4x.
∴ His savings at the end of five years
= x(1 + 0.99 + 0.992 + 0.993 + 0.994)
= 20000(1 – 0.995)/0.01
≈  2000000(1 – 0.95) ≈ 98020
∴ The man saves approximately Rs. 98,020 at the end of five years.


Example: The sum of the first, third, fifth, seventh, …, thirteenth terms of a G.P. is 14.87 and the sum of the second, fourth, sixth, …, fourteenth terms of a G.P. is 15.02. If the fifth term is 2.08, find the common ratio.

Solution:
Let a and r be the first term and common ratio of the G.P.
Let S1 = a + ar2 + ar4 + … + ar12> = a(1 + r2 + r4 + r6 + …+ r12)
And S2 = ar + ar3 + ar5 + … + ar13 = ar(1 + r2 + r4 + …+ r12)
∴ S2/S1 = r
∴ r = 15.02/14.87 ≈1.01
Here, the fact that the fifth term is 2.08 is not used in the solution at all.


Example: Find the sum of 2n terms of the series 1, 1.5, 3, 2.25, 5, 3.375, 7,…

Solution:
The given series can be broken into two different series:
S1 = 1, 3, 5, 7, …
S2 = 1.5, 2.25, 3.375, …
S1 is an A.P. with first term = 1 and common difference = 2
∴ Sum of n terms of S1 = n/2 × [2 + 2(n – 1)] = n2
∴ S2 is a G.P. with first term = 1.5 and common ratio = 1.5
∴ Sum of n terms of S2 = 1.5((1.5)n - 1)/(1.5 - 1)
= (1.5((1.5)n - 1))/0.5
= 3(1.5n - 1)
Sum of series = n2 + 3(1.5n – 1)


Sum of an infinite G.P.

We know the sum of a GP is a(rn-1)r-1

If the GP consists of infinite terms we can still calculate the sum of these infinite terms if -1 < r < 1.

when, -1 < r < 1 and n is infinite, then rn ≈ 0, hence the sum of infinite GP will be:

S = a1 - r


Example: Determine the sum of the infinite series in Geometric Progression:
1/3 - 1/9 + 1/27 - 1/81 + ⋯

(a) ¼
(b) infinite
(c) 2
(d) 4/3

Solution:
The given series is an infinite Geometric Progression with the common ratio 1/3.
∴ The required sum ‘s’ would be given by,
s = a/(1 - r)
where a = first term of the series = 1/3 and r = common ratio = -1/3
∴ s = (1/3)/(1 + 1/3) = 1/4
Hence, option (a).


Example: A ping pong ball is dropped from a 45 metres high multi-storey building. The ball bounces back three fifth of the distance each time before coming to rest. The total distance traversed by the ball is:

(a) 150 m
(b) 180 m
(c) 175 m
(d) None of the above

Solution:
The distance traversed by the ball is
45 + 2 × 3/5 × (45) + 2 × 3/5 × (3/5 × (45)) + 2 × 3/5 × (3/5 × (3/5 × (45))) + ⋯
= 45 + 2 × 45 × [3/5 + (3/5)2 + (3/5)3 + ⋯]
= 45 + 2 × 45 × [(3/5)/(1 - 3/5)]
= 180 m
Hence, option (b).


Geometric Mean

If n terms a1, a2, …, an are in G.P., then the Geometric Mean G of these n terms is given by
G = a1×a2×...×ann

If three terms are in G.P. then the middle term is the Geometric mean of the other two terms. If a, b and c are in G.P. (a, c > 0 or a, c < 0), then b is the geometric mean of a and c, and is given by b = ac or b2 = ac.
If n is even, the geometric mean of the terms of the G.P. is equal to the geometric mean of kth term fron the beginning and kth term from the end.
If n is odd, the geometric mean of the terms of the G.P. is equal to the n+12th term of the G.P.


Remember:
  • If each term of a G.P. is multiplied or divided by the some non-zero quantity, then the resulting sequence is also a G.P. with the common ratio remaining the same.
  • The reciprocals of the terms of a given G.P. also form a G.P., where the common ratio is the reciprocal of that of the earlier G.P.
  • In a finite G.P., the product of two terms equidistant from the first and the last terms is same as the product of the first and the last term.

 

HARMONIC PROGRESSION

Quantities are said to be in Harmonic Progression (H.P.) when their reciprocals are in A.P. In general if a, a + d, a + 2d, a + 3d, … are successive terms of an arithmetic progression, then 1a, 1a + d, 1a + 2d and 1a + 3d, … are in harmonic progression.

The nth term (Tn) of a harmonic progression is given by
Tn = 1a + (n-1)d


Example: The third term of a H.P. is 1/3 and the sixth term is 1/9. Find the 31st term of the H.P.

Solution:
For the corresponding A.P. the third term is 3 and the sixth term is 9.
Hence, a + 2d = 3 and a + 5d = 9
∴ d = 2 and a = 1
Hence the 31st term of this A.P. = a + (31 – 1)d = 1 + 30 × 2 = 59
Hence for the corresponding H.P., the 31st term is 1/59.


Harmonic Mean

The Harmonic Mean of n numbers is the reciprocal of the arithmetic mean of the reciprocals of these n numbers.
The harmonic mean can be derived using the concept of arithmetic mean.
If a, b and c are in H.P., it implies that b is the harmonic mean of a and c.
1a, 1b and 1c are in A.P. Thus we have
1a - 1b = 1b - 1c
2b = 1a + 1c
∴ b = 2aca + c
Hence the harmonic mean for two numbers is given by
b = 2aca + c
In general, the harmonic mean of n numbers a1, a2, a3, a4, …, an is
H = n1a1+1a2+1a3+...+1an


Example: If three positive real numbers a, b and c (c > a) are in Harmonic Progression, then
log (a + c) + log (a – 2b + c) is equal to:

(a) 2 log (c – b)
(2) 2 log (a – c)
(3) 2 log (c – a)
(4) log a + log b + log c

Solution:
As a, b, c are in H.P.,
b = 2ac/(a+c)
∴ log (a + c) + log (a – 2b + c) = log [(a + c)2 – 2b(a + c)]
= log [(a + c)2 – 4ac)]
= log (a – c)2>
= 2 log(c – a) … ( c > a)
Hence, option (c).


SOME IMPORTANT RESULTS
  • There is no general formula for the sum of quantities in H.P. Questions on H.P. are solved by making use of the properties of the corresponding A.P.
  • The concept of harmonic mean is used when the quantities are specified in rates like distance covered per hour or speed (km/hr, m/s) or production per hour (20 units per hour, 56 units per hour), etc. Thus if a and b are two rates, then the average rate is given by the harmonic mean 2ab/(a + b).

A common application of harmonic mean is to find the average speed. If a person travels a particular distance at speed a and then the same distance at the speed b, then his average speed is the harmonic mean of a and b.


Example: Kishan travels a distance of 300 km from Rampur to Lakhangaam at 30 km per hour in a truck and then to Shyamnagar at 50 km per hour in a car. Find his average speed from Rampur to Shyamnagar, if the distance between Lakhangaam and Shyamnagar is 300 km.

Solution:
As the distance between Rampur and Lakhangaam is equal to the distance between Lakhangaam and Shyamnagar, Kishan’s average speed is the harmonic mean of 30 and 50.
∴ Kishan’s average speed = 2 × 30 × 5030 + 50 = 37.5 km per hour.


Note: If a person travels at two different speeds for the same amount of time, the average speed is the arithmetic mean of the two different speeds.


Arithmetic, Geometric and Harmonic Mean Relationship

Relation Between Arithmetic, Geometric And Harmonic Means
Let A, G and H represent the Arithmetic, Geometric and Harmonic means of two positive or two negative quantities a and b. Then,
A = (a + b)/2
G = √ab
H = 2ab/(a + b)

∴ A × H = ab = G2

⇒ A × H = G2

Thus G is the geometric mean of A and H.

Also, we know that A > G for two unequal quantities.
But as G is the geometric mean of A and H, G lies between A and H.

∴ Arithmetic Mean > Geometric Mean > Harmonic Mean

Note: AM = GM = HM when the two quantities are equal.


Example: The arithmetic mean of two numbers exceeds its geometric mean by 18. The geometric mean is 2.125 times the harmonic mean. Find the difference between the two numbers.

Solution:
Let p and q be the two numbers.
A = G + 18
G = 2.125H = 17H/8
But, G2> = A × H
28964H2 = (17H/8 + 18) × H
∴ H = 128/17
∴ A = 34 and G = 16
(p – q)2 = (p + q)2 – 4pq
∴ (p – q)2 = (2A)2 – 4G2
∴ (p – q)2 = 682 – 4 × 162
∴ (p – q)2 = 3600
∴ (p – q) = 60
∴ Difference between the two numbers = 60


Example: If the product of n positive real numbers is unity, then their sum is necessarily

(a) a multiple of n
(b) equal to n + 1/n
(c) never less than n
(d) a positive integer

Solution:
Let a1, a2, a3, …, an be n positive real numbers.
Now, a × a2 × a3 × … × an = 1
We know that, A.M. ≥ G.M.
a1 + a2 + a3 + … + an ≥ n × (a1 × a2 × a3 × … × an)1/n
∴ a1 + a2 + a3 + … + an ≥ n × (1)1/n
∴ a1 + a2 + a3 + … + an ≥ n
Hence, option (c).


Example: p, q, r and s are in G.P. Are logn(pn – qn), logn(qn – rn) and logn(rn – sn) also terms of a progression?

Solution:
p, q, r and s are in G.P. Let the common ratio be k.
∴ q = pk, r = pk2, s = pk3
∴ pn – qn = pn(1 – kn),
qn – rn = pnkn(1 – kn),
rn – sn = pnk2n(1 – kn)
∴ (pn – qn), (qn – rn) and (rn – sn) form a G.P. with common ratio kn.
∴ (qn – rn)2 = (pn – qn)(rn – sn)
∴ 2 logn(qn – rn) = logn(pn – qn) + logn(rn – sn)
∴ log(pn – qn), log(qn – rn) and log(rn – sn) are in A.P.

 

Miscellaneous Series

Arithmetic Geometric Progression

In an AGP, the numerators of each term are in Arithmetic Progression and the denominators are in Geometric Progression.


Example: What is the approximate sum of 50 terms of the series given below?
5/2 + 7/4 + 9/8 + 11/16 + ⋯

Solution:
We see that the numerators of the terms in the given series are in A.P., with a common difference of 2. Thus the numerator of the 50th term is 5 + 49 × 2 = 103.
The denominators are in G.P. with common ratio = 2. The denominator of the 50th term is 250.
S50 = 5/2 + 7/4 + 9/8 + ⋯ + 103/250  …(1)
∴ ½S50 = 5/4 + 7/8 + 9/16 + ⋯ + 101/250 + 103/251  …(2)
Subtracting (2) from (1),
½S50 = 5/2 + (7-5)/4 + (9-7)/8 + ⋯ + (103-101)/250 - 103/251
∴ ½S50 = 5/2 - 103/251 + 2(1/4 + 1/8 + 1/16 + ⋯ + 1/250)
∴ ½S50 = 5/2 - 103/251 + 2((249 - 1)/250)
∴ S50 = 5 - 103/250 + 2((249 - 1)/249 )
(249 - 1)/249 ≈ 1 and 103/250 ≈0
∴ S50 ≈ 5 - 0 + 2 = 7


Example: Let S = 2x + 5x2 + 9x3 + 14x4 + 20x5 .... infinity

(1) (x(2-x))/(1-x)3
(2) ((2-x))/(1-x)3
(3) (x(2-x))/(1-x)2
(4) None of these

Solution:
S = 2x + 5x2 + 9x3 + 14x4 + 20x5 .... infinity …(1)
Multiplying both sides by x,
xS = 2x2 + 5x3 + 9x4    +...           ...(2)
Subtracting (2) from (1),
(1 – x)S = 2x + 3x2 + 4x3...      ...(3)
Again multiplying (3) with x. x(1 – x)S = 2x2 + 3x3 + 4x4...      ...(4)
Subtracting (4) from (3),
(1 – x)2S = 2x + x2 + x3...
(1 – x)2S = x + x/(1 - x) (x < 1)
S = (x(2-x))/(1-x)3
Hence, option (a).


Series of Natural Numbers

Remember:
  • Sum of first n natural numbers is given by Sn = 1 + 2 + 3 + ... + n = n(n + 1)2
  • Sum of squares of first n natural numbers is given by Sn2 = 12 + 22 + 32 + ... + n2 = n(n + 1)(2n + 1)6
  • Sum of cubes of first n natural numbers is given by Sn3 = 13 + 23 + 33 + ... + n3 = n(n+1)22

Example: A student finds the sum 1 + 2 + 3 + ... as his patience runs out. He found the sum as 575. When the teacher declared the result wrong, the student realized that he missed a number. What was the number the student missed?

(a) 16
(2) 18
(3) 14
(4) 20

Solution:
Sum of first n natural numbers = Sn

Sum given by student = 575

S10 = (10 × 11)/2 = 55

S20 = (20 × 21)/2 = 210

S30 = (30 × 31)/2 = 465

S40 = (40 × 41)/2 = 820

∴ The student stopped counting somewhere between 30 and 40.

Consider S35 = (36 × 35)/2 = 630

The student stopped somewhere before 35.

∴ S31 = 496, S32 = 528, S33 = 561 and S34 = 595

But the student gave 575 as the sum, so the student missed the number 20.

Hence, option (d).


Example: Ginny is adding the numbers from 1 to 50 for her homework. She makes a mistake and wrongly adds the reverse of a two-digit number instead of the number. As a result, her sum is 1347. Which is the number that she adds the reverse of?

Solution:
Sum of numbers from 1 to 50 = 50 × 51/2 = 1275
Ginny’s sum = 1347
Let the number that Ginny wrongly adds be 10x + y
∴ Ginny’s sum = 1275 – (10x + y) + (10y + x)
∴ 1347 = 1275 – 9(x – y)
∴ 9(y – x) = 72
∴ y – x = 8
∴ y = 9 and x = 1
∴ Ginny wrongly adds the reverse of 19.


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