# Concept: Progression

CONTENTS

**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 T_{1}, T_{2}, T_{3}, …, T_{n}.

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

T_{1} = 1 + 2 × 0 = 1

T_{2} = 1 + 2 × 1 = 3

T_{3} = 3 + 2 × 2 = 7

T_{4} = 7 + 2 × 3 = 13

T_{5} = 13 + 2 × 4 = 21

T_{6} = 21 + 2 × 5 = 31

Thus the n^{th} term of the sequence can be written as

T_{n} = T_{n-1} + 2(n – 1)

We can see that the n^{th} term depends on the previous term as well as its position in the sequence.

Hence,

T_{7} = 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,

AP_{1} ⇒ 1, 2, 3, 4,… where a = 1 and d = 1

AP_{2} ⇒ 3, 7, 11, 15,… where a = 3 and d = 4

AP_{3} ⇒ 8, 2, -4, -10,… where a = 8 and d = -6

^{th}term of an AP

The first, second, third,… n^{th} terms of an A.P. are denoted by T_{1}, T_{2}, T_{3}, …, T_{n} respectively

T_{1} = a = a + (1 - 1)d

T_{2} = T_{1} + d = a + d = a + (2 – 1)d

T_{3} = T_{2} + d = a + 2d = a + (3 – 1)d and so on.

Continuing thus, the n^{th} term of an A.P. is,

T_{n} = a + (n – 1)d

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

**Solution**:

T_{n} = a + (n – 1)d

Here, a = -3, d = -9 – (-3) = -6 and n = 15

∴ T_{15} = -3 + (15 - 1) × (-6)

∴ T_{15} = -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**:

T_{9} = a + (9 - 1)d = a + 8d

T_{5} = a + (5 – 1)d = a + 4d

T_{9} – T_{5} = 32

⇒ 4d = 32

⇒ d = 8

Also, T_{9} + T_{5} = 2a + 12d

⇒ 114 = 2(a + 6d)

⇒ a + 6d = 57

∴ T_{8} = a + 7d = a + 6d + d = 57 + 8 = 65

**Example**: The 54^{th} and the 4^{th} terms of an A.P. are -61 and 64 respectively. Find the 23^{rd} 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 23^{rd} 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 T_{n} be the n^{} term of the A.P. Then, the sum of n terms of the A.P. is

**S _{n} = $\frac{\mathrm{n}}{2}$(2a + (n - 1)d)**

Explanation:

T_{n} = a + (n – 1)d

Let the sum of n terms of the A.P. be denoted by S_{n}. Then,

S_{n} = a + (a + d) + (a + 2d) + … + (T_{n} – 2d) + (T_{n} – d) + T_{n} …(1)

The same equation can be written as

S_{n} = T_{n} + (T_{n} – d) + (T_{n} – 2d) + … + (a + 2d) + (a + d) + a …(2)

Adding (1) and (2),

∴ 2S_{n} = (a + T_{n}) + (a + T_{n}) + (a + T_{n}) … n times

∴ 2S_{n} = n(a + T_{n})

∴ S_{n} = $\frac{\mathrm{n}}{2}$ × [a + T_{n}] …(3)

∴ S_{n} = $\frac{\mathrm{n}}{2}$ × [a + a + (n – 1)d]

∴ S_{n} = $\frac{\mathrm{n}}{2}$ [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**:

T_{6} = a + 5d = 38

T_{8} = a + 7d = 52

Solving the two equations, d = 7 and a = 3

∴ The sum of 12 terms of the A.P. is

S_{12} = $\frac{12}{3}$ × [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

The average or the arithmetic mean of n terms of an A. P.

= $\frac{{S}_{n}}{n}$ = $\frac{n}{2}$ × $\frac{(a+{T}_{n})}{n}$

∴ The average of n terms of an A.P.

= $\frac{(a+{T}_{n})}{2}$ …(1)

∴ The average of n terms of an A.P.

= $\frac{(a+d+{T}_{n}-d)}{2}$ …(2)

∴ The average of n terms of an A.P.

= $\frac{(a+2d+{T}_{n}-2d)}{2}$ …(3)

Continuing thus, we see that the average of the terms of an A.P. is equal to the average of its first
and n^{th} 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 k^{th} term from the beginning and the k^{th} 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

$\frac{\mathrm{(n\; +\; 1)}}{2}$^{th} 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. = $\frac{\mathrm{(9\; +\; 1)}}{2}$^{th} term

∴ The 5^{th} 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**:

T_{14} = a + 13d = 25

T_{15} = a + 14d = 32

∴ d = 7 and a = -66

T_{30} = - 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.

S_{n} = n/2 × [2a + (n – 1)d]

∴ 585 = n/2 × [22 + (n – 1)4]

∴ (n – 15)(2n + 39) = 0

∴ n = 15

∴ The 8_{th} 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 r_{th} 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.

- 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 n^{th} term of a sequence is T_{n} = 3n + 2,

Then, T_{n + 1} = 3n + 5

T_{n + 1} – T_{n} = 3

The sequence of numbers given by Tn = 3n + 2 is an A.P.

But, if T_{n} = 3n^{2} + 2n – 1, then the difference

T_{n + 1} – T_{n} = 6n + 5, which is not independent of n.

∴ The sequence of numbers given by T_{n} = 3n^{2} + 2n – 1 is not an A.P.

The number of elements in an arithmetic series from n_{1} to n_{2}, with a step size (or common difference) of m is 1 + (n_{2} - n_{1})/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**:

∵ S_{16} = S_{24}

∴ S_{(16 + 24)} = 0

∴ S_{40} = 0

**Example**: The sum of 3^{rd} and 15^{th} elements of an arithmetic progression is equal to the
sum of 6^{th}, 11^{th} and 13^{th} 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.

∵ T_{3} + T_{15} = T_{6} + T_{11} + T_{13}

∴ (a + 2d) + (a + 14d) = (a + 5d) + (a + 10d) + (a + 12d)

∴ a + 11d = 0

But, T_{12} = a + 11d

∴ The 12^{th} 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.

The common terms of any two APs will also be in AP, whose common difference will be equal to LCM of common difference of original APs.

The highest terms of this series of common terms will be less than or equal to the lower of the highest terms of the original APs.

**Example**: Find the number of common terms of series 6, 12, 18, ...., 72 and 4, 8, 12, ..., 96.

**Solution**:

AP_{1}: 6, 12, 18, ..., 72 ⇒ Common difference = 6

AP_{2}: 4, 8, 12, ..., 96 ⇒ Common difference = 4

The common terms of these two APs will also be in AP whose common difference will be LCM(6, 4) = 12.

Also, the first common term is 12.

∴ The series of common terms is 12, 24, 36, ...

Now, n^{th} terms of this series can be written as 12 + (n - 1) × 12 = 12n

Also, 12n should be less than or equal to the lower of the highest terms of the original APs i.e., 72.

⇒ 12n ≤ 72

⇒ n ≤ 6

∴ There are 6 terms common to both these APs.

**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

^{th}Term of a G.P.

If T_{1}, T_{2}, T_{3}, …, T_{n} denote the terms of a G.P., then

T_{1} = a = ar^{1 – 1}

T_{2} = ar = ar^{2 - 1}

T_{} = ar^{2}> = ar^{3 -1}

Continuing thus, the n^{th} term of the geometric progression is given by,

T_{n} = ar^{n – 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 5^{th} term = ar^{5 – 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 × ar^{2} × ar^{3} × ar^{4} = 28

∴ a^{5}r^{10} = 28

∴ (ar^{2})^{5} = 28

∴ ar^{2} = $\sqrt[5]{28}$

∴ The third term = $\sqrt[5]{28}$

**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

∴ a^{3} = 216

∴ The second term = a = 6

Also, a/r + a + ar = 21

∴ 6r^{2} + 6r + 6 = 21r

∴ 6r^{2} – 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 ar^{2} = 6 × 2^{2} = 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, ar^{2} + ar^{3} = 48

∴ ar^{2}(1 + r) = 48… (2)

Dividing (2) by (1),

$\frac{a{r}^{2}(1+r)}{a(1+r)}$ = $\frac{48}{12}$ = 4

∴ r^{2} = 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).

The sum of n terms of a G.P. is a$\frac{({r}^{n}-1)}{r-1}$

Explanation:

Let S_{n} be the sum of n terms of a G.P.

∴ S_{n} = a + ar + ar^{2} + ar^{3} + ….+ ar^{n – 2} + ar^{n – 1}

∴ r × S_{n} = ar + ar^{2} + ar^{2} + … + ar^{n – 1} + ar^{n}

∴ rS_{n} – S_{n} = ar^{n} – a

∴ S_{n}(r – 1) = a(r^{n}> – 1)

∴S_{n} = a$\frac{({r}^{n}-1)}{r-1}$ …(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

∴S_{5} = 1/5 × $\frac{({\mathrm{5/2}}^{5}-1)}{\mathrm{5/2}-1}$

∴S_{5} ≈ 12.88

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

**Solution**:

S_{n} = 9.01 + 99.001 + 999.0001 + … upto n terms

∴ S_{n} = (9 + 99 + 999 + … upto n terms) + (0.01 + 0.001 + 0.0001 + … upto n terms)

∴ S_{n} = [(10 1) + (100 1) + (1000 1) + … upto n terms] + (0.01 + 0.001 + 0.0001 + … upto n terms)

Let S_{1} = (10 - 1) + (100 – 1) + (1000 – 1) + … upto n terms

and S_{2} = 0.01 + 0.001 + 0.0001 + …upto n terms

∴ S_{1} = (10 + 100 + 1000 + …) – n

∴S_{1} = 10 × $\frac{{10}^{n}-1}{10-1}$ - n

∴S_{2} = 0.01$\frac{{0.1}^{n}-1}{0.1-1}$ = 0.01 × $\frac{1-0.{1}^{n}}{0.9}$

∴S_{n} = 10 × $\frac{{10}^{n}-1}{10-1}$ + 0.01 × $\frac{1-0.{1}^{n}}{0.9}$ - 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 + 0.99^{2} + 0.99^{3} + 0.99^{4})

= 20000(1 – 0.99^{5})/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 S_{1} = a + ar^{2} + ar^{4} + … + ar^{12}> = a(1 + r^{2} + r^{4} + r^{6} + …+ r^{12})

And S_{2} = ar + ar^{3} + ar^{5} + … + ar^{13} = ar(1 + r^{2} + r^{4} + …+ r^{12})

∴ S_{2}/S_{1} = 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:

S_{1} = 1, 3, 5, 7, …

S_{2} = 1.5, 2.25, 3.375, …

S_{1} is an A.P. with first term = 1 and common difference = 2

∴ Sum of n terms of S_{1} = n/2 × [2 + 2(n – 1)] = n^{2}

∴ S_{2} is a G.P. with first term = 1.5 and common ratio = 1.5

∴ Sum of n terms of S_{2} = 1.5((1.5)^{n} - 1)/(1.5 - 1)

= (1.5((1.5)^{n} - 1))/0.5

= 3(1.5^{n} - 1)

Sum of series = n^{2} + 3(1.5^{n} – 1)

We know the sum of a GP is a$\frac{({r}^{n}-1)}{r-1}$

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 r^{n} ≈ 0, hence the sum of infinite GP will be:

**S _{∞} = $\frac{\mathrm{a}}{\mathrm{1\; -\; r}}$**

**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).

If n terms a_{1}, a_{2}, …, an are in G.P., then the Geometric Mean G of these n terms is given by

G = $\sqrt[n]{{a}_{1}\times {a}_{2}\times ...\times {a}_{n}}$

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 = $\sqrt{ac}$ or b^{2} = ac.

If n is even, the geometric mean of the terms of the G.P. is equal to the geometric mean of k^{th} term fron the beginning and k^{th} term from the end.

If n is odd, the geometric mean of the terms of the G.P. is equal to the ${\left(\frac{n+1}{2}\right)}^{th}$ term of the G.P.

- 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 $\frac{1}{\mathrm{a}}$, $\frac{1}{\mathrm{a\; +\; d}}$, $\frac{1}{\mathrm{a\; +\; 2d}}$ and $\frac{1}{\mathrm{a\; +\; 3d}}$, … are in harmonic progression.

The n^{th} term (T_{n}) of a harmonic progression is given by

T_{n} = $\frac{1}{\mathrm{a\; +\; (n-1)d}}$

**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.

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.

⇒ $\frac{1}{\mathrm{a}}$, $\frac{1}{\mathrm{b}}$ and $\frac{1}{\mathrm{c}}$ are in A.P. Thus we have

⇒ $\frac{1}{\mathrm{a}}$ - $\frac{1}{\mathrm{b}}$ = $\frac{1}{\mathrm{b}}$ - $\frac{1}{\mathrm{c}}$

∴ $\frac{2}{\mathrm{b}}$ = $\frac{1}{\mathrm{a}}$ + $\frac{1}{\mathrm{c}}$

∴ b = $\frac{\mathrm{2ac}}{\mathrm{a\; +\; c}}$

Hence the harmonic mean for two numbers is given byb = $\frac{\mathrm{2ac}}{\mathrm{a\; +\; c}}$

In general, the harmonic mean of n numbers a_{1}, a_{2}, a_{3}, a_{4}, …, a_{n} is

H = $\frac{n}{{\displaystyle \frac{1}{{a}_{1}}}+{\displaystyle \frac{1}{{a}_{2}}}+{\displaystyle \frac{1}{{a}_{3}}}+...+{\displaystyle \frac{1}{{a}_{n}}}}$

**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).

- 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 = $\frac{\mathrm{2\; \times \; 30\; \times \; 50}}{\mathrm{30\; +\; 50}}$ = 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**

Let A, G and H represent the Arithmetic, Geometric and Harmonic means of a_{1}, a_{2}, ..., a_{n}. Then,

A = (a_{1} + a_{2} + ... + a_{2})/n

G = $\sqrt[\mathrm{n}]{{\mathrm{a}}_{1}\times {\mathrm{a}}_{2}\times ...\times {\mathrm{a}}_{\mathrm{n}}}$

H = $\frac{\mathrm{n}}{{\displaystyle \frac{1}{{\mathrm{a}}_{1}}}+{\displaystyle \frac{1}{{\mathrm{a}}_{2}}}+...+{\displaystyle \frac{1}{{\mathrm{a}}_{\mathrm{n}}}}}$

⇒ AM ≥ GM ≥ HM

Equality exists when all number are equal i.e., a_{1} = a_{2} = ... = a_{n}

**Relationship between AM, GM & HM of 2 numbers**

AM = (a + b)/2

GM = √ab

H = 2ab/(a + b)

∴ A × H = ab = G^{2}

⇒ A × H = G^{2}

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, G^{2}> = A × H

∴ $\frac{289}{64}$H^{2} = (17H/8 + 18) × H

∴ H = 128/17

∴ A = 34 and G = 16

(p – q)^{2} = (p + q)^{2} – 4pq

∴ (p – q)^{2} = (2A)^{2} – 4G^{2}

∴ (p – q)^{2} = 68^{2} – 4 × 16^{2}

∴ (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 a_{1}, a_{2}, a_{3}, …, a_{n} be n positive real numbers.

Now, a_{} × a_{2} × a_{3} × … × a_{n} = 1

We know that, A.M. ≥ G.M.

a_{1} + a_{2} + a_{3} + … + a_{n} ≥ n × (a_{1} × a_{2} × a_{3} × … × a_{n})^{1/n}

∴ a_{1} + a_{2} + a_{3} + … + a_{n} ≥ n × (1)^{1/n}

∴ a_{1} + a_{2} + a_{3} + … + a_{n} ≥ n

Hence, option (c).

**Example**: p, q, r and s are in G.P. Are log_{n}(p^{n} – q^{n}), log_{n}(q^{n} – r^{n}) and log_{n}(r^{n} – s^{n}) also terms of a progression?

**Solution**:

p, q, r and s are in G.P. Let the common ratio be k.

∴ q = pk, r = pk^{2}, s = pk^{3}

∴ p^{n} – q^{n} = p^{n}(1 – k^{n}),

q^{n} – r^{n} = p^{n}k^{n}(1 – k^{n}),

r^{n} – s^{n} = p^{n}k^{2n}(1 – k^{n})

∴ (p^{n} – q^{n}), (q^{n} – r^{n}) and (r^{n} – s^{n}) form a G.P. with common ratio k^{n}.

∴ (q^{n} – r^{n})^{2} = (p^{n} – q^{n})(r^{n} – s^{n})

∴ 2 log_{n}(q^{n} – r^{n}) = log_{n}(p^{n} – q^{n}) + log_{n}(r^{n} – s^{n})

∴ log(p^{n} – q^{n}), log(q^{n} – r^{n}) and log(r^{n} – s^{n}) are in A.P.

**Miscellaneous Series**

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 50^{th} term is 5 + 49 × 2 = 103.

The denominators are in G.P. with common ratio = 2. The denominator of the 50^{th} term is 2^{50}.

S_{50} = 5/2 + 7/4 + 9/8 + ⋯ + 103/2^{50} …(1)

∴ ½S_{50} = 5/4 + 7/8 + 9/16 + ⋯ + 101/2^{50} + 103/2^{51} …(2)

Subtracting (2) from (1),

½S_{50} = 5/2 + (7-5)/4 + (9-7)/8 + ⋯ + (103-101)/2^{50} - 103/2^{51}

∴ ½S_{50} = 5/2 - 103/2^{51} + 2(1/4 + 1/8 + 1/16 + ⋯ + 1/2^{50})

∴ ½S_{50} = 5/2 - 103/2^{51} + 2((2^{49} - 1)/2^{50})

∴ S_{50} = 5 - 103/2^{50} + 2((2^{49} - 1)/2^{49} )

(2^{49} - 1)/2^{49} ≈ 1 and 103/2^{50} ≈0

∴ S_{50} ≈ 5 - 0 + 2 = 7

**Example**: Let S = 2x + 5x^{2} + 9x^{3} + 14x^{4} + 20x^{5} .... 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 + 5x^{2} + 9x^{3} + 14x^{4} + 20x^{5} .... infinity …(1)

Multiplying both sides by x,

xS = 2x^{2} + 5x^{3} + 9x^{4} +... ...(2)

Subtracting (2) from (1),

(1 – x)S = 2x + 3x^{2} + 4x^{3}... ...(3)

Again multiplying (3) with x.
x(1 – x)S = 2x^{2} + 3x^{3} + 4x^{4}... ...(4)

Subtracting (4) from (3),

(1 – x)2S = 2x + x^{2} + x^{3}...

(1 – x)2S = x + x/(1 - x) (x < 1)

S = (x(2-x))/(1-x)^{3}

Hence, option (a).

- Sum of first n natural numbers is given by S
_{n}= 1 + 2 + 3 + ... + n = $\frac{\mathrm{n(n\; +\; 1)}}{2}$ - Sum of squares of first n natural numbers is given by S
_{n2}= 1^{2}+ 2^{2}+ 3^{2}+ ... + n^{2}= $\frac{\mathrm{n(n\; +\; 1)(2n\; +\; 1)}}{6}$ - Sum of cubes of first n natural numbers is given by S
_{n3}= 1^{3}+ 2^{3}+ 3^{3}+ ... + n^{3}= ${\left(\frac{n(n+1)}{2}\right)}^{2}$

**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 = S_{n}

Sum given by student = 575

S_{10} = (10 × 11)/2 = 55

S_{20} = (20 × 21)/2 = 210

S_{30} = (30 × 31)/2 = 465

S_{40} = (40 × 41)/2 = 820

∴ The student stopped counting somewhere between 30 and 40.

Consider S_{35} = (36 × 35)/2 = 630

The student stopped somewhere before 35.

∴ S_{31} = 496, S_{32} = 528, S_{33} = 561 and S_{34} = 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.