CBSE Class 11 Applied Mathematics Sequences and Series Notes Set 01

Sequences and Series

Sequence

A succession of numbers arranged in a definite order according to a certain given rule is called a sequence.
The number occurring at the n^th place of a sequence is called its n^th term or the general term, to be denoted by a_n. A sequence is said to be finite or infinite according as the number of terms in it is finite or infinite respectively.
A sequence can be regarded as a function whose domain is the set of natural numbers or some subset of it of the type {1, 2, 3, …, k}. Sometimes we use the functional notation a(n) for a_n.

Series

By adding the terms of a sequence, we get a series. Let a₁, a₂, …, a_n be a given sequence. Then the expression a₁ + a₂ + … + a_n is called the series associated with the given sequence. Series are often represented in compact form, called sigma notation. Thus the series a₁ + a₂ + … + a_n is abbreviated as \( \sum_{k=1}^{n} a_k \).

Example 1. Write the first three terms in each of the following sequences defined by: (i) a_n = 2n + 5 (ii) \( a_n = \frac{n - 3}{4} \)
Answer:
(i) Here a_n = 2n + 5. Substituting n = 1, 2, 3, we get a₁ = 2(1) + 5 = 7, a₂ = 9, a₃ = 11.
Therefore, the required terms are 7, 9 and 11.
(ii) Here \( a_n = \frac{n - 3}{4} \). Thus \( a_1 = \frac{1 - 3}{4} = -\frac{1}{2} \), \( a_2 = -\frac{1}{4} \), a₃ = 0
Hence, the first three terms are \( -\frac{1}{2}, -\frac{1}{4} \) and 0.

Arithmetic Progression (AP)

Progression and Arithmetic Progression

Progression: Sequences following patterns are called progressions.
Arithmetic Progression (AP): It is a sequence in which each term except the first one differs from its preceding term by a constant. This constant difference is called the common difference of the AP.
In an AP we usually denote the first term by 'a', the common difference by 'd' and the n^th term by T_n.

Example 2. Show that the sequence log a, \( \log\left(\frac{a^2}{b}\right), \log\left(\frac{a^3}{b^2}\right), \log\left(\frac{a^4}{b^3}\right), \ldots \) forms an AP. Find its common difference.
Answer:
By symmetry, we find that \( T_n = \log\left(\frac{a^n}{b^{n-1}}\right) \) and \( T_{n-1} = \log\left(\frac{a^{n-1}}{b^{n-2}}\right) \)
\( T_n - T_{n-1} = \log\left(\frac{a^n}{b^{n-1}}\right) - \log\left(\frac{a^{n-1}}{b^{n-2}}\right) = \log\left\{\frac{a^n}{b^{n-1}} \times \frac{b^{n-2}}{a^{n-1}}\right\} = \log\left(\frac{a}{b}\right) \) = constant
Hence the given sequence is an AP with common difference \( \log\left(\frac{a}{b}\right) \).

Example 3. Show that the sequence defined by T_n = 3n² + 2 is not an AP.
Answer:
We have T_n = 3n² + 2 and T_n−1 = 3(n − 1)² + 2 = 3n² − 6n + 5
T_n − T_n−1 = (3n² + 2) − (3n² − 6n + 5) = 6n − 3
This shows that (T_n − T_n−1) is not independent of n and therefore it is not constant.
Hence, the given sequence is not an AP.

General Term of an AP

Theorem: The n^th term of an AP with first term 'a' and common difference 'd' is given by T_n = a + (n − 1)d.
Proof: Let us consider an AP with first term 'a' and common difference 'd'. Then the given AP is a, (a + d), (a + 2d), …
In this AP, we have
First term T₁ = a = (a + 0 × d) = a + (1 − 1)d
Second term T₂ = a + d = a + (2 − 1)d
Third term T₃ = a + 2d = a + (3 − 1)d
…
n^th term T_n = a + (n − 1)d
Hence, T_n = a + (n − 1)d.

Some Properties of an AP

(i) If a constant is added to each term of an AP, then the resulting progression is an AP.
(ii) If a constant is subtracted from each term of an AP, then the resulting progression is an AP.
(iii) If each term of an AP is multiplied by the same non-zero number, then the resulting progression is an AP.
(iv) If each term of an AP is divided by the same non-zero number, then the resulting progression is an AP.

Example 4. Show that the progression log a, log (ab), log (ab²), … is an AP. Find its general term and the 10^th term.
Answer:
We have log (ab) − log a = log a + log b − log a = log b
log (ab²) − log (ab) = log a + 2 log b − log a − log b = log b
∴ (T₂ − T₁) = (T₃ − T₂) = … = log b = constant. So the given progression is an AP.
Let A be the first term and d be the common difference of this AP. Then A = log a and d = log b.
General term T_n = A + (n − 1)d = log a + (n − 1) log b
10^th term, T₁₀ = log a + (10 − 1) log b = log a + 9 log b = log a + log b⁹ = log (ab⁹)

Example 5. Find the 10^th common term between the arithmetic series 3 + 7 + 11 + 15 + … and 1 + 6 + 11 + 16 + …
Answer:
Clearly, the first common term of the two series is 11.
LCM of the common differences of the two series = LCM (4, 5) = 20
Let us now consider an AP in which a = 11 and d = 20. Every term of this AP is a common term of the two given series.
So, the required 10^th common term = 11 + (10 − 1) × 20 = 11 + 180 = 191
Hence, 191 is the 10^th common term of the two given series.

Theorem: If the n^th term of a progression is a linear expression in n, then it is an AP.
Proof: Let the n^th term of a progression be a linear expression in n. Then T_n = an + b … (i), where a and b are constants.
Replacing n by (n − 1) in (i), we get T_n−1 = a(n − 1) + b … (ii)
On subtracting (ii) from (i), we get [T_n − T_n−1] = a, which is constant.
This shows that the difference between any two consecutive terms of the given progression is constant. Hence, the given progression is an AP.

Sum of n Terms of an AP

Let us consider an AP containing n terms with the first term 'a', the common difference 'd' and the last term l. Let the sum of these n terms be S_n. Then
S_n = a + (a + d) + (a + 2d) + … + (l − 2d) + (l − d) + l … (i)
Writing the above series in reverse order, we get
S_n = l + (l − d) + (l − 2d) + … + (a + 2d) + (a + d) + a … (ii)
Adding the corresponding terms of (i) and (ii), we get
2S_n = [(a + l) + (a + l) + … to n terms] = n(a + l)
∴ \( S_n = \frac{n}{2}(a + l) \)
But l = n^th term = a + (n − 1)d
∴ \( S_n = \frac{n}{2}[a + a + (n - 1)d] = \frac{n}{2}[2a + (n - 1)d] \)
Hence \( S_n = \frac{n}{2}[2a + (n - 1)d] \)

Example 6. Find the sum of all two-digit numbers which when divided by 4 yield 1 as remainder.
Answer:
Required sum = 13 + 17 + 21 + 25 + … + 97
This is an AP in which a = 13, d = (17 − 13) = 4 and l = 97.
Let the number of terms be n. Then T_n = 97 ⇒ a + (n − 1)d = 97 ⇒ 13 + (n − 1) × 4 = 97 ⇒ n = 22
∴ Required sum \( = \frac{n}{2}(a + l) = \frac{22}{2} \times (13 + 97) = 11 \times 110 = 1210 \)
Hence the required sum is 1210.

Example 7. The difference between any two consecutive interior angles of a polygon is 5°. If the smallest angle is 120°, find the number of sides of the polygon.
Answer:
Let the number of sides of the polygon be n. Then the sum of its interior angles = (2n − 4) right angles = (n − 2) × 180°
Since the difference between any two consecutive interior angles of the polygon is constant, its angles are in AP. In this AP, we have a = 120 and d = 5.
Now S_n = (n − 2) × 180 ⇒ \( \frac{n}{2}\{2 \times 120 + (n - 1) \times 5\} = (n - 2) \times 180 \)
⇒ \( \frac{235n}{2} + \frac{5n^2}{2} = 180n - 360 \) ⇒ 5n² − 125n + 720 = 0
⇒ n² − 25n + 144 = 0 ⇒ (n − 9)(n − 16) = 0 ⇒ n = 9 or n = 16
But when n = 16, we have last angle = {120 + (16 − 1) × 5}° = 195°, which is not possible (an interior angle of a polygon is less than 180°).
∴ n = 9. Hence the number of sides of the given polygon = 9.

Arithmetic Mean

Arithmetic Mean (AM)

If a, A, b are in AP, then we say that A is the arithmetic mean (AM) between a and b.
Insertion of a single arithmetic mean between a and b: Let a and b be two given numbers and let A be the arithmetic mean between a and b. Then a, A, b are in AP.
∴ A − a = b − A ⇒ 2A = a + b ⇒ \( A = \frac{a + b}{2} \)
Hence, the arithmetic mean between a and b is \( \frac{a + b}{2} \).

Insertion of n Arithmetic Means between a and b

Let a and b be two given numbers and let A₁, A₂, …, A_n be the n arithmetic means between a and b. Then a, A₁, A₂, …, A_n, b are in AP.
In this AP, we have first term = a, last term = b, number of terms = (n + 2). Let the common difference be d. Then
b = T_n+2 ⇒ b = a + (n + 2 − 1)d ⇒ \( d = \frac{b - a}{n + 1} \)
∴ \( A_1 = (a + d) = \left(a + \frac{b - a}{n + 1}\right) \), \( A_2 = (a + 2d) = \left\{a + \frac{2(b - a)}{n + 1}\right\} \), …, \( A_n = (a + nd) = \left\{a + \frac{n(b - a)}{n + 1}\right\} \)
These are the required n arithmetic means between a and b.

Example 8. If \( \frac{a^n + b^n}{a^{n-1} + b^{n-1}} \) is the AM between a and b, then find the value of n.
Answer:
We know that the AM between a and b is \( \frac{a + b}{2} \).
∴ \( \frac{a^n + b^n}{a^{n-1} + b^{n-1}} = \frac{a + b}{2} \)
⇒ 2aⁿ + 2bⁿ = aⁿ + aⁿ⁻¹b + bⁿ⁻¹a + bⁿ
⇒ aⁿ + bⁿ = aⁿ⁻¹b + bⁿ⁻¹a ⇒ aⁿ − aⁿ⁻¹b = bⁿ⁻¹a − bⁿ
⇒ aⁿ⁻¹(a − b) = bⁿ⁻¹(a − b) ⇒ aⁿ⁻¹ = bⁿ⁻¹ (taking a ≠ b)
⇒ \( \left(\frac{a}{b}\right)^{n-1} = 1 = \left(\frac{a}{b}\right)^0 \) ⇒ n − 1 = 0 ⇒ n = 1
Hence, the required value of n is 1.

Example 9. If n arithmetic means are inserted between two numbers, prove that the sum of the means equidistant from the beginning and the end is constant.
Answer:
Let A₁, A₂, …, A_n be n AMs between a and b. Then a, A₁, A₂, …, A_n, b are in AP. Let d be the common difference.
m^th mean from the beginning + m^th mean from the end = (m + 1)^th term from the beginning + (m + 1)^th term from the end
= {a + (m + 1 − 1)d} + {b − (m + 1 − 1)d} = (a + b) = constant
Hence, the sum of the means equidistant from the beginning and the end is constant.

Example 10. If a, b, c are in AP, show that (i) \( \frac{1}{bc}, \frac{1}{ac}, \frac{1}{ab} \) are in AP (ii) \( a\left(\frac{1}{b} + \frac{1}{c}\right), b\left(\frac{1}{c} + \frac{1}{a}\right), c\left(\frac{1}{a} + \frac{1}{b}\right) \) are in AP.
Answer:
(i) Since a, b, c are in AP, the terms obtained by dividing each term of this AP by abc are also in AP. Consequently \( \frac{1}{bc}, \frac{1}{ac}, \frac{1}{ab} \) are in AP.
(ii) a, b, c are in AP ⇒ \( \frac{1}{bc}, \frac{1}{ca}, \frac{1}{ab} \) are in AP (dividing each term by abc)
⇒ \( \frac{ab + bc + ca}{bc}, \frac{ab + bc + ca}{ca}, \frac{ab + bc + ca}{ab} \) are in AP (multiplying each term by ab + bc + ca)
⇒ \( \frac{ab + bc + ca}{bc} - 1, \frac{ab + bc + ca}{ca} - 1, \frac{ab + bc + ca}{ab} - 1 \) are in AP
⇒ \( \frac{ab + ca}{bc}, \frac{ab + bc}{ca}, \frac{bc + ca}{ab} \) are in AP
⇒ \( a\left(\frac{1}{b} + \frac{1}{c}\right), b\left(\frac{1}{c} + \frac{1}{a}\right), c\left(\frac{1}{a} + \frac{1}{b}\right) \) are in AP

Example 11. If \( \frac{b + c - a}{a}, \frac{c + a - b}{b}, \frac{a + b - c}{c} \) are in AP, prove that \( \frac{1}{a}, \frac{1}{b}, \frac{1}{c} \) are in AP.
Answer:
\( \frac{b + c - a}{a}, \frac{c + a - b}{b}, \frac{a + b - c}{c} \) are in AP
⇒ \( \left(\frac{b + c - a}{a} + 2\right), \left(\frac{c + a - b}{b} + 2\right), \left(\frac{a + b - c}{c} + 2\right) \) are in AP
⇒ \( \frac{a + b + c}{a}, \frac{a + b + c}{b}, \frac{a + b + c}{c} \) are in AP
⇒ \( \frac{1}{a}, \frac{1}{b}, \frac{1}{c} \) are in AP (dividing each term by a + b + c)

Example 12. If a, b, c are in AP, prove that (a − c)² = 4(b² − ac).
Answer:
Since a, b, c are in AP, we have \( b = \frac{1}{2}(a + c) \)
RHS = \( 4(b^2 - ac) = 4\left\{\frac{(a + c)^2}{4} - ac\right\} = (a + c)^2 - 4ac = (a - c)^2 \) = LHS
Hence (a − c)² = 4(b² − ac).

Geometric Progression (GP)

Geometric Progression and Geometric Series

A sequence a₁, a₂, …, a_n, … is called a geometric progression if each term is non-zero and \( \frac{a_k}{a_{k-1}} = r \) (constant) for all k > 1.
The constant ratio is called its common ratio. In a GP we usually denote the first term by a, the common ratio by r and the n^th term by T_n.
Geometric series: If a₁, a₂, …, a_n, … is a GP, then the sum a₁ + a₂ + … + a_n + … is called a geometric series.

General Term of a GP

Let us consider a GP in which first term = a and common ratio = r. Then the GP is a, ar, ar², …
Its first term, T₁ = a = ar¹⁻¹
Second term, T₂ = ar = ar²⁻¹
…
n^th term, T_n = arⁿ⁻¹
Hence T_n = arⁿ⁻¹.

Example 13. If a, b, c are three consecutive terms of an AP and x, y, z are three consecutive terms of a GP, then prove that x^b−c · y^c−a · z^a−b = 1.
Answer:
It is given that a, b, c are in AP, so 2b = a + c; and x, y, z are in GP, so \( y = \sqrt{xz} \).
∴ \( x^{b-c} \cdot y^{c-a} \cdot z^{a-b} = x^{b-c} \cdot (\sqrt{xz})^{c-a} \cdot z^{a-b} = x^{b-c} \cdot x^{\frac{c-a}{2}} \cdot z^{\frac{c-a}{2}} \cdot z^{a-b} \)
\( = x^{b - c + \frac{c - a}{2}} \cdot z^{\frac{c - a}{2} + a - b} = x^{\frac{2b - c - a}{2}} \cdot z^{\frac{c + a - 2b}{2}} \)
\( = x^{\frac{a + c - c - a}{2}} \cdot z^{\frac{c + a - c - a}{2}} \) [∵ 2b = a + c]
= x⁰ · z⁰ = 1

Example 14. If a, b, c, d are in GP, prove that (a² + b² + c²)(b² + c² + d²) = (ab + bc + cd)².
Answer:
Let r be the common ratio of the GP a, b, c, d. Then b = ar, c = ar² and d = ar³.
LHS = (a² + b² + c²)(b² + c² + d²) = (a² + a²r² + a²r⁴)(a²r² + a²r⁴ + a²r⁶) = a⁴r²(1 + r² + r⁴)²
RHS = (ab + bc + cd)² = (a²r + a²r³ + a²r⁵)² = a⁴r²(1 + r² + r⁴)²
Hence, (a² + b² + c²)(b² + c² + d²) = (ab + bc + cd)².

Example 15. If a, b, c are in AP; b, c, d are in GP and \( \frac{1}{c}, \frac{1}{d}, \frac{1}{e} \) are in AP, then show that a, c, e are in GP.
Answer:
a, b, c are in AP ⇒ 2b = a + c … (i)
b, c, d are in GP ⇒ c² = bd … (ii)
\( \frac{1}{c}, \frac{1}{d}, \frac{1}{e} \) are in AP ⇒ \( \frac{2}{d} = \frac{1}{c} + \frac{1}{e} = \frac{c + e}{ce} \) ⇒ \( d = \frac{2ce}{c + e} \) … (iii)
Now c² = bd ⇒ \( c^2 = \frac{(a + c)}{2} \cdot \frac{2ce}{(c + e)} \) ⇒ \( c = \frac{(a + c)e}{c + e} \)
⇒ c(c + e) = (a + c)e ⇒ c² + ce = ae + ce ⇒ c² = ae
⇒ a, c, e are in GP. Hence proved.

Example 16. Find all sequences which are simultaneously AP and GP.
Answer:
Let a₁, a₂, …, a_n, … be a sequence which is both an AP as well as a GP. Consider three consecutive terms a_n, a_n+1, a_n+2 of this AP and GP.
Then 2a_n+1 = a_n + a_n+2 … (i)
And, if r is the common ratio of the GP, then a_n = a₁rⁿ⁻¹, a_n+1 = a₁rⁿ and a_n+2 = a₁rⁿ⁺¹
Putting these values in (i), we get 2a₁rⁿ = a₁rⁿ⁻¹ + a₁rⁿ⁺¹ ⇒ r² − 2r + 1 = 0 ⇒ (r − 1)² = 0 ⇒ r = 1
Putting r = 1, we get a_n = a₁, a_n+1 = a₁ and a_n+2 = a₁. Thus, we get a constant sequence a₁, a₁, a₁, …
Hence, a (non-zero) constant sequence is the only sequence which is both an AP and a GP.

n^th Term from the End of a GP

Let a be the first term, r be the common ratio and l be the last term of a given GP. Then
2^nd term from the end \( = \frac{l}{r} = \frac{l}{r^{2-1}} \)
3^rd term from the end \( = \frac{l}{r^2} = \frac{l}{r^{3-1}} \)
…
Hence, the n^th term from the end \( = \frac{l}{r^{n-1}} \).

Convenient Choice of Terms in a GP

For solving problems on GP, it is always convenient to take:
(i) 3 numbers in GP as \( \frac{a}{r} \), a, ar
(ii) 4 numbers in GP as \( \frac{a}{r^3}, \frac{a}{r} \), ar, ar³
(iii) 5 numbers in GP as \( \frac{a}{r^2}, \frac{a}{r} \), a, ar, ar²
(iv) the terms as a, ar, ar², … when their product is not given

Example 17. If the product of three numbers in GP is 216 and the sum of their products in pairs is 156, find the numbers.
Answer:
Let the required numbers be \( \frac{a}{r} \), a, ar.
Then \( \frac{a}{r} \times a \times ar = 216 \) ⇒ a³ = 216 = 6³ ⇒ a = 6
and \( \frac{a}{r} \times a + a \times ar + \frac{a}{r} \times ar = 156 \) ⇒ \( a^2\left(\frac{1}{r} + r + 1\right) = 156 \)
⇒ 6²(1 + r + r²) = 156r ⇒ 36(r² + r + 1) = 156r ⇒ 3(r² + r + 1) = 13r
⇒ 3r² − 10r + 3 = 0 ⇒ (3r − 1)(r − 3) = 0 ⇒ r = 3 or \( r = \frac{1}{3} \)
So, the required numbers are 2, 6, 18 or 18, 6, 2.

Sum of n Terms of a GP

Let us consider a GP with the first term a and the common ratio r. Then
S_n = a + ar + … + arⁿ⁻¹ … (i)
Case 1: If r = 1, we have S_n = a + a + … to n terms = na.
Case 2: If r ≠ 1, we have rS_n = ar + ar² + … + arⁿ⁻¹ + arⁿ … (ii)
On subtracting (ii) from (i), we get (1 − r)S_n = (a − arⁿ) = a(1 − rⁿ)
∴ \( S_n = \frac{a(1 - r^n)}{1 - r} \) or \( S_n = \frac{a(r^n - 1)}{r - 1} \)
We use \( S_n = \frac{a(1 - r^n)}{1 - r} \) when r < 1 and \( S_n = \frac{a(r^n - 1)}{r - 1} \) when r > 1.

Example 18. Find the sum of the sequence 7, 77, 777, 7777, … to n terms.
Answer:
This is not a GP; however, we can relate it to a GP by writing the terms as
S_n = 7 + 77 + 777 + 7777 + … to n terms \( = \frac{7}{9}[9 + 99 + 999 + \ldots \text{ to n terms}] \)
\( = \frac{7}{9}[(10 - 1) + (10^2 - 1) + (10^3 - 1) + \ldots \text{ to n terms}] \)
\( = \frac{7}{9}[(10 + 10^2 + \ldots \text{ n terms}) - (1 + 1 + 1 + \ldots \text{ n terms})] \)
\( = \frac{7}{9}\left[\frac{10(10^n - 1)}{10 - 1} - n\right] = \frac{7}{9}\left[\frac{10(10^n - 1)}{9} - n\right] = \frac{7}{81}\left[10^{n+1} - 9n - 10\right] \)

Example 19. A person has 2 parents, 4 grandparents, 8 great grandparents and so on. Find the number of his ancestors during the ten generations preceding his own.
Answer:
Here a = 2, r = 2 and n = 10.
Using the sum formula \( S_n = \frac{a(r^n - 1)}{r - 1} \), we have S₁₀ = 2(2¹⁰ − 1) = 2046
Hence, the number of ancestors preceding the person is 2046.

Geometric Mean

Geometric Mean (GM)

Let a and b be two given numbers. We say that G is the geometric mean (GM) between a and b if a, G, b are in GP.
If a, G, b are in GP, then \( \frac{G}{a} = \frac{b}{G} \) ⇒ G² = ab ⇒ \( G = \sqrt{ab} \)
Remarks:
(i) The GM between two positive numbers is positive.
(ii) The GM between two negative numbers is negative.
(iii) The GM between two numbers of opposite signs does not exist.

Example 20. Find two positive numbers a and b whose AM and GM are 34 and 16 respectively.
Answer:
We have \( \frac{a + b}{2} = 34 \) and \( \sqrt{ab} = 16 \) ⇒ a + b = 68 and ab = 256
∴ \( (a - b) = \sqrt{(a + b)^2 - 4ab} = \sqrt{(68)^2 - 4 \times 256} = \sqrt{3600} = 60 \)
a + b = 68 and a − b = 60 ⇒ a = 64, b = 4 (or, interchanging, a = 4, b = 64)
Hence the required numbers are 64 and 4.

n Geometric Means between Two Given Numbers

Let a and b be two given numbers. We say that G₁, G₂, …, G_n are n geometric means between a and b if a, G₁, G₂, …, G_n, b are in GP. This GP contains (n + 2) terms. Let the common ratio of this GP be r. Then
T_n+2 = b ⇒ arⁿ⁺¹ = b ⇒ \( r = \left(\frac{b}{a}\right)^{\frac{1}{n+1}} \)
∴ \( G_1 = ar = a\left(\frac{b}{a}\right)^{\frac{1}{n+1}},\ G_2 = ar^2 = a\left(\frac{b}{a}\right)^{\frac{2}{n+1}},\ G_3 = ar^3 = a\left(\frac{b}{a}\right)^{\frac{3}{n+1}}, \ldots, G_n = ar^n = a\left(\frac{b}{a}\right)^{\frac{n}{n+1}} \)

Example 21. Insert three numbers between 1 and 256 so that the resulting sequence is a GP.
Answer:
Let G₁, G₂, G₃ be three numbers between 1 and 256 such that 1, G₁, G₂, G₃, 256 is a GP.
Therefore 256 = 1 × r⁴ ⇒ r = ±4
For r = 4, we have G₁ = ar = 4, G₂ = ar² = 16, G₃ = ar³ = 64. Similarly, for r = −4, the numbers are −4, 16 and −64.
Hence, we can insert 4, 16, 64 (or −4, 16, −64) between 1 and 256 so that the resulting sequence is a GP.

Relationship between AM and GM

Let A and G be the AM and GM of two given positive real numbers a and b respectively. Then \( A = \frac{a + b}{2} \) and \( G = \sqrt{ab} \).
Thus we have \( A - G = \frac{a + b}{2} - \sqrt{ab} = \frac{a + b - 2\sqrt{ab}}{2} = \frac{(\sqrt{a} - \sqrt{b})^2}{2} \geq 0 \)
⇒ A ≥ G. Hence AM ≥ GM (equality holds when a = b).

Example 22. Find the value of n so that \( \frac{a^{n+1} + b^{n+1}}{a^n + b^n} \) may be the geometric mean between a and b, where a ≠ b.
Answer:
\( \frac{a^{n+1} + b^{n+1}}{a^n + b^n} \) is the GM between a and b ⇒ \( \frac{a^{n+1} + b^{n+1}}{a^n + b^n} = \sqrt{ab} = a^{\frac{1}{2}}b^{\frac{1}{2}} \)
⇒ \( a^{n+1} + b^{n+1} = (a^n + b^n)a^{\frac{1}{2}}b^{\frac{1}{2}} = a^{n+\frac{1}{2}}b^{\frac{1}{2}} + a^{\frac{1}{2}}b^{n+\frac{1}{2}} \)
⇒ \( a^{n+1} - a^{n+\frac{1}{2}}b^{\frac{1}{2}} = a^{\frac{1}{2}}b^{n+\frac{1}{2}} - b^{n+1} \)
⇒ \( a^{n+\frac{1}{2}}\left(a^{\frac{1}{2}} - b^{\frac{1}{2}}\right) = b^{n+\frac{1}{2}}\left(a^{\frac{1}{2}} - b^{\frac{1}{2}}\right) \)
⇒ \( a^{n+\frac{1}{2}} = b^{n+\frac{1}{2}} \) [∵ \( a^{\frac{1}{2}} - b^{\frac{1}{2}} \neq 0 \), since a ≠ b]
⇒ \( \left(\frac{a}{b}\right)^{n+\frac{1}{2}} = 1 = \left(\frac{a}{b}\right)^0 \) ⇒ \( n + \frac{1}{2} = 0 \) ⇒ \( n = -\frac{1}{2} \)
Hence \( n = -\frac{1}{2} \).

Example 23. If a, b, c, d are four distinct positive numbers in GP, then prove that a + d > b + c.
Answer:
a, b, c, d are in GP ⇒ a, b, c are in GP
Since AM > GM for distinct positive numbers, \( \frac{a + c}{2} > b \) ⇒ a + c > 2b … (i)
Again a, b, c, d are in GP ⇒ b, c, d are in GP ∴ \( \frac{b + d}{2} > c \) ⇒ b + d > 2c … (ii)
From (i) and (ii), we get a + c + b + d > 2b + 2c
Hence a + d > b + c.

Example 24. If x ∈ R, find the minimum value of 3^x + 3^1−x.
Answer:
We know that AM ≥ GM. ∴ For all x ∈ R, we have \( \frac{3^x + 3^{1-x}}{2} \geq \sqrt{3^x \times 3^{1-x}} \)
⇒ \( \frac{3^x + 3^{1-x}}{2} \geq \sqrt{3} \) ⇒ \( 3^x + 3^{1-x} \geq 2\sqrt{3} \)
Equality holds when 3^x = 3^1−x, i.e. \( x = \frac{1}{2} \).
Hence the minimum value of 3^x + 3^1−x is \( 2\sqrt{3} \).

Infinite Geometric Series

Sum of an Infinite GP

Let us consider an infinite GP with first term a and common ratio r, where −1 < r < 1, i.e. |r| < 1. The sum of n terms of a GP is given by
\( S_n = \frac{a(1 - r^n)}{1 - r} \) ⇒ \( S_n = \frac{a}{1 - r} - \frac{ar^n}{1 - r} \) … (i)
Since |r| < 1, when n increases rⁿ decreases; thus when n → ∞, rⁿ → 0.
∴ \( \lim_{n \to \infty} \frac{ar^n}{1 - r} = 0 \)
Hence, the sum of an infinite GP is given by
\( \lim_{n \to \infty} S_n = \lim_{n \to \infty}\left\{\frac{a}{1 - r} - \frac{ar^n}{1 - r}\right\} = \frac{a}{1 - r} - 0 = \frac{a}{1 - r} \)
This gives \( S = \frac{a}{1 - r} \), when |r| < 1.
If |r| ≥ 1, then the sum of an infinite GP does not exist (it is not a finite number).

Example 25. Use a geometric series to express 0.555… = \( 0.\overline{5} \) as a rational number.
Answer:
We have \( 0.\overline{5} = 0.555\ldots = 0.5 + 0.05 + 0.005 + \ldots \) to ∞
\( = \frac{5}{10} + \frac{5}{10^2} + \frac{5}{10^3} + \ldots \) to ∞
This is an infinite GP with \( a = \frac{5}{10} \) and \( r = \frac{1}{10} \). ∴ Sum \( = \frac{5/10}{1 - \frac{1}{10}} = \frac{5}{9} \)
Hence \( 0.\overline{5} = \frac{5}{9} \).

Sum of n Terms of Special Series

Sum of First n Natural Numbers

To prove: \( 1 + 2 + 3 + \ldots + n = \frac{1}{2}n(n + 1) \), i.e. \( \sum n = \frac{1}{2}n(n + 1) \)
Proof: Consider the series (1 + 2 + 3 + … + n). This is an arithmetic series in which a = 1, d = 1 and l = n.
∴ \( S_n = \frac{n}{2}(1 + n) \) [∵ \( S_n = \frac{n}{2}(a + l) \)], or \( \sum n = \frac{1}{2}n(n + 1) \).

Sum of the Squares of First n Natural Numbers

To prove: \( 1^2 + 2^2 + 3^2 + \ldots + n^2 = \frac{1}{6}n(n + 1)(2n + 1) \)
Proof: Let S_n = 1² + 2² + … + n². Consider the identity k³ − (k − 1)³ = 3k² − 3k + 1 … (i)
Putting k = 1, 2, 3, …, (n − 1), n successively in (i), we get
1³ − 0³ = 3 · 1² − 3 · 1 + 1
2³ − 1³ = 3 · 2² − 3 · 2 + 1
3³ − 2³ = 3 · 3² − 3 · 3 + 1
…
(n − 1)³ − (n − 2)³ = 3(n − 1)² − 3(n − 1) + 1
n³ − (n − 1)³ = 3n² − 3n + 1
Adding columnwise, we get n³ − 0³ = 3(1² + 2² + … + n²) − 3(1 + 2 + 3 + … + n) + n
⇒ \( n^3 = 3\sum n^2 - 3 \cdot \frac{1}{2}n(n + 1) + n \)
⇒ \( 3\sum n^2 = n^3 + \frac{3}{2}n(n + 1) - n = \frac{1}{2}(2n^3 + 3n^2 + n) \)
⇒ \( \sum n^2 = \frac{1}{6}(2n^3 + 3n^2 + n) = \frac{1}{6}n(2n^2 + 3n + 1) = \frac{1}{6}n(n + 1)(2n + 1) \)
Hence \( \sum n^2 = \frac{1}{6}n(n + 1)(2n + 1) \).

Sum of the Cubes of First n Natural Numbers

To prove: \( 1^3 + 2^3 + \ldots + n^3 = \left\{\frac{1}{2}n(n + 1)\right\}^2 \), i.e. \( \sum n^3 = \left\{\frac{1}{2}n(n + 1)\right\}^2 \)
Proof: Let S_n = 1³ + 2³ + … + n³. Consider the identity k⁴ − (k − 1)⁴ = 4k³ − 6k² + 4k − 1
Putting k = 1, 2, 3, …, (n − 1), n successively, we get
1⁴ − 0⁴ = 4 · 1³ − 6 · 1² + 4 · 1 − 1
2⁴ − 1⁴ = 4 · 2³ − 6 · 2² + 4 · 2 − 1
…
(n − 1)⁴ − (n − 2)⁴ = 4(n − 1)³ − 6(n − 1)² + 4(n − 1) − 1
n⁴ − (n − 1)⁴ = 4n³ − 6n² + 4n − 1
Adding columnwise, we get (n⁴ − 0⁴) = 4(1³ + 2³ + … + n³) − 6(1² + 2² + … + n²) + 4(1 + 2 + … + n) − (1 + 1 + … to n terms)
⇒ \( n^4 = 4\sum n^3 - 6\sum n^2 + 4\sum n - n \)
⇒ \( n^4 = 4\sum n^3 - 6 \cdot \frac{1}{6}n(n + 1)(2n + 1) + 4 \cdot \frac{1}{2}n(n + 1) - n \)
⇒ \( 4\sum n^3 = n^4 + n(n + 1)(2n + 1) - 2n(n + 1) + n \)
⇒ \( 4\sum n^3 = n^4 + 2n^3 + 3n^2 + n - 2n^2 - 2n + n = n^4 + 2n^3 + n^2 \)
⇒ \( \sum n^3 = \frac{1}{4}n^2(n^2 + 2n + 1) = \frac{1}{4}n^2(n + 1)^2 = \left\{\frac{1}{2}n(n + 1)\right\}^2 \)
⇒ \( \sum n^3 = \left(\sum n\right)^2 \)

Example 26. Find the sum 1² + (1² + 2²) + (1² + 2² + 3²) + … to n terms.
Answer:
We have \( T_k = (1^2 + 2^2 + \ldots + k^2) = \frac{1}{6}k(k + 1)(2k + 1) = \frac{1}{6}(2k^3 + 3k^2 + k) \)
∴ \( S_n = \sum_{k=1}^{n} T_k = \frac{1}{6}\sum_{k=1}^{n}(2k^3 + 3k^2 + k) = \frac{1}{3}\sum_{k=1}^{n} k^3 + \frac{1}{2}\sum_{k=1}^{n} k^2 + \frac{1}{6}\sum_{k=1}^{n} k \)
\( = \frac{1}{3}\left\{\frac{1}{2}n(n + 1)\right\}^2 + \frac{1}{2} \cdot \frac{1}{6}n(n + 1)(2n + 1) + \frac{1}{6} \cdot \frac{1}{2}n(n + 1) \)
\( = \frac{1}{12}n^2(n + 1)^2 + \frac{1}{12}n(n + 1)(2n + 1) + \frac{1}{12}n(n + 1) \)
\( = \frac{1}{12}n(n + 1)[n(n + 1) + (2n + 1) + 1] = \frac{1}{12}n(n + 1)(n^2 + 3n + 2) = \frac{1}{12}n(n + 1)^2(n + 2) \)

Example 27. Find the sum of n terms of the series \( \frac{1}{2 \times 5} + \frac{1}{5 \times 8} + \frac{1}{8 \times 11} + \ldots \)
Answer:
We have \( T_k = \frac{1}{(k^{th} \text{ term of } 2, 5, 8, \ldots) \times (k^{th} \text{ term of } 5, 8, 11, \ldots)} = \frac{1}{\{2 + (k - 1)3\} \times \{5 + (k - 1)3\}} \)
\( = \frac{1}{(3k - 1)(3k + 2)} = \frac{1}{3}\left\{\frac{1}{3k - 1} - \frac{1}{3k + 2}\right\} \)
Putting k = 1, 2, 3, …, n successively, we get \( T_1 = \frac{1}{3}\left(\frac{1}{2} - \frac{1}{5}\right) \), \( T_2 = \frac{1}{3}\left(\frac{1}{5} - \frac{1}{8}\right) \), …, \( T_n = \frac{1}{3}\left(\frac{1}{3n - 1} - \frac{1}{3n + 2}\right) \)
Adding columnwise, we get \( S_n = (T_1 + T_2 + \ldots + T_n) = \frac{1}{3}\left(\frac{1}{2} - \frac{1}{3n + 2}\right) = \frac{1}{3} \cdot \frac{3n}{2(3n + 2)} \)
∴ \( S_n = \frac{n}{2(3n + 2)} \)