A sequence is defined as an ordered list of numbers arranged according to a definite rule. Each number in the list is called a term of the sequence, and the position of a term is crucial, since the same set of numbers arranged differently would form a different sequence. This emphasis on order distinguishes sequences from general sets.

Mathematically, a sequence is viewed as a function whose domain is the set of natural numbers (or a subset of them) and whose range consists of real numbers. If the sequence is denoted by \(\left\{a_n\right\}\), then \(a_n\) represents the general term or \(n^{th}\) term, which gives the value of the sequence at position \(n\). It connects algebraic expressions with numerical patterns. Sequences may be finite or infinite.
A finite sequence has a limited number of terms, such as the marks obtained in five tests, while an infinite sequence continues indefinitely, like the natural numbers \(1,2,3,\ldots\).

A series is formed when the terms of a sequence are added together according to a specific order. If a sequence is denoted by \(\left\{a_1,a_2,a_3,\ldots\right\}\), then the corresponding series is written as \[a_1+a_2+a_3+\cdots\] Thus, while a sequence emphasizes arrangement, a series emphasizes summation.

Like sequences, series may be finite or infinite. A finite series has a definite number of terms and a fixed sum, such as the total of the first ten natural numbers. An infinite series, on the other hand, involves endlessly many terms, and its sum may or may not exist in a meaningful sense

A geometric progression is defined as a sequence in which the ratio of any term to its immediately preceding term remains constant. This constant is called the common ratio and is usually denoted by \(r\). If the first term is \(a\), then the GP takes the form \[a,\;ar,\; ar^2,\; ar^3,\;\ldots\] This simple structure allows students to clearly identify a GP and distinguish it from other types of sequences.

n-th term

The n-th term of a GP is given by \[a_n=ar^{n-1}\] which enables the direct determination of any term without listing all previous terms. This formula reflects the multiplicative nature of the progression and highlights the exponential growth or decay inherent in GPs.

The sum of the first n terms

The sum of the first nterms, denoted by \(S_n\), is derived algebraically and expressed as

Geometric progressions are closely linked to real-life phenomena involving repeated multiplication, such as compound interest, population growth, radioactive decay, and depreciation of assets.

The geometric mean between two positive numbers aand bis defined as the number Gsuch that \(a,\;G,\;b\) form a geometric progression. By definition of a GP, the ratio of successive terms is equal, which leads to \[\frac{G}{a}=\frac{b}{G}\]

From this relation, we can derives the standard formula for the geometric mean: \[G=\sqrt{ab}\]

This definition highlights that the geometric mean depends on the product of the two numbers rather than their sum.

If ngeometric means are inserted between \(a\) and \(b\), then all the numbers together form a geometric progression with first term \(a\), last term \(b\), and total number of terms n+2. The common ratio is determined using

after which the required means are obtained systematically. This process reinforces the link between geometric means and the general structure of a GP.

Conceptually, the geometric mean represents a balanced multiplicative average and is especially meaningful in situations involving ratios, rates of growth, and proportional change.

The relationship between Arithmetic Mean (A.M.) and Geometric Mean (G.M.) as an important theoretical result that connects two fundamental measures of central tendency. This relationship not only has algebraic significance but also provides insight into the comparative behavior of numbers. For any two positive real numbers aand b, the arithmetic mean is defined as \[A=\frac{a+b}{2}\] while the geometric mean is defined as \[G=\sqrt{ab}\]

AM–GM inequality

The AM–GM inequalitystates that \[A\geq G\] that is, the arithmetic mean of two positive numbers is always greater than or equal to their geometric mean.

An important special case is also discussed: equality holds if and only if a=b. This condition provides a precise criterion for when the arithmetic and geometric means coincide and reinforces the idea of balance between the two numbers.

Write the first three terms in each of the following sequences defined by the following:
(i) \(a_n = 2n + 5\),
(ii) \(a_n=\dfrac{n-3}{4}\)

Solution

We are given two sequences and asked to write the first three terms of each by substituting \(n = 1, 2, 3\) into the general term.

\[ \begin{aligned} (i)\quad a_{n} &= 2n + 5 \\ a_{1} &= 2(1) + 5 = 7 \\ a_{2} &= 2(2) + 5 = 9 \\ a_{3} &= 2(3) + 5 = 11 \end{aligned} \]

This sequence increases by a constant difference of \(2\), so the first three terms are \(7, 9, 11\).

\[ \begin{aligned} (ii)\quad a_{n} &= \frac{n - 3}{4} \\ a_{1} &= \frac{1 - 3}{4} = \frac{-2}{4} = -\frac{1}{2} \\ a_{2} &= \frac{2 - 3}{4} = \frac{-1}{4} \\ a_{3} &= \frac{3 - 3}{4} = 0 \end{aligned} \]

Thus, the first three terms of the second sequence are \(-\frac{1}{2}, -\frac{1}{4}, 0\).

What is the 20th term of the sequence defined by
\(a_n = (n – 1) (2 – n) (3 + n)\)

Solution

We are given the general term of the sequence and asked to find the 20th term by substituting \(n = 20\) into the formula.

\[ \begin{aligned} a_{n} &= (n - 1)(2 - n)(3 + n) \\ a_{20} &= (20 - 1)(2 - 20)(3 + 20) \\ &= 19 \cdot (-18) \cdot 23 \\ &= -7866 \end{aligned} \]

Therefore, the 20th term of the given sequence is \(-7866\).

Let the sequence an be defined as follows:
\(a_1 = 1, a_n = a_{n – 1} + 2 \text{ for } n \gt 2\)
Find first five terms and write corresponding series.

Solution

The sequence is defined recursively with the first term given as \(a_{1} = 1\), and each subsequent term obtained by adding \(2\) to the previous term. Using this rule, we compute the first five terms step by step.

\[ \begin{aligned} a_{1} &= 1 \\ a_{n} &= a_{n-1} + 2 \\ a_{2} &= a_{(2-1)} + 2 \\ &= a_{1} + 2 \\ &= 1 + 2 = 3 \\ a_{3} &= a_{(3-1)} + 2 \\ &= a_{2} + 2 = 3 + 2 = 5 \\ a_{4} &= a_{(4-1)} + 2 \\ &= a_{3} + 2 \\ &= 5 + 2 = 7 \\ a_{5} &= a_{(5-1)} + 2 \\ &= a_{4} + 2 \\ &= 7 + 2 = 9 \end{aligned} \]

Thus, the first five terms are \(1,\; 3,\; 5,\; 7,\; 9\)

Find the \(10^{th}\) and \(n^{th}\) terms of the G.P. 5, 25,125,… .

Solution

The given sequence \(5, 25, 125, \ldots\) is a geometric progression because each term is obtained by multiplying the previous term by a constant ratio.

GP: 5, 25, 125

\[ \begin{aligned} a &= 5 \\ r &= \frac{25}{5} = 5 \\ a_{n} &= ar^{\,n-1} \\ a_{10} &= 5 \cdot (5)^{10-1} \\ &= 5 \times 5^{9} \\ &= 5^{10} \end{aligned} \]

Thus, the \(n^{\text{th}}\) term of the G.P. is \(a_{n} = 5 \cdot 5^{\,n-1} = 5^{n}\), and the \(10^{\text{th}}\) term is \(5^{10}\).

Which term of the G.P., 2,8,32, ... up to \(n\) terms is 131072?

Solution

The given sequence \(2, 8, 32, \ldots\) is a geometric progression because each term is obtained by multiplying the previous term by a constant ratio.

GP: 2, 8, 32 …

\[ \begin{aligned} A &= 2, \quad r = \frac{8}{2} = 4 \\ a_{n} &= a \cdot r^{\,n-1} \\ 131072 &= 2 \cdot r^{\,n-1} \\ \Rightarrow 2 \cdot 4^{\,n-1} &= 131072 \\ \Rightarrow 4^{\,n-1} &= \frac{131072}{2} \\ \Rightarrow 4^{\,n-1} &= 65536 \\ \Rightarrow (n-1)\log 4 &= \log 65536 \\ \Rightarrow (n-1) &= \frac{\log_{10} 65536}{\log_{10} 4} = 8 \\ \Rightarrow (n-1) &= 8 \\ \Rightarrow n &= 8 + 1 = 9 \end{aligned} \]

Therefore, \(131072\) is the \(9^{\text{th}}\) term of the given G.P.

In a G.P., the \(3^{rd}\) term is 24 and the \(6^{th}\) term is 192.Find the \(10^{th}\) term.

Solution

The given sequence is a geometric progression in which the 3rd term is \(24\) and the 6th term is \(192\). Using the general term of a G.P., we substitute the known values to determine the common ratio and the first term.

\[ \begin{aligned} \text{3rd term of GP} &= 24 \\ \text{6th term of GP} &= 192 \\ a_{3} &= a \cdot r^{3-1} \\ 24 &= ar^{2} \\ a_{6} &= a \cdot r^{6-1} \\ 192 &= ar^{5} \\ \frac{192}{24} &= \frac{ar^{5}}{ar^{2}} \\ 8 &= r^{3} \\ \Rightarrow r &= \sqrt[3]{8} \\ &= 2 \\ 24 &= ar^{2} \\ 24 &= a \cdot (2)^{2} \\ \Rightarrow a &= \frac{24}{4} \\ &= 6 \\ 10 - 1 \\ a_{10} &= a \cdot r^{10-1} \\ &= 6 \cdot 2^{9} \\ &= 6 \cdot 512 \\ &= 3072 \end{aligned} \]

Therefore, the \(10^{\text{th}}\) term of the given geometric progression is \(3072\).

Find the sum of first n terms and the sum of first 5 terms of the geometric series \(1+\dfrac{2}{3}+\dfrac{4}{9}\ldots\)

Solution

The given series \(1 + \dfrac{2}{3} + \dfrac{4}{9} + \cdots\) is a geometric progression since each term is obtained by multiplying the previous term by a constant ratio.

GP = \(1 + \dfrac{2}{3} + \dfrac{4}{9} + \cdots\)

Here, the first term \(a = 1\) and the common ratio \(r = \dfrac{2}{3}\)

\[ \begin{aligned} S_{n} &= \frac{a(1 - r^{n})}{1 - r} \end{aligned} \]

Now, substituting \(n = 5\) to find the sum of the first five terms

\[ \begin{aligned} S_{5} &= \frac{a(r^{n} - 1)}{r - 1} \\ &= \frac{1 \cdot (r^{5} - 1)}{r - 1} \\ &= \frac{r^{5} - 1}{r - 1} \\ &= \frac{\left(\frac{2}{3}\right)^{5} - 1}{\frac{2}{3} - 1} \\ &= 3\left[1 - \left(\frac{2}{3}\right)^{5}\right] \\ &= 3\left[1 - \frac{32}{243}\right] \\ &= 3\left[\frac{243 - 32}{243}\right] \\ &= \frac{3 \times 211}{243} \\ &= \frac{211}{81} \end{aligned} \]

Thus, the sum of the first \(n\) terms is \(S_{n} = \dfrac{1 - \left(\frac{2}{3}\right)^{n}}{1 - \frac{2}{3}}\), and the sum of the first five terms is \(\dfrac{211}{81}\).

How many terms of the G.P. \(3,\; \dfrac{3}{2},\;\dfrac{3}{4},\;\ldots\) are needed to give the sum \(\dfrac{3069}{512}\)

Solution

The given sequence \(3,\; \dfrac{3}{2},\; \dfrac{3}{4},\; \ldots\) is a geometric progression because each term is obtained by multiplying the previous term by a constant ratio.

GP: \(3,\; \dfrac{3}{2},\; \dfrac{3}{4},\; \ldots\)

Here, the first term \(a = 3\) and the common ratio \(r = \dfrac{1}{2}\)

\[ \begin{aligned} S_{n} &= \frac{a(1 - r^{n})}{1 - r} \\ \frac{3069}{512} &= \frac{3\left[1 - \left(\frac{1}{2}\right)^{n}\right]}{1 - \frac{1}{2}} \\ \frac{3069}{512} &= 6\left[1 - \left(\frac{1}{2}\right)^{n}\right] \\ \frac{3069}{512 \cdot 6} &= 1 - \left(\frac{1}{2}\right)^{n} \\ \frac{3069}{3072} &= 1 - \left(\frac{1}{2}\right)^{n} \\ \left(\frac{1}{2}\right)^{n} &= 1 - \frac{3069}{3072} \\ &= \frac{3072 - 3069}{3072} \\ &= \frac{3}{3072} \\ &= \frac{1}{1024} \\ \left(\frac{1}{2}\right)^{n} &= \left(\frac{1}{2}\right)^{10} \\ n &= 10 \end{aligned} \]

Therefore, \(10\) terms of the given geometric progression are needed to obtain the sum \(\dfrac{3069}{512}\).

The sum of first three terms of a G.P. is \(\dfrac{13}{12}\) and their product is – 1. Find the common ratio and the terms.

Solution

The sum of the first three terms of a geometric progression is \(\dfrac{13}{12}\) and their product is \(-1\). Let the three terms of the G.P. be \(\dfrac{a}{r},\; a,\; ar\).

\[ \begin{aligned} a_{1} &= \frac{a}{r} \\ a_{2} &= a \\ a_{3} &= ar \\ \frac{a}{r} + a + ar &= \frac{13}{12} \\ \frac{a}{r} \cdot a \cdot ar &= -1 \\ a^{3} &= -1 \\ a &= -1 \end{aligned} \]

Substituting \(a = -1\) into the sum equation:

\[ \begin{aligned} \frac{a}{r} + a + ar &= \frac{13}{12} \\ \frac{-1}{r} - 1 - r &= \frac{13}{12} \end{aligned} \]

Multiplying both sides by \(12r\) to remove denominators:

\[ \begin{aligned} -12 - 12r - 12r^{2} &= 13r \\ 12r^{2} + 25r + 12 &= 0 \end{aligned} \]

Now factorising the quadratic equation:

\[ \begin{aligned} 12r^{2} + 25r + 12 &= 0 \\ 12r^{2} + 16r + 9r + 12 &= 0 \\ 4r(3r + 4) + 3(3r + 4) &= 0 \\ (3r + 4)(4r + 3) &= 0 \end{aligned} \]

Solving for \(r\):

\[ \begin{aligned} 3r + 4 &= 0 \quad \Rightarrow \quad r = -\frac{4}{3} \\ 4r + 3 &= 0 \quad \Rightarrow \quad r = -\frac{3}{4} \end{aligned} \]

Hence, the corresponding three terms of the G.P. are:

\[ \begin{aligned} \frac{a}{r},\; a,\; ar &= \frac{4}{3},\; -1,\; \frac{3}{4} \\ \text{or} \quad \frac{3}{4},\; -1,\; \frac{4}{3} \end{aligned} \]

Therefore, the common ratio is \(-\dfrac{4}{3}\) or \(-\dfrac{3}{4}\), and the corresponding terms are \(\left(\dfrac{4}{3}, -1, \dfrac{3}{4}\right)\) or \(\left(\dfrac{3}{4}, -1, \dfrac{4}{3}\right)\).

Find the sum of the sequence 7, 77, 777, 7777, ... to n terms.

Solution

We are required to find the sum of the sequence \(7, 77, 777, 7777, \ldots\) up to \(n\) terms. Each term consists of repeated digit \(7\), so we rewrite the series in a convenient algebraic form.

\[ \begin{aligned} S_{n} &= 7 + 77 + 777 + \cdots \text{ up to } n \text{ terms} \\ S_{n} &= 7\left(1 + 11 + 111 + \cdots \text{ up to } n \text{ terms}\right) \\ S_{n} &= \frac{7}{9}\left(9 + 99 + 999 + \cdots \text{ up to } n \text{ terms}\right) \\ S_{n} &= \frac{7}{9}\left[(10 - 1) + (100 - 1) + (1000 - 1) + \cdots \text{ up to } n \text{ terms}\right] \\ S_{n} &= \frac{7}{9}\left[(10 + 10^{2} + 10^{3} + \cdots + 10^{n}) - (1 + 1 + 1 + \cdots + 1)\right] \\ S_{n} &= \frac{7}{9}\left[\frac{10(10^{n} - 1)}{10 - 1} - n\right] \\ S_{n} &= \frac{7}{9}\left[\frac{10(10^{n} - 1)}{9} - n\right] \\ S_{n} &= \frac{7}{81}\left[10(10^{n} - 1) - 9n\right] \\ S_{n} &= \frac{7}{81}\left[10^{n+1} - 10 - 9n\right] \end{aligned} \]

Therefore, the sum of the first \(n\) terms of the sequence is \[ S_{n} = \frac{7}{81}\left(10^{n+1} - 10 - 9n\right) \]

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.

Solution

The number of ancestors follows a geometric progression because each generation has twice as many ancestors as the previous one. The sequence is \(2, 4, 8, \ldots\), where the first term represents the number of parents.

2, 4, 8, …

Here, the first term \(A = 2\) and the common ratio \(r = 2\)

\[ \begin{aligned} S_{n} &= \frac{A(r^{n} - 1)}{r - 1} \\ S_{n} &= \frac{2\left(2^{10} - 1\right)}{2 - 1} \\ S_{n} &= 2\left(2^{10} - 1\right) \\ &= 2(1024 - 1) \\ &= 2 \times 1023 \\ &= 2046 \end{aligned} \]

Therefore, the total number of ancestors in the ten generations preceding his own is \(2046\).

Insert three numbers between 1 and 256 so that the resulting sequence is a G.P.

Solution

Solution.

We are required to insert three numbers between \(1\) and \(256\) such that the entire sequence forms a geometric progression. Since there will be five terms in total, let the sequence be

\[ \begin{aligned} 1,\; a,\; b,\; c,\; 256 \end{aligned} \]

Let the common ratio be \(r\). Then the terms of the G.P. can be written as

\[ \begin{aligned} 1 &= 1 \\ a &= r \\ b &= r^2 \\ c &= r^3 \\ 256 &= r^4 \end{aligned} \]

From the last term, we get

\[ \begin{aligned} r^4 &= 256 \\ r &= \sqrt[4]{256} \\ r &= 4 \end{aligned} \]

Now substitute \(r = 4\) to find the inserted numbers

\[ \begin{aligned} a &= 4 \\ b &= 4^2 = 16 \\ c &= 4^3 = 64 \end{aligned} \]

Therefore, the three numbers to be inserted are \(4,\; 16,\; 64\), and the resulting G.P. is \(1,\; 4,\; 16,\; 64,\; 256\).

If A.M. and G.M. of two positive numbers \(a\) and \(b\) are 10 and 8, respectively, find the numbers.

Solution

Solution.

Let the two positive numbers be \(a\) and \(b\). We are given that their Arithmetic Mean (A.M.) is \(10\) and their Geometric Mean (G.M.) is \(8\).

Using the formula for A.M., we write

\[ \begin{aligned} \frac{a + b}{2} = 10 \end{aligned} \]

\[ \begin{aligned} a + b = 20 \end{aligned} \]

Using the formula for G.M., we write

\[ \begin{aligned} \sqrt{ab} = 8 \end{aligned} \]

\[ \begin{aligned} ab = 64 \end{aligned} \]

Now, the two numbers are roots of the quadratic equation

\[ \begin{aligned} x^2 - (a + b)x + ab = 0 \end{aligned} \]

Substituting \(a + b = 20\) and \(ab = 64\), we get

\[ \begin{aligned} x^2 - 20x + 64 = 0 \end{aligned} \]

Solving this quadratic equation

\[ \begin{aligned} x &= \frac{20 \pm \sqrt{20^2 - 4 \cdot 64}}{2} \\ &= \frac{20 \pm \sqrt{400 - 256}}{2} \\ &= \frac{20 \pm \sqrt{144}}{2} \\ &= \frac{20 \pm 12}{2} \end{aligned} \]

\[ \begin{aligned} x = 16 \quad \text{or} \quad x = 4 \end{aligned} \]

Therefore, the two numbers are \(16\) and \(4\).