Arithmetic Progression

We begin with some important special sequences and series, like arithmetic progression. geometric progression, harmonic progression, arithmetico-geometric progression etc Then, we will be looking at some general sequences and series. Our main goal will be finding the closed formulae for series if possible.

Definition of Arithmetic Progression

An arithmetic progression (AP) is a sequence of numbers \(a_1, a_2, a_3, \ldots, a_n, \ldots\) where each term after the first is found by adding a constant difference, \(d\), to the previous term. This can be defined recursively as:

\( a_1 = a \) (the first term),

\( a_n = a_{n-1} + d \) for all \( n > 1 \), where \( a_n \) is the \(n\)-th term, \( a_{n-1} \) is the term before \( a_n \), and \( d \) is the common difference of the arithmetic progression.

Example

Given the first term \( a_1 = 5 \) and the common difference \( d = 3 \), the sequence of the arithmetic progression (AP) can be defined recursively as:

\[ a_1 = 5 \]

\[ a_n = a_{n-1} + 3 \]

Using this recursive relationship, the first few terms of the sequence are:

\[ \begin{align*} &a_1 = 5 \\ &a_2 = a_1 + 3 = 5 + 3 = 8 \\ &a_3 = a_2 + 3 = 8 + 3 = 11 \\ &a_4 = a_3 + 3 = 11 + 3 = 14 \\ \vdots \\ \end{align*} \]

Continuing this pattern, the sequence is 5, 8, 11, 14, 17, ..., where each term is 3 more than the previous term.

Explicit Formula for the nth term

Given an arithmetic progression \(a_1, a_2, a_3, \ldots, a_n, \ldots\) with the first term \(a_1 = a\) and common difference \(d\), the recursive definition of the sequence is given by:

\[ a_n = a_{n-1} + d \quad \text{for all } n > 1. \]

We observe the differences between successive terms:

\[\begin{align*} a_2 - a_1 &= d, \\ a_3 - a_2 &= d, \\ &\ \vdots \\ a_n - a_{n-1} &= d. \end{align*}\]

Summing these equations, we have:

\[\begin{align*} (a_2 - a_1) + (a_3 - a_2) + \cdots + (a_n - a_{n-1}) &= d + d + \cdots + d \\ a_n - a_1 &= d(n - 1). \end{align*}\]

From the telescoping nature of the series, all intermediate terms cancel, leaving:

\[ a_n = a + (n - 1)d. \]

This expression provides the explicit formula for the \(n\)-th term of an arithmetic progression.

Sum of n terms of a sequence in AP

The sum of the first \(n\) terms of an arithmetic progression, denoted \(S_n\), is given by \(S_n = \frac{n}{2}(2a + (n-1)d)\), where \(a\) is the first term and \(d\) is the common difference of the progression.

Proof:

Let an arithmetic progression be given by \(a_1, a_2, a_3, \ldots, a_n\), with the first term \(a_1 = a\) and the common difference \(d\). We seek to find an expression for the sum of the first \(n\) terms, \(S_n\).

Consider the sequence in its original order and in reverse:

\[ S_n = a_1 + a_2 + a_3 + \ldots + a_{n-1} + a_n \]

\[ S_n = a_n + a_{n-1} + a_{n-2} + \ldots + a_2 + a_1 \]

Adding these two expressions for \(S_n\) term by term, we observe that:

\[ 2S_n = (a_1 + a_n) + (a_2 + a_{n-1}) + (a_3 + a_{n-2}) + \ldots + (a_{n-1} + a_2) + (a_n + a_1) \]

From the formula for the \(n\)-th term of an arithmetic progression, \(a_n = a + (n-1)d\), we find that each pair of terms in the summation, such as \(a_1 + a_n\), \(a_2 + a_{n-1}\), and so on, sums to \(2a + (n-1)d\). This is because:

\[ a_k + a_{n-k+1} = a + (k-1)d + a + (n-k)d = 2a + (n-1)d \]

Therefore, the sum of each pair of terms in the sequence is constant, and with \(n\) terms in total, we have \(n/2\) such pairs. Hence, the sum of the first \(n\) terms is:

\[ 2S_n = n(2a + (n-1)d) \]

\[ S_n = \frac{n}{2}(2a + (n-1)d) \]

This completes the proof, demonstrating that the sum of the first \(n\) terms of an arithmetic progression can be calculated using the formula \(S_n = \frac{n}{2}(2a + (n-1)d)\).

Reversal Method

If you have read the proof above, you would see that we reversed the series and added with the original series. This ideas of reversing works whenever the sum the \(kth\) term from the beginning and the \(kth\) term from the end is constant, that is, independednt of \(k\) for all \(k\). In simple words, whever we have the pattern \(a_1+a_n=a_2+a_{n-1}=a_3+a_{n-2}=...=constant\). For example, consider the following pronlem:

Problem: Given a series \(S_n = \frac{9^{1/n}}{9^{1/n}+3} + \frac{9^{2/n}}{9^{2/n}+3} + \ldots + \frac{9^{(n-1)/n}}{9^{(n-1)/n}+3}\), demonstrate that the sum of the \(k\)-th term from the beginning and the \(k\)-th term from the end equals 1, and then use this property to calculate the sum of the series.

Solution:

Consider the series \(S_n = \frac{9^{1/n}}{9^{1/n}+3} + \frac{9^{2/n}}{9^{2/n}+3} + \ldots + \frac{9^{(n-1)/n}}{9^{(n-1)/n}+3}\). Let \(T_k = \frac{9^{k/n}}{9^{k/n}+3}\) represent the \(k\)-th term from the beginning and \(T_{n-k} = \frac{9^{(n-k)/n}}{9^{(n-k)/n}+3}\) the \(k\)-th term from the end.

Simplify \(T_{n-k}\) by multipling numerator and denominator by \(9^{k/n}\), yielding \(T_{n-k} = \frac{3}{9^{k/n}+3}\).

Adding \(T_k\) and \(T_{n-k}\) gives:

\[T_k + T_{n-k} = \frac{9^{k/n}}{9^{k/n}+3} + \frac{3}{9^{k/n}+3} = 1\]

This demonstrates that the sum of the \(k\)-th term from the beginning and the \(k\)-th term from the end is constant and equals 1 for any \(k\).

To find \(S_n\), reverse the series and add to the original:

Considering the reversal leads to a series where each pair of terms from the beginning and the end sums to 1, and since there are \(n-1\) such pairs, the total sum of the series and its reverse combined is \(2S_n = n-1\).

\[S_n = T_1 + T_2 + T_3 + \ldots + T_{n-1}\]

\[S_n = T_{n-1} + \ldots + T_3 + T_2 + T_1\]

Add original and reversed:

\[2S_n = (1 + 1 + 1 + \ldots + 1) = n-1\]

Thus, the sum \(S_n\):

\[S_n = \frac{n-1}{2}\]

An alternative way to write the sum of AP

The sum of the first \(n\) terms of an arithmetic progression can alternatively be expressed as:

\[S_n = \frac{n}{2}(a + l)\]

where \(a\) is the first term of the sequence and \(l\) represents the last term, or the \(n\)-th term.

This formula provides a direct relationship between the sum of the arithmetic series and its first and last terms, highlighting the symmetry inherent in the progression. The factor \(\frac{n}{2}\) accounts for the average of the first and last terms, multiplied by the total number of terms, \(n\), thereby giving the total sum of the series. This expression is particularly useful because it allows for the calculation of the sum without needing to directly use the common difference \(d\), provided the first and last terms of the sequence are known.

For example:

To find the sum of the first \(n\) natural numbers using the formula \(S_n = \frac{n}{2}(a + l)\), where \(a\) is the first term of the sequence and \(l\) is the last term (or the \(n\)-th term): In the sequence of the first \(n\) natural numbers, \(a = 1\) (the first natural number), and \(l = n\) (the \(n\)-th natural number, also the last term in this sequence).

Substituting these values into the formula, we get:

\[ S_n = \frac{n}{2}(1 + n) \]

\[ S_n = \frac{n(n + 1)}{2} \]

Example

Problem:

Calculate the sum of the first \(n\) terms of the sequence \(1 + 3 + 5 + \ldots\) up to \(n\) terms, where the sequence consists of odd numbers starting from 1.

Solution:

The given sequence is an arithmetic progression (AP) with the first term \(a = 1\) and a common difference \(d = 2\) (since each term increases by 2).

To find the sum of the first \(n\) terms, \(S_n\), of this AP, we can use the formula for the sum of an AP:

\[ S_n = \frac{n}{2} \left[2a + (n - 1)d\right] \]

Given \(a = 1\) and \(d = 2\), substituting these values gives:

\[ \begin{align*} S_n &= \frac{n}{2} \left[2 \cdot 1 + (n - 1) \cdot 2\right] \\ S_n &= \frac{n}{2} \left[2 + 2n - 2\right] \\ S_n &= \frac{n}{2} \left[2n\right] \\ S_n &= n^2 \end{align*} \]

Therefore, the sum of the first \(n\) terms of the sequence of odd numbers starting from 1 is \(n^2\).

The general form of the nth term

The general term of an arithmetic progression (AP) can be succinctly represented as a linear polynomial in \(n\), that is, \(a_n = pn + q\), where \(p\) and \(q\) are constants. This formulation emerges naturally from the conventional AP expression \(a_n = a + (n-1)d\), with \(a\) as the initial term and \(d\) as the common difference.

To derive \(a_n = pn + q\), we observe:

\[a_n = a + (n-1)d = nd + (a - d) = pn + q\]

where \(p = d\), the coefficient of n is the common difference of AP, and \(q = a - d\). In this framework, evaluating at \(n = 1\) yields the first term:

\[a_1 = p + q\]

Moreover, the coefficient \(p\) is equal to the common difference \(d\).

For the arithmetic progression given by \(a_n = 3n - 4\), we can identify the common difference and the first term by comparing this expression to the general form \(a_n = pn + q\).

In this example, \(p = 3\) and \(q = -4\). Thus, the common difference \(d\) of this AP is \(3\).

To find the first term \(a_1\), substitute \(n = 1\) into the formula:

\[a_1 = 3(1) - 4 = 3 - 4 = -1\]

Therefore, for the AP \(a_n = 3n - 4\), the common difference is \(3\) and the first term is \(-1\).

General Form of the sum of an AP

For an arithmetic progression with first term \(a\) and common difference \(d\), the sum of the first \(n\) terms, \(S_n\), is traditionally given by:

\[S_n = \frac{n}{2}(2a + (n - 1)d).\]

This sum can also be represented as a quadratic polynomial without a constant term:

\[S_n = pn^2 + qn,\]

where \(p = \frac{d}{2}\) and \(q = a - \frac{d}{2}\). Deriving this form involves rearranging the original sum formula:

\[S_n = \frac{n}{2}(2a + nd - d) = \frac{d}{2}n^2 + (a - \frac{d}{2})n.\]

Thus, the quadratic representation \(S_n = pn^2 + qn\) directly relates to the progression's common difference \(d\) and first term \(a\).

Relation between the general term and the sum coefficients

Problem Statement:

Given an arithmetic progression (AP) with the \(n\)-th term represented as \(a_n = pn + q\) and the sum of the first \(n\) terms expressed as \(S_n = Pn^2 + Qn\), where \(p\), \(q\), \(P\), and \(Q\) are constants, prove the following relationships:

\[P = \frac{p}{2}\]

\[Q = q + \frac{p}{2}\]

Assume that the common difference of the AP is \(d\), and the first term is \(a\), with the known relations

\(p = d\), \(q = a - d\), \(P = \frac{d}{2}\), and \(Q = a - \frac{d}{2}\).

Solution:

Proving \(P = \frac{p}{2}\)

Given that \(P = \frac{d}{2}\) and \(p = d\), we directly substitute \(d\) from \(p = d\) into \(P = \frac{d}{2}\) to get:

\[2P = 2\left(\frac{d}{2}\right) = d = p\]

Thus, we have proven that \(P = \frac{p}{2}\).
Proving \(Q = q + \frac{p}{2}\):

From the given, \(q = a - d\) and \(Q = a - \frac{d}{2}\). Since \(p = d\), we can substitute \(d\) with \(p\) in the expression for \(Q\):

\[Q = a - \frac{p}{2}\]

To find \(a\) in terms of \(q\) and \(p\), we rearrange \(q = a - d = a - p\) to get \(a = q + p\). Substituting this expression for \(a\) into \(Q\):

\[Q = (q + p) - \frac{p}{2} = q + \frac{p}{2}\]

This proves that \(Q = q + \frac{p}{2}\), establishing the relationship between \(q\) and \(Q\) as well as between \(p\) and \(P\).

Hence, the relationships \(P = \frac{p}{2}\) and \(Q = q + \frac{p}{2}\) accurately describe the connections between the coefficients of the \(n\)-th term and the sum formula of an arithmetic progression in terms of \(p\), \(q\), \(P\), and \(Q\). \(\blacksquare\)

We can also write the relationship as:

\[p=2P\]

\[q=Q-P\]

Example

Problem:

Given the \(n\)-th term of an arithmetic progression \(t_n = 2n + 3\), find the sum of the first \(n\) terms, \(S_n\), using the relationships \(p = 2P\) and \(q = Q - P\) derived from the general form of an arithmetic progression sum \(S_n = Pn^2 + Qn\).

Solution:

Step 1: Identify \(p\) and \(q\) from \(t_n\)

From \(t_n = 2n + 3\), we identify:

\(p = 2\), the coefficient of \(n\), representing the common difference \(d\).
\(q = 3\), the constant term, representing \(a - d\), where \(a\) is the first term.

Step 2: Use the relationships to find \(P\) and \(Q\)

Given the relationships:

\(p = 2P\)
\(q = Q - P\)

Substitute \(p = 2\) into \(p = 2P\) to find \(P\):

\(2 = 2P\)
\(P = 1\)

Substitute \(q = 3\) and \(P = 1\) into \(q = Q - P\) to find \(Q\):

\(3 = Q - 1\)
\(Q = 4\)

Step 3: Apply \(P\) and \(Q\) to find \(S_n\)

With \(P = 1\) and \(Q = 4\), substitute into \(S_n = Pn^2 + Qn\):

\(S_n = 1n^2 + 4n\)
\(S_n = n^2 + 4n\)

Conclusion:

The sum of the first \(n\) terms of the arithmetic progression with the \(n\)-th term \(t_n = 2n + 3\) is \(S_n = n^2 + 4n\).

Example

Problem:

Given the sum of the first \(n\) terms of an arithmetic progression as \(S_n = 2n^2 - 3n\), determine the \(n\)-th term, \(t_n\), utilizing the direct formulae \(p = 2P\) and \(q = Q - P\).

Solution:

From \(S_n = 2n^2 - 3n\), we identify: - \(P = 2\), the coefficient of \(n^2\), - \(Q = -3\), the coefficient of \(n\).

Apply the Direct Formulae:

Find \(p\) using \(p = 2P\): \(p = 2 \times 2 = 4\)
Find \(q\) using \(q = Q - P\): \(q = -3 - 2 = -5\)

Derive \(t_n\):

Given the general formula for the \(n\)-th term of an AP is \(t_n = pn + q\), substitute \(p = 4\) and \(q = -5\):

\[t_n = 4n - 5\]

Conclusion:

The \(n\)-th term of the arithmetic progression, given \(S_n = 2n^2 - 3n\), is \(t_n = 4n - 5\).\(\blacksquare\)

You can also derive this using \(t_n=S_n-S_{n-1}\)

Ratio of nth terms and sums

Let us understand this from the following examples. There are three methods we can use to solve this problem. Observe each one of them. The last one is obviously the simplest one and you should use that method. Expalantion is given below the example.

Example

Problem:

Given the sum of the first \(n\) terms of an arithmetic progression (AP) as \(S_n = 2n^2 - 3n\), determine the \(n\)-th term of the AP, \(t_n\).

Solution 1:Solution 2:Solution 3:

Given the ratio of the sums of \(n\) terms of two arithmetic progressions (APs) as \(\frac{3n+2}{5n-1}\), we are tasked with finding the ratio of their respective \(n\)-th terms.

Step 1: Assume Sums of \(n\) Terms for Each AP For the first AP, let the sum of \(n\) terms be \(S_{n1} = (3n+2)\lambda\), and for the second AP, \(S_{n2} = (5n-1)\lambda\), where \(\lambda\) is a proportionality constant.

Step 2: Identify \(P\) and \(Q\) for Each AP Given: - \(P_1 = 3\lambda\) and \(Q_1 = 2\lambda\) for the first AP, - \(P_2 = 5\lambda\) and \(Q_2 = -\lambda\) for the second AP.

Step 3: Derive \(p\) and \(q\) Using Relationships Utilizing \(p = 2P\) and \(q = Q - P\):

For the first AP:
- \(p_1 = 2P_1 = 2(3\lambda) = 6\lambda\)
- \(q_1 = Q_1 - P_1 = 2\lambda - 3\lambda = -\lambda\)
For the second AP:
- \(p_2 = 2P_2 = 2(5\lambda) = 10\lambda\)
- \(q_2 = Q_2 - P_2 = -\lambda - 5\lambda = -6\lambda\)

Step 4: Express \(n\)-th Terms and Find Their Ratio The \(n\)-th term for each AP is \(t_{n1} = p_1n + q_1\) and \(t_{n2} = p_2n + q_2\), substituting the derived values:

\(t_{n1} = 6\lambda n - \lambda\)
\(t_{n2} = 10\lambda n - 6\lambda\)

The ratio of the \(n\)-th terms is:

\[ \frac{t_{n1}}{t_{n2}} = \frac{6\lambda n - \lambda}{10\lambda n - 6\lambda} \]

Canceling \(\lambda\) and simplifying gives:

\[ \frac{t_{n1}}{t_{n2}} = \frac{6n - 1}{10n - 6} \]

This ratio represents the relationship between the \(n\)-th terms of the two APs, independent of the proportionality constant \(\lambda\).

Let's correctly address the problem with the appropriate assumption for \(S_n\) for each AP.

Step 1: Assume Sums of \(n\) Terms for Each AP

For the first AP, let the sum of \(n\) terms be \(S_{n1} = n(3n + 2)\lambda\), and for the second AP, \(S_{n2} = n(5n - 1)\lambda\).

Step 2: Evaluate \(t_n\) for Each AP Using \(t_n = S_n - S_{n-1}\)

For the first AP:

\[ \begin{align*} t_{n1} &= S_{n1} - S_{n1-1}\\ &= n(3n + 2)\lambda - (n-1)(3(n-1) + 2)\lambda\\ &= [3n^2 + 2n - 3(n^2 - 2n + 1) - 2(n-1)]\lambda\\ &= [3n^2 + 2n - 3n^2 + 6n - 3 - 2n + 2]\lambda\\ &= (6n - 1)\lambda \end{align*} \]

For the second AP:

\[\begin{align*} t_{n2} &= S_{n2} - S_{n2-1} \\ &= n(5n - 1)\lambda - (n-1)(5(n-1) - 1)\lambda \\ &= [5n^2 - n - 5(n^2 - 2n + 1) + (n-1)]\lambda \\ &= [5n^2 - n - 5n^2 + 10n - 5 + n - 1]\lambda \\ &= (10n - 6)\lambda \end{align*}\]

Step 3: Take the Ratio of \(t_{n1}\) to \(t_{n2}\) and Cancel \(\lambda\) The ratio of the \(n\)-th terms of the two APs is: - \(\frac{t_{n1}}{t_{n2}} = \frac{(6n - 1)\lambda}{(10n - 6)\lambda} = \frac{6n - 1}{10n - 6}\)

Replace \(n\) with \(2n-1\) in \(\frac{3n+2}{5n-1}\):

\[ \text{Ratio of the \(n\)-th terms} = \frac{3(2n-1)+2}{5(2n-1)-1} \]

\[ = \frac{6n - 3 + 2}{10n - 5 - 1} \]

\[ = \frac{6n - 1}{10n - 6} \]

Hence, the ratio of the respective \(n\)-th terms of the two APs is \(\frac{6n - 1}{10n - 6}\).

Ratio of nth terms when ratio of sums is known (Better than the method in previous examples)

For two arithmetic progressions with the \(n\)-th terms denoted by \(t_n\) and \(t_n'\) respectively, the ratio of the \(n\)-th terms is equal to the ratio of the sums of their first \((2n-1)\) terms, that is, \(t_n/t_n' = S_{2n-1}/S_{2n-1}'\).

Proof:

Given two arithmetic progressions with their \(n\)-th terms defined as \(t_n = a + (n-1)d\) and \(t_n' = a' + (n-1)d'\), where \(a\) and \(a'\) are the first terms, and \(d\) and \(d'\) are the common differences of the first and second sequences respectively.

The sum of the first \(m\) terms of an arithmetic progression is given by \(S_m = \frac{m}{2}(2a + (m-1)d)\), and similarly for the second sequence, \(S_m' = \frac{m}{2}(2a' + (m-1)d')\).

Consider the ratio of sums \(S_m/S_m'\):

\[ \frac{S_m}{S_m'} = \frac{\frac{m}{2}[2a + (m-1)d]}{\frac{m}{2}[2a' + (m-1)d']} \]
Cancel \(m/2\) from 2 to simplify:

\[ \frac{S_m}{S_m'} = \frac{a + \frac{m-1}{2}d}{a' + \frac{m-1}{2}d'} \]

This expression resembles the formula for the \(n\)-th term of an AP, where the \(n\)-th term \(t_n = a + (n-1)d\). By observing that if we set \(\frac{m-1}{2} = n-1\), which implies \(m = 2n - 1\), we align the ratio of sums with the structure of the \(n\)-th term ratio.

Conclusion: Hence, for \(m = 2n - 1\), the ratio of sums of \(m\) terms of two APs directly parallels the ratio of their \(n\)-th terms, leading to:

\[ \frac{t_n}{t_n'} = \frac{a + (n-1)d}{a' + (n-1)d'} = \frac{S_{2n-1}}{S_{2n-1}'} \]

This derivation demonstrates that by choosing \(m = 2n - 1\), we can use the sum formula for APs to effectively calculate the ratio of \(n\)-th terms, streamlining the process without directly calculating each \(n\)-th term.

Ratio of sum of n terms of two AP's when the ratio of their nth terms is known

Given two arithmetic progressions (APs) with sums of the first \(n\) terms denoted by \(S_n\) and \(S_n'\), respectively, the ratio of these sums \(S_n/S_n'\) is equal to the ratio of their \(\left(\frac{n+1}{2}\right)\)-th terms, i.e., \(S_n/S_n' = t_{\left(\frac{n+1}{2}\right)}/t_{\left(\frac{n+1}{2}\right)}'\).

Proof:

From the previously established theorem that \(t_n/t_n' = S_{2n-1}/S_{2n-1}'\), we infer a relationship between the terms of two APs and the sums of their sequences.

To prove the current theorem, let us consider the expression \(S_n/S_n' = t_{\left(\frac{n+1}{2}\right)}/t_{\left(\frac{n+1}{2}\right)}'\) and utilize the insights from the previous theorem.

Relation from the Previous Theorem:

The previous theorem establishes that for any integer \(m\), the ratio of the \(m\)-th terms of two APs is equal to the ratio of the sums of their first \((2m-1)\) terms. Mathematically, if \(m = \frac{n+1}{2}\), then \(2m-1 = n\).
Applying the Previous Theorem:

Given that \(m = \frac{n+1}{2}\), we directly apply the previous theorem's result:

\[t_m/t_m' = S_{2m-1}/S_{2m-1}'\]

Substituting \(m = \frac{n+1}{2}\) into \(2m-1 = n\), we get:

\[t_{\left(\frac{n+1}{2}\right)}/t_{\left(\frac{n+1}{2}\right)}' = S_n/S_n'\]

We have a similar theorem on an AP about ratios:

Important Properties of AP

Three terms in AP

If three numbers \(a\), \(b\), and \(c\) are in Arithmetic Progression (AP), then \(b = \frac{a + c}{2}\).

In an Arithmetic Progression, the difference between consecutive terms is constant. This constant difference is known as the common difference, denoted as \(d\). When three numbers \(a\), \(b\), and \(c\) are in AP, it means that the difference between \(b\) and \(a\) is the same as the difference between \(c\) and \(b\). Mathematically, this is expressed as:

\[b - a = c - b\]

Rearranging the terms to solve for \(b\), we get:

\[2b = a + c\]

Dividing both sides by 2 gives us the average of \(a\) and \(c\):

\[b = \frac{a + c}{2}\]

Thus, \(b\) is the arithmetic mean of \(a\) and \(c\). This result implies that in an AP, the middle term is always the average of the terms on either side of it, maintaining a uniform progression throughout the sequence.

kth Term from the End in an AP:

For an arithmetic progression (AP) with \(n\) terms, first term \(a\), and common difference \(d\), the \(k\)-th term from the end is given by \(a + (n-k)d\).

Proof:

General Formula for the \(n\)-th Term: The \(n\)-th term of an AP, \(a_n\), is defined as \(a_n = a + (n-1)d\), where \(a\) is the first term, \(d\) is the common difference, and \(n\) is the term's position in the sequence.
Position from the End: The \(k\)-th term from the end means that there are \(k-1\) terms after it. Since there are \(n\) terms in total, the position of this term from the start is \(n - (k-1)\) or \(n-k+1\).
Applying the Formula: To find the value of the term at this position, we substitute \(n-k+1\) for \(n\) in the general formula:

\[a_{(n-k+1)} = a + [(n-k+1)-1]d\]

\[a_{(n-k+1)} = a + (n-k)d\]

Conclusion: This proves that the \(k\)-th term from the end of an AP is calculated as \(a + (n-k)d\), effectively demonstrating how the structure of an AP allows for straightforward calculation of terms based on their position from either end of the sequence.

Constant Sum Property of Equidistant Terms in Arithmetic Progressions

In an Arithmetic Progression (AP) consisting of terms \(a_1, a_2, a_3, \ldots, a_n\), the property of equidistant terms can be expressed as follows:

\[a_1 + a_n = a_2 + a_{n-1} = a_3 + a_{n-2} = \ldots = a_k + a_{n-k+1} = \ldots\]

that is, the sum of the first and last term, \(a_1 + a_n\), is equal to the sum of the second term and the second last term, \(a_2 + a_{n-1}\), and so on. This pattern continues such that the sum of terms equidistant from the beginning and end of the sequence is constant.

Proof:

The \(n\)-th term of an AP, \(a_n\), can be expressed as: [a_n = a + (n-1)d]

Consider two terms, \(a_k\) and \(a_{n-k+1}\), which are equidistant from the start and end of the sequence: - The \(k\)-th term, \(a_k = a + (k-1)d\) - The \((n-k+1)\)-th term, \(a_{n-k+1} = a + (n-k)d\)

The sum of these two terms is:

\[a_k + a_{n-k+1} = [a + (k-1)d] + [a + (n-k)d]\]

\[= 2a + (n-1)d\]

Notice that the sum \(2a + (n-1)d\) does not depend on \(k\); it depends only on \(a\), \(d\), and \(n\), which are constant for the entire sequence. Thus, for any \(k\), the sum of terms equidistant from the start and end of the AP is the same, and it is equal to the sum of the first and last term of the progression.

Conclusion: This proves that in an AP with first term \(a\) and common difference \(d\), the sum of pairs of terms equidistant from both ends is constant, demonstrating a fundamental symmetry in the structure of arithmetic progressions.

Average of n Terms of an Arithmetic Progression

In an arithmetic progression (AP) with \(n\) terms, the average of these \(n\) terms is equal to the average of the first term (\(a_1\)) and the last term (\(a_n\)). Mathematically, this property is expressed as:

\[\frac{a_1 + a_2 + a_3 + \ldots + a_n}{n} = \frac{a_1 + a_n}{2}\]

This means that the average of all the terms in an AP is the same as the average of its first and last terms, effectively representing the midpoint value of the sequence.

Proof:

Sum of the First \(n\) Terms of an AP: The sum of the first \(n\) terms of an AP is given by \(S_n = \frac{n}{2}(a_1 + a_n)\), where \(a_1\) is the first term, \(a_n\) is the \(n\)-th (last) term, and \(n\) is the total number of terms.
Average of the First \(n\) Terms: The average of these \(n\) terms, denoted as \(\bar{a}\), is found by dividing the total sum \(S_n\) by \(n\):

\[\bar{a} = \frac{S_n}{n} = \frac{\frac{n}{2}(a_1 + a_n)}{n}\]
Simplification: Simplify the expression by canceling \(n\) in the numerator and the denominator:

\[\bar{a} = \frac{1}{2}(a_1 + a_n)\]

Conclusion: This simplification proves that the average of \(n\) terms of an AP is equal to the average of its first and last terms, \(\frac{a_1 + a_n}{2}\).'

Arithmetic Operations and Their Effects on Arithmetic Progressions

Arithmetic Progressions (APs) exhibit several fundamental properties that remain consistent under various arithmetic operations. These properties highlight the structural integrity and predictable nature of APs:

Addition/Subtraction:
- If a constant is added to or subtracted from each term of an AP, the resulting sequence remains an AP with the same common difference.
- Example: Given an AP \(a, a+d, a+2d, \ldots\), adding or subtracting a constant \(c\) yields a new AP \(a+c, a+d+c, a+2d+c, \ldots\), with the same common difference \(d\).
Multiplication:
- If each term of an AP is multiplied by a constant, the resulting sequence is also an AP. The common difference of the new AP is the original common difference multiplied by that constant.
- Example: Multiplying each term of an AP \(a, a+d, a+2d, \ldots\) by a constant \(k\) results in \(ak, ak+kd, ak+2kd, \ldots\), with a new common difference of \(kd\).
Division:
- If each term of an AP is divided by the same non-zero constant, the resulting sequence is an AP. The common difference of the new AP is the original common difference divided by that constant.
- Example: Dividing each term of an AP \(a, a+d, a+2d, \ldots\) by a non-zero constant \(k\) forms a new AP \(\frac{a}{k}, \frac{a+d}{k}, \frac{a+2d}{k}, \ldots\), with a new common difference of \(\frac{d}{k}\).
Combining APs:
- If two sequences are APs with the same common difference, then the sequence formed by adding or subtracting corresponding terms of these APs is also an AP with the same common difference.
- Example: Given two APs \(a_1, a_1+d, a_1+2d, \ldots\) and \(a_2, a_2+d, a_2+2d, \ldots\), adding corresponding terms yields a new AP \(a_1+a_2, a_1+a_2+2d, a_1+a_2+4d, \ldots\), with the same common difference \(d\).

Combinig APs by multiplication and Division

Multiplying or dividing the corresponding terms of two APs does not generally result in a sequence that maintains the properties of an AP. This is because the operation of multiplication or division on the terms from different APs disrupts the uniformity of the common difference, which is fundamental to the definition of an AP.

For instance, consider two APs: \(a, a+d, a+2d, \ldots\) and \(b, b+d', b+2d', \ldots\). Multiplying the \(n\)-th terms of these APs yields \(ab, (a+d)(b+d'), (a+2d)(b+2d'), \ldots\), where the difference between successive terms varies and does not remain constant. Similarly, dividing corresponding terms also leads to a sequence where the ratio of successive terms is not constant, deviating from the characteristic feature of an AP.

Simplified Assumptions for Sums in AP

Sometimes we have to assume a sequence of three three, four, five or more numbers in AP. We follow the following thumb rulesq while making assumptions:

Three Numbers in AP: Prefer \(a-d\), \(a\), \(a+d\) over \(a\), \(a+d\), \(a+2d\).

Four Numbers in AP: Use \(a-3d\), \(a-d\), \(a+d\), \(a+3d\). Common difference effectively \(2d\), centered around \(a\).

Five Numbers in AP: Assume \(a-2d\), \(a-d\), \(a\), \(a+d\), \(a+2d\). Symmetry with \(a\) as central term simplifies calculations.

While it's possible to assume three numbers in an AP as \(a\), \(a+d\), \(a+2d\), choosing \(a-d\), \(a\), \(a+d\) is often more practical, especially when the sum of the numbers is known. This assumption leverages the symmetry around the central term \(a\) for more straightforward calculations and problem-solving.

Example

Problem: Find four terms in an Arithmetic Progression (AP) whose sum is 36, and the sum of their squares is 404, using the assumption \(a-3d\), \(a-d\), \(a+d\), and \(a+3d\).

Solution:

Step 1: Set Up the Terms Based on the Given Assumption

The four terms of the AP are assumed as \(a-3d\), \(a-d\), \(a+d\), and \(a+3d\).

Step 2: Apply the Given Conditions

The sum of the terms is 36: \((a-3d) + (a-d) + (a+d) + (a+3d) = 4a = 36\).
The sum of their squares is 404: \((a-3d)^2 + (a-d)^2 + (a+d)^2 + (a+3d)^2 = 404\).

Step 3: Solve for \(a\)

From the sum condition:

\[4a = 36 \Rightarrow a = 9\]

Step 4: Calculate \(d\) Using the Sum of Squares

Substitute \(a = 9\) into the sum of squares condition and simplify:

\[\begin{align*} (9-3d)^2 + (9-d)^2 + (9+d)^2 + (9+3d)^2 &= 404 \\ 81 - 54d + 9d^2 + 81 - 18d + d^2 + 81 + 18d + d^2 + 81 + 54d + 9d^2 &= 404 \\ 4(81) + 20d^2 &= 404 \\ 324 + 20d^2 &= 404 \\ 20d^2 &= 80 \\ d^2 &= 4 \\ d &= \pm 2 \end{align*}\]

Step 5: Find the Four Terms

Using \(a = 9\) and \(d = 2\) (choosing the positive difference for simplicity):

\(a-3d = 9 - 3(2) = 3\)
\(a-d = 9 - 2 = 7\)
\(a+d = 9 + 2 = 11\)
\(a+3d = 9 + 3(2) = 15\)

Conclusion: The four terms in the AP that meet the given conditions are \(3, 7, 11, 15\), with a sum of 36 and a sum of their squares of 404.

Subsequences in an Arithmetic Progression (AP) Formed by Selecting Terms at Regular Intervals

Given an arithmetic progression (AP) with terms \(a_1, a_2, a_3, a_4, \ldots\), and a common difference \(d\), if we select terms from this AP at regular intervals \(k\), then the selected terms also form an AP. The common difference of the new AP is \(kd\).

Definition of Interval

In this context, an interval \(k\) refers to the fixed number of positions we move forward in the sequence to select the next term. For example, if \(k=2\), we select every second term from the original AP, such as \(a_1, a_3, a_5, \ldots\), skipping one term between selections.

Proof:

Consider the original AP defined by the sequence \(a_1, a_2, a_3, \ldots\), where the \(n^{th}\) term (\(a_n\)) is given by:

\[ a_n = a_1 + (n-1)d \]

Here, \(a_1\) is the first term, \(d\) is the common difference, and \(n\) is the term's position in the sequence.

When selecting terms at intervals \(k\), we choose terms \(a_1, a_{1+k}, a_{1+2k}, \ldots\). The \(m^{th}\) term of this new sequence, denoted as \(a'_m\), corresponds to the term \(a_{1+(m-1)k}\) in the original AP. Therefore, \(a'_m\) can be expressed as:

\[ a'_m = a_1 + \left(1+(m-1)k - 1\right)d = a_1 + (m-1)kd \]

This equation shows that the selected terms form an AP with the first term \(a_1\) and a common difference of \(kd\), proving that any subsequence formed by selecting terms at regular intervals from an AP is itself an AP with a modified common difference.

Common Terms of Two Integer Arithmetic Progressions Form an Arithmetic Progression

Given two integer arithmetic progressions (APs) \(A_1, A_2, A_3, \ldots\) with common difference \(A\) and \(B_1, B_2, B_3, \ldots\) with common difference \(B\), if there exist common terms between these two sequences, then these common terms themselves form an arithmetic progression with a common difference equal to the least common multiple (LCM) of \(A\) and \(B\).

How to find the first common term?

Usually you are asked problems where the first term can be found by observation via writing first few terms of both AP's.

The general way of finding the first common term involves some Number Theory concepts.

Given two APs:

\(a, a+d, a+2d, a+3d, \ldots\)
\(b, b+e, b+2e, b+3e, \ldots\)

Suppose the first common term is the \((m+1)^{th}\) term in the first AP and the \((n+1)^{th}\) term in the second AP, leading to the equation \(a + md = b + ne\).

Rearranging gives:

\[ md - ne = b - a \]

This is a linear Diophantine equation in \(m\) and \(n\), with coefficients being integers.

To find the first common term, we need to solve \(md - ne = b - a\) for \(m\) and \(n\). A solution exists if and only if the greatest common divisor (gcd) of \(d\) and \(-e\) (or equivalently, \(gcd(d, e)\) since gcd is indifferent to the sign) divides \(b-a\).

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