Sequence and Series
Sequences and Series
Arithmetic Progression (AP)
General Term, Summation Dynamics, and Arithmetic MeansTopic 1
An Arithmetic Progression (AP) is a sequence of numbers in which the difference between consecutive terms remains a constant value, known as the common difference ($d$). The general term is defined as $T_n = a + (n-1)d$, where $a$ is the first term. The sum of the first $n$ terms is given by $S_n = \frac{n}{2}[2a + (n-1)d] = \frac{n}{2}(a + l)$, where $l$ is the terminal term. A key structural link is $T_n = S_n - S_{n-1}$.
The Single Arithmetic Mean (AM) between two quantities $a$ and $b$ is $\frac{a+b}{2}$. When inserting $n$ distinct AMs ($A_1, A_2, \dots, A_n$) between $a$ and $b$, the total sequence forms an AP of $n+2$ terms, where the common difference becomes $d = \frac{b-a}{n+1}$, and $A_r = a + r\left(\frac{b-a}{n+1}\right)$. A core property of any AP is that the sum of terms equidistant from the beginning and end remains constant: $a_m + a_n = a_p + a_q$ whenever $m+n = p+q$.
The sum of the first $n$ terms of a sequence is given by $S_n = 3n^2 + 5n$. Prove that the sequence is an AP and find its $15^{\text{th}}$ term.
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We isolate the general term $T_n$ using the fundamental recurrence relationship $T_n = S_n - S_{n-1}$:
\[ T_n = (3n^2 + 5n) - [3(n-1)^2 + 5(n-1)] \]
Expand the bracketed terms systematically:
\[ T_n = 3n^2 + 5n - [3(n^2 - 2n + 1) + 5n - 5] \]
\[ T_n = 3n^2 + 5n - [3n^2 - 6n + 3 + 5n - 5] = 3n^2 + 5n - 3n^2 + x - 2 \]
Simplifying the linear expression yields:
\[ T_n = 6n + 2 \]
Since the general term $T_n$ is a linear expression in $n$, the first difference $T_n - T_{n-1} = (6n+2) - (6(n-1)+2) = 6$, which is a constant independent of $n$. This proves the sequence forms a valid AP with common difference $d = 6$.
To find the $15^{\text{th}}$ term ($T_{15}$), substitute $n = 15$ into our linear formula:
\[ T_{15} = 6(15) + 2 = 90 + 2 = 92 \]
Final Answer: $92$.
Insert 11 arithmetic means between 23 and 71, and find the value of the $5^{\text{th}}$ arithmetic mean ($A_5$).
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Let $a = 23$ and $b = 71$. We are instructed to insert $n = 11$ AMs between these limits.
Step 1: Calculate the common difference $d$ of the resulting progression:
\[ d = \frac{b - a}{n + 1} = \frac{71 - 23}{11 + 1} = \frac{48}{12} = 4 \]
Step 2: The $5^{\text{th}}$ arithmetic mean $A_5$ corresponds exactly to the $6^{\text{th}}$ term ($T_6$) of the total expanded AP:
\[ A_5 = a + 5d \]
Substitute the values of $a$ and $d$:
\[ A_5 = 23 + 5(4) = 23 + 20 = 43 \]
Final Answer: $43$.
If the sum of the first $p$ terms of an AP equals the sum of the first $q$ terms ($p \neq q$), prove that the sum of the first $(p+q)$ terms is exactly zero.
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Write out the structural identities using the standard sum formula: \[ S_p = S_q \implies \frac{p}{2}[2a + (p-1)d] = \frac{q}{2}[2a + (q-1)d] \] Cancel the common denominator 2 and expand the bracketed terms: \[ 2ap + p(p-1)d = 2aq + q(q-1)d \] Rearrange all terms to the left-hand side to factor the expression: \[ 2a(p - q) + [p^2 - p - (q^2 - q)]d = 0 \] \[ 2a(p - q) + [(p^2 - q^2) - (p - q)]d = 0 \] Factor out the common linear binomial $(p - q)$, which is valid since $p \neq q \implies p - q \neq 0$: \[ (p - q)[2a + (p + q - 1)d] = 0 \] Divide by $(p-q)$ to find the underlying core condition: \[ 2a + (p + q - 1)d = 0 \] Now write out the formula for the sum of the first $(p+q)$ terms: \[ S_{p+q} = \frac{p+q}{2} \cdot \left[ 2a + (p + q - 1)d \right] \] Substitute our core condition into the brackets: \[ S_{p+q} = \frac{p+q}{2} \cdot (0) = 0 \] Final Answer: Proved.
Geometric Progression (GP)
General Term, Infinite Progressions, and Mean InequalitiesTopic 1
A Geometric Progression (GP) is a sequence where the ratio of any term to its preceding term remains a constant multiplier, known as the common ratio ($r$). The general term is defined as $T_n = a r^{n-1}$. The sum of the first $n$ terms is given by $S_n = \frac{a(r^n-1)}{r-1}$ for $r \neq 1$, and dynamically drops to $S_n = na$ if $r = 1$. When the common ratio is bounded within the open interval $|r| < 1$, the sum of an infinite number of terms converges to the limiting value: $S_\infty = \frac{a}{1-r}$.
The Single Geometric Mean (GM) between two positive numbers $a$ and $b$ is $\sqrt{ab}$. When inserting $n$ GMs ($G_1, G_2, \dots, G_n$) between $a$ and $b$, the total product of these inserted means equals the $n^{\text{th}}$ power of their single GM: $\prod_{i=1}^n G_i = (\sqrt{ab})^n$.
A core algebraic property for positive real numbers is the AM--GM Inequality: $\frac{a+b}{2} \ge \sqrt{ab}$, where equality holds true if and only if the constituent numbers are perfectly identical ($a = b$).
Express the recurring pure decimal value $0.474747\dots$ as a simplified rational fraction using infinite geometric progression algorithms.
Show solution
- First term $a = \frac{47}{100}$
- Common ratio $r = \frac{47/10000}{47/100} = \frac{1}{100}$
Find the minimum value of the structural expression $E = 4^x + 4^{1-x}$ for all $x \in \mathbb{R}$ using mean inequality properties.
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Notice that both terms ($4^x$ and $4^{1-x}$) are strictly positive real numbers for any real value of $x$. This allows us to apply the AM--GM inequality directly: \[ \frac{4^x + 4^{1-x}}{2} \ge \sqrt{4^x \cdot 4^{1-x}} \] Simplify the product inside the radical using standard rules of indices: \[ 4^x \cdot 4^{1-x} = 4^{x + 1 - x} = 4^1 = 4 \] Substitute this back into the inequality: \[ \frac{4^x + 4^{1-x}}{2} \ge \sqrt{4} \implies \frac{4^x + 4^{1-x}}{2} \ge 2 \] Multiply both sides of the inequality by 2 to isolate our target expression: \[ 4^x + 4^{1-x} \ge 4 \] Therefore, the minimum value that the expression can achieve is 4. This minimum is reached when the two terms are identical: $4^x = 4^{1-x} \implies x = 1-x \implies 2x = 1 \implies x = 0.5$. Final Answer: $4$.
Insert 3 geometric means between 2 and 162, and find the value of the second inserted mean ($G_2$).
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Let $a = 2$ and $b = 162$. We are instructed to insert $n = 3$ geometric means between these boundaries.
Step 1: Calculate the common ratio $r$ of the resulting geometric progression:
\[ r = \left(\frac{b}{a}\right)^{\frac{1}{n+1}} = \left(\frac{162}{2}\right)^{\frac{1}{3+1}} = (81)^{\frac{1}{4}} \]
Since $81 = 3^4$, the common ratio is:
\[ r = (3^4)^{1/4} = 3 \]
Step 2: The second geometric mean $G_2$ corresponds to the third term ($T_3$) of the total expanded sequence:
\[ G_2 = a \cdot r^2 \]
Substitute the values of $a$ and $r$:
\[ G_2 = 2 \cdot (3)^2 = 2 \cdot 9 = 18 \]
Final Answer: $18$.
Harmonic Progressions & Mixed Sequences
Harmonic Means, Extended Inequalities, and AGP SumsTopic 1
A Harmonic Progression (HP) is defined as a sequence where the reciprocals of the terms form an Arithmetic Progression (AP). Thus, if $H_n$ is an HP, then $\frac{1}{H_n} = \frac{1}{a} + (n-1)d$. The Single Harmonic Mean (HM) between two numbers $a$ and $b$ is $H = \frac{2ab}{a+b}$. Combining our mean definitions fields the complete classical inequality chain for positive real numbers: \[ \mathbf{\text{AM} \ge \text{GM} \ge \text{HM} \implies \frac{a+b}{2} \ge \sqrt{ab} \ge \frac{2ab}{a+b}} \] where the equality condition holds true if and only if $a=b$.
An Arithmetico-Geometric Progression (AGP) is a mixed sequence formed by multiplying corresponding terms of an AP and a GP together. It takes the general form: $a, (a+d)r, (a+2d)r^2, \dots$. We find the sum of an AGP using the multiply-by-$r$-and-subtract method, which shifts terms along the sequence to isolate a clean geometric series. The sum of an infinite convergent AGP ($|r| < 1$) is given by the formula $S_\infty = \frac{a}{1-r} + \frac{dr}{(1-r)^2}$.
Find the sum to infinity of the Arithmetico-Geometric Progression: \[ S_\infty = 1 + \frac{4}{3} + \frac{7}{9} + \frac{10}{27} + \dots \]
Show solution
Let us first write out the series clearly to identify the underlying progression components:
\[ S_\infty = 1 \cdot 1 + 4 \cdot \left(\frac{1}{3}\right) + 7 \cdot \left(\frac{1}{9}\right) + 10 \cdot \left(\frac{1}{27}\right) + \dots \]
The arithmetic components are $1, 4, 7, 10, \dots$, which form an AP with first term $a = 1$ and common difference $d = 3$. The geometric components are $1, \frac{1}{3}, \frac{1}{9}, \frac{1}{27}, \dots$, which form a GP with common ratio $r = \frac{1}{3}$. Since $|r| = \frac{1}{3} < 1$, the infinite series converges.
Let us use the explicit formula $S_\infty = \frac{a}{1-r} + \frac{dr}{(1-r)^2}$ to calculate the sum:
\[ S_\infty = \frac{1}{1 - \frac{1}{3}} + \frac{3 \cdot \left(\frac{1}{3}\right)}{\left(1 - \frac{1}{3}\right)^2} \]
Simplify each component fraction independently:
\[ \text{First term} = \frac{1}{\frac{2}{3}} = \frac{3}{2} \]
\[ \text{Second term} = \frac{1}{\left(\frac{2}{3}\right)^2} = \frac{1}{\frac{4}{9}} = \frac{9}{4} \]
Sum the values using a common denominator:
\[ S_\infty = \frac{3}{2} + \frac{9}{4} = \frac{6}{4} + \frac{9}{4} = \frac{15}{4} \]
Final Answer: $\frac{15}{4}$.
If the AM between two positive numbers is 5 and their HM is 3.2, find the value of their Geometric Mean (GM) and determine the numbers.
Show solution
Step 1: Apply the fundamental relationship that connects the three classical means: $\text{GM}^2 = \text{AM} \times \text{HM}$. Substitute the given values into the equation: \[ \text{GM}^2 = 5 \times 3.2 = 16 \implies \text{GM} = \sqrt{16} = 4 \] Step 2: Set up equations for the individual numbers $a$ and $b$ using the definitions of AM and GM: \[ \text{AM} = \frac{a+b}{2} = 5 \implies a + b = 10 \] \[ \text{GM} = \sqrt{ab} = 4 \implies ab = 16 \] Step 3: Substitute $b = 10 - a$ into the product equation to form a quadratic equation: \[ a(10 - a) = 16 \implies 10a - a^2 = 16 \implies a^2 - 10a + 16 = 0 \] Factor the quadratic equation: \[ (a - 8)(a - 2) = 0 \implies a = 8 \quad \text{or} \quad a = 2 \] This gives the paired numbers solution: $\{2, 8\}$. Final Answer: $\text{GM} = 4$; the numbers are $2$ and $8$.
If $a, b, c$ form a valid Harmonic Progression, prove that $\frac{a}{c} = \frac{a-b}{b-c}$.
Show solution
Since $a, b, c$ form an HP, their reciprocals $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ must form an AP by definition. Equate the common differences between consecutive terms: \[ \frac{1}{b} - \frac{1}{a} = \frac{1}{c} - \frac{1}{b} \] Simplify the expressions on both sides over common denominators: \[ \frac{a - b}{ab} = \frac{b - c}{bc} \] Rearrange the fractions to group the difference terms together: \[ \frac{a - b}{b - c} = \frac{ab}{bc} \] Cancel out the common factor $b$ from the numerator and denominator on the right-hand side: \[ \frac{a - b}{b - c} = \frac{a}{c} \] This completes the proof. Final Answer: Proved.
Special Series & Summation Methods
Standard Sums, Telescoping Series, and Method of DifferencesTopic 1
Evaluating advanced series requires a mastery of specialized summation methods. First, memorize the standard polynomial sums for the first $n$ natural numbers: \[ \sum_{k=1}^n k = \frac{n(n+1)}{2}, \quad \sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}, \quad \sum_{k=1}^n k^3 = \left[\frac{n(n+1)}{2}\right]^2 \] The Telescoping Series method (Difference Method) is a powerful technique for simplifying complex fractions or products. It involves splitting the general term $T_k$ into a difference of two consecutive functional terms: $T_k = f(k) - f(k-1)$. When summing these terms together, intermediate values cancel out in a cascading fashion, leaving only the first boundary term and the last boundary term: $\sum_{k=1}^n T_k = f(n) - f(0)$.
The $V_n$ Method applies this strategy specifically to products or fractions of terms in an AP, allowing us to find closed-form sums quickly for competitive exams.
Find the sum of the first $n$ terms of the series whose general $k^{\text{th}}$ term is defined by $T_k = k(k+1)(k+2)$.
Show solution
Let us expand the general term into a standard polynomial form before applying our summation rules: \[ T_k = k(k^2 + 3k + 2) = k^3 + 3k^2 + 2k \] Now write the total sum $S_n$ using linearity properties to separate the individual polynomial sums: \[ S_n = \sum_{k=1}^n T_k = \sum_{k=1}^n k^3 + 3 \cdot \sum_{k=1}^n k^2 + 2 \cdot \sum_{k=1}^n k \] Substitute the standard formulas for each polynomial sum: \[ S_n = \left[\frac{n(n+1)}{2}\right]^2 + 3 \cdot \left[\frac{n(n+1)(2n+1)}{6}\right] + 2 \cdot \left[\frac{n(n+1)}{2}\right] \] Simplify the constant fractional multipliers: \[ S_n = \frac{n^2(n+1)^2}{4} + \frac{n(n+1)(2n+1)}{2} + n(n+1) \] Factor out the common term $\frac{n(n+1)}{4}$ from all three components: \[ S_n = \frac{n(n+1)}{4} \left[ n(n+1) + 2(2n+1) + 4 \right] \] Expand and simplify the expression inside the brackets: \[ n^2 + n + 4n + 2 + 4 = n^2 + 5n + 6 \] Factor the resulting quadratic expression: \[ n^2 + 5n + 6 = (n+2)(n+3) \] Substitute this back to find the final formatted product expression for the sum: \[ S_n = \frac{n(n+1)(n+2)(n+3)}{4} \] This matches the standard result derived using the $V_n$ method. Final Answer: \(\frac{n(n+1)(n+2)(n+3)}{4}\).
Find the sum to $n$ terms of the infinite telescoping series: \[ S_n = \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \frac{1}{3 \cdot 4} + \dots + \frac{1}{n(n+1)} \]
Show solution
Identify the general term of the series: \[ T_k = \frac{1}{k(k+1)} \] Split the fraction into a difference of two partial fractions by rewriting the numerator $1$ as $(k+1) - k$: \[ T_k = \frac{(k+1) - k}{k(k+1)} = \frac{k+1}{k(k+1)} - \frac{k}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1} \] This matches the telescoping form $T_k = f(k) - f(k+1)$, where $f(k) = \frac{1}{k}$. Write out the terms of the sum sequentially: \[ S_n = \left(1 - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{4}\right) + \dots + \left(\frac{1}{n} - \frac{1}{n+1}\right) \] Notice that the intermediate terms cancel out in a cascading fashion: $-\frac{1}{2}$ cancels with $+\frac{1}{2}$, $-\frac{1}{3}$ cancels with $+\frac{1}{3}$, and so on. After all cancellations, only the very first and the very last terms remain: \[ S_n = 1 - \frac{1}{n+1} \] Simplify the fraction over a common denominator: \[ S_n = \frac{(n+1) - 1}{n+1} = \frac{n}{n+1} \] Notice that as $n \to \infty$, the sum converges to $\lim_{n \to \infty} \frac{n}{n+1} = 1$. Final Answer: \(\frac{n}{n+1}\).
Evaluate the value of the squared series sum $S = 1^2 + 2^2 + 3^2 + \dots + 10^2$.
Show solution
This problem requires calculating the sum of the squares of the first $n = 10$ natural numbers. Apply the standard polynomial sum formula: \[ \sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6} \] Substitute $n = 10$ into the formula: \[ S = \frac{10 \cdot (10+1) \cdot (2(10)+1)}{6} = \frac{10 \cdot 11 \cdot 21}{6} \] Simplify the numbers before multiplying: divide 21 and 6 by 3, then divide 10 and 2 by 2: \[ S = \frac{10 \cdot 11 \cdot 7}{2} = 5 \cdot 11 \cdot 7 = 55 \cdot 7 = 385 \] Final Answer: $385$.
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