Chapter 8 — Sequences and Series

Theory Insight

A Geometric Progression (G.P.) is a sequence where each term is obtained by multiplying the previous term by a constant ratio \(r\).

General form: \[ a,\; ar,\; ar^2,\; ar^3,\; \ldots \]

The nth term of a G.P. is: \[ a_n = a r^{n-1} \]

Key observation: If terms are decreasing, then \(0 < r < 1\).

The shrinking circles visually indicate a decreasing G.P. where each term is multiplied by a factor less than 1.

Solution Roadmap

Step 1: Identify first term \(a\)
Step 2: Compute common ratio \(r = \dfrac{a_2}{a_1}\)
Step 3: Use formula \(a_n = ar^{n-1}\)
Step 4: Substitute required value of \(n\)

Solution

Given G.P.: \(\dfrac{5}{2}, \dfrac{5}{4}, \dfrac{5}{8}, \ldots\)

First term: \[ a = \dfrac{5}{2} \]

Common ratio: \[ r = \frac{5/4}{5/2} = \frac{1}{2} \]

Using formula: \[ a_n = a r^{n-1} \]

General term: \[ a_n = \frac{5}{2} \left(\frac{1}{2}\right)^{n-1} = \frac{5}{2^n} \]

20th term: \[ a_{20} = \frac{5}{2^{20}} \]

Final Answer

\[ a_n = \frac{5}{2^n}, \quad a_{20} = \frac{5}{2^{20}} \]

Exam Significance

For Board Exams: This is a direct formula-based question testing your ability to identify \(a\) and \(r\) correctly. Common mistakes include incorrect ratio calculation.

For JEE/NEET: This concept is foundational. It is frequently used in:

• Sum of G.P.
• Infinite series
• Growth and decay models
• Logarithmic transformations

Mastery here ensures faster solving in advanced mixed problems.

Theory Insight

In a Geometric Progression (G.P.), each term is related to another using powers of the common ratio.

General term: \[ a_n = a r^{n-1} \]

A powerful transformation used in exams: \[ a_n = a_m \cdot r^{(n-m)} \]

This avoids finding the first term explicitly and is highly efficient in competitive exams.

Increasing circle sizes represent exponential growth when \(r > 1\).

Solution Roadmap

Step 1: Use relation between two terms of G.P.
Step 2: Apply \(a_n = a_m \cdot r^{(n-m)}\)
Step 3: Substitute given values directly
Step 4: Simplify using powers of 2

Solution

Given: \[ a_8 = 192,\quad r = 2 \]

Using relation between terms: \[ \begin{aligned} a_{12} &= a_8 \cdot r^{12-8}\\ a_{12} &= 192 \cdot 2^4 \\ a_{12} &= 192 \cdot 16 \\ &= 3072 \end{aligned} \]

Final Answer

\[ a_{12} = 3072 \]

Exam Significance

For Board Exams: This question checks your understanding of the nth term formula and manipulation of exponents.

For JEE/NEET: The transformation \(a_n = a_m \cdot r^{(n-m)}\) is extremely important for:

• Time-saving calculations
• Multi-step G.P. problems
• Logarithmic and exponential equations

Direct substitution without finding the first term is a key competitive advantage.

Theory Insight

In a Geometric Progression, terms follow an exponential pattern:

\[ a_n = a r^{n-1} \]

A key structural property:

Terms equally spaced in a G.P. satisfy: \[ (\text{middle term})^2 = (\text{left term}) \times (\text{right term}) \]

This is analogous to the concept of geometric mean.

The middle term acts as a geometric mean of the surrounding terms.

Solution Roadmap

Step 1: Express each given term using \(a_n = ar^{n-1}\)
Step 2: Write algebraic forms of \(p, q, s\)
Step 3: Compute \(q^2\) and \(ps\)
Step 4: Compare both expressions

Solution

Given: \[ a_5 = p,\quad a_8 = q,\quad a_{11} = s \]

Using \(a_n = ar^{n-1}\):

\[ a_5 = ar^4 = p \] \[ a_8 = ar^7 = q \] \[ a_{11} = ar^{10} = s \]

Now,

\[ q^2 = (ar^7)^2 = a^2 r^{14} \]

And,

\[ ps = (ar^4)(ar^{10}) = a^2 r^{14} \]

Hence, \[ q^2 = ps \]

Final Result

\[ \boxed{q^2 = ps} \]

Exam Significance

For Board Exams: This is a standard proof question testing algebraic manipulation and understanding of G.P. structure.

For JEE/NEET: This result is extremely useful in:

• Finding missing terms in G.P.
• Geometric mean problems
• Logarithmic transformations
• Sequence-based proofs

Recognizing this pattern instantly can save significant time in objective questions.

Theory Insight

In a G.P., each term is expressed as: \[ a_n = a r^{n-1} \]

When a condition relates two different terms (like square, product, ratio), always convert both into powers of \(r\). This converts the problem into an algebraic equation.

Key idea: \[ a_m = (a_n)^k \;\Rightarrow\; \text{convert both sides into powers of } r \]

The highlighted term shows how one term can depend algebraically on another.

Solution Roadmap

Step 1: Write expressions for \(a_2\) and \(a_4\)
Step 2: Apply given condition
Step 3: Solve for common ratio \(r\)
Step 4: Substitute into \(a_7 = ar^6\)

Solution

Given: \[ a = -3 \]

Second term: \[ a_2 = ar = -3r \]

Fourth term: \[ a_4 = ar^3 = -3r^3 \]

According to the condition: \[ a_4 = (a_2)^2 \]

\[ -3r^3 = (-3r)^2 \]

\[ -3r^3 = 9r^2 \]

Rearranging: \[ -3r^3 - 9r^2 = 0 \]

\[ -3r^2(r + 3) = 0 \]

Possible values: \[ r = 0 \quad \text{or} \quad r = -3 \]

If \(r = 0\), the G.P. becomes trivial after first term, which is not meaningful in this context.

Hence, \[ r = -3 \]

Now, \[ a_7 = ar^6 = -3 \cdot (-3)^6 \]

\[ (-3)^6 = 729 \]

\[ a_7 = -3 \cdot 729 = -2187 \]

Final Answer

\[ a_7 = -2187 \]

Exam Significance

For Board Exams: Tests your ability to translate word conditions into algebraic equations.

For JEE/NEET: This pattern appears frequently in:

• Exponent comparison problems
• Functional equations in sequences
• Mixed algebra + sequence questions

Critical skill: Converting terms into powers of \(r\) and solving efficiently.

Theory Insight

To find the position of a term in a G.P., use: \[ a_n = a r^{n-1} \]

The most efficient method is to convert all quantities into the same base (like powers of 2 or 3), then compare exponents.

Key exam idea: \[ \text{Match powers} \;\Rightarrow\; \text{equate exponents} \]

Each jump represents multiplication by a constant ratio.

Solution Roadmap

Step 1: Identify \(a\) and \(r\)
Step 2: Write \(a_n\)
Step 3: Convert RHS into same base
Step 4: Equate powers and solve for \(n\)

Solution

(a)

\[ a = 2,\quad r = \sqrt{2} \]

\[ a_n = 2(\sqrt{2})^{n-1} \]

Given \(a_n = 128 = 2^7\):

\[ 2(\sqrt{2})^{n-1} = 2^7 \]

\[ (\sqrt{2})^{n-1} = 2^6 = (2^{1/2})^{12} \]

\[ n-1 = 12 \Rightarrow n = 13 \]

(b)

\[ a = \sqrt{3},\quad r = \sqrt{3} \]

\[ a_n = (\sqrt{3})^n \]

Given \(a_n = 729 = 3^6 = (3^{1/2})^{12}\):

\[ (\sqrt{3})^n = (\sqrt{3})^{12} \]

\[ n = 12 \]

(c)

\[ a = \frac{1}{3},\quad r = \frac{1}{3} \]

\[ a_n = \left(\frac{1}{3}\right)^n \]

Given: \[ \frac{1}{19683} = \frac{1}{3^9} = \left(\frac{1}{3}\right)^9 \]

\[ n = 9 \]

Final Answers

(a) \(13^{th}\) term
(b) \(12^{th}\) term
(c) \(9^{th}\) term

Exam Significance

For Board Exams: Tests your ability to manipulate exponents and identify G.P. structure correctly.

For JEE/NEET: This is a high-frequency pattern used in:

• Logarithmic equations
• Exponential growth/decay problems
• Mixed sequence problems

Fast base-conversion and exponent comparison is a major time-saving skill.

Theory Insight

Three numbers \(a,\;b,\;c\) are in G.P. if: \[ b^2 = ac \]

This is the fastest method and avoids dealing with ratios.

Alternative condition: \[ \frac{b}{a} = \frac{c}{b} \]

Both approaches lead to the same result, but the first is preferred in exams.

The middle term behaves as the geometric mean of the first and third terms.

Solution Roadmap

Step 1: Apply condition \(b^2 = ac\)
Step 2: Substitute given values
Step 3: Solve resulting equation
Step 4: State all valid solutions

Solution

Given numbers: \[ a = -\frac{2}{7},\quad b = x,\quad c = -\frac{7}{2} \]

Using condition: \[ b^2 = ac \]

\[ x^2 = \left(-\frac{2}{7}\right)\left(-\frac{7}{2}\right) \]

\[ x^2 = 1 \]

\[ x = \pm 1 \]

Hence, both values satisfy the G.P. condition.

Alternative Method (Ratio Approach)

\[ \frac{x}{-\frac{2}{7}} = \frac{-\frac{7}{2}}{x} \]

Cross-multiplying:

\[ x^2 = 1 \]

\[ x = \pm 1 \]

Final Answer

\[ x = 1 \quad \text{or} \quad x = -1 \]

Exam Significance

For Board Exams: Direct application of geometric mean property.

For JEE/NEET: This concept is frequently used in:

• Finding missing terms in sequences
• Logarithmic transformations
• Quadratic formation from sequences

Always prefer \(b^2 = ac\) for fastest solving in objective questions.

Theory Insight

The sum of first \(n\) terms of a G.P. is: \[ S_n = \frac{a(1 - r^n)}{1 - r}, \quad r \neq 1 \]

When \(0 < r < 1\), the term \(r^n\) becomes extremely small for large \(n\), which allows approximation: \[ S_n \approx \frac{a}{1 - r} \]

Rapid shrinking shows why higher powers of \(r\) become negligible.

Solution Roadmap

Step 1: Identify \(a\) and \(r\)
Step 2: Apply sum formula
Step 3: Simplify expression
Step 4: Use approximation if needed

Solution

Given: \[ a = 0.15,\quad r = \frac{0.015}{0.15} = 0.1 = \frac{1}{10},\quad n = 20 \]

Using formula: \[ S_{20} = \frac{0.15(1 - (0.1)^{20})}{1 - 0.1} \]

\[ S_{20} = \frac{0.15(1 - 10^{-20})}{0.9} \]

\[ \begin{aligned} S_{20} &= \frac{0.15 \times 10}{9} (1 - 10^{-20})\\ &= \frac{1.5}{9}(1 - 10^{-20})\\ &= \frac{1}{6}(1 - 10^{-20}) \end{aligned} \]

Since \(10^{-20}\) is extremely small,

\[ S_{20} \approx \frac{1}{6} \]

Final Answer

\[ S_{20} = \frac{1}{6}(1 - 10^{-20}) \approx \frac{1}{6} \]

Exam Significance

For Board Exams: Focus on correct substitution and simplification.

For JEE/NEET: Key takeaway is recognizing when a G.P. behaves like an infinite series.

• Used in convergence problems
• Important in calculus (limits)
• Appears in physics decay models

Spotting negligible terms quickly saves time in objective exams.

Theory Insight

Sum of first \(n\) terms of a G.P.: \[ S_n = \frac{a(r^n - 1)}{r - 1}, \quad r \neq 1 \]

When radicals are involved, simplify ratios first: \[ \frac{\sqrt{a}}{\sqrt{b}} = \sqrt{\frac{a}{b}} \]

Rationalization is often required for final simplification.

Increasing size indicates \(r > 1\), hence growth in the series.

Solution Roadmap

Step 1: Identify \(a\)
Step 2: Compute \(r\) using radicals
Step 3: Apply sum formula
Step 4: Rationalize denominator

Solution

First term: \[ a = \sqrt{7} \]

Common ratio: \[ r = \frac{\sqrt{21}}{\sqrt{7}} = \sqrt{3} \]

Using sum formula: \[ S_n = \frac{\sqrt{7}\left((\sqrt{3})^n - 1\right)}{\sqrt{3} - 1} \]

Rationalizing denominator:

\[ S_n = \frac{\sqrt{7}\left((\sqrt{3})^n - 1\right)(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)} \]

\[ S_n = \frac{\sqrt{7}(\sqrt{3}+1)\left((\sqrt{3})^n - 1\right)}{2} \]

Final Answer

\[ S_n = \frac{\sqrt{7}(\sqrt{3}+1)\left((\sqrt{3})^n - 1\right)}{2} \]

Exam Significance

For Board Exams: Focus on correct handling of surds and rationalization.

For JEE/NEET: This type is important for:

• Radical-based G.P. problems
• Exponential simplifications
• Algebraic manipulation speed

Quick simplification of roots into powers gives a major time advantage.

Theory Insight

Sum of first \(n\) terms of a G.P.: \[ S_n = \frac{a(1 - r^n)}{1 - r}, \quad r \neq 1 \]

When signs alternate, the common ratio is negative.

Important identity: \[ (-a)^n = \begin{cases} a^n, & \text{if } n \text{ is even} \ -a^n, & \text{if } n \text{ is odd} \end{cases} \]

This plays a crucial role in simplification and MCQs.

Alternating colors represent sign changes due to negative ratio.

Solution Roadmap

Step 1: Identify first term \(a_1\)
Step 2: Find common ratio \(r\)
Step 3: Apply sum formula
Step 4: Simplify carefully with signs

Solution

First term: \[ a_1 = 1 \]

Common ratio: \[ r = \frac{-a}{1} = -a \]

Using sum formula: \[ S_n = \frac{1(1 - (-a)^n)}{1 - (-a)} \]

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

Final Answer

\[ S_n = \frac{1 - (-a)^n}{1 + a}, \quad (a \ne -1) \]

Exam Significance

For Board Exams: Focus on correct substitution and sign handling.

For JEE/NEET: This pattern is frequently tested in:

• Alternating series problems
• Parity-based simplifications (even/odd \(n\))
• Advanced algebraic manipulations

Recognizing \((-a)^n\) behavior quickly can significantly reduce solving time.

Theory Insight

Sum of first \(n\) terms of a G.P.: \[ S_n = \frac{a(1 - r^n)}{1 - r}, \quad r \neq 1 \]

When powers follow a pattern like: \[ x^3,\; x^5,\; x^7,\; \ldots \] observe that exponents increase linearly. This helps identify the common ratio quickly.

Key observation: \[ x^{2k+1} \Rightarrow \text{common ratio } = x^2 \]

Increasing size reflects multiplication by \(x^2\) at each step.

Solution Roadmap

Step 1: Identify first term \(a\)
Step 2: Compute common ratio \(r\)
Step 3: Apply sum formula
Step 4: Simplify power expression

Solution

First term: \[ a = x^3 \]

Common ratio: \[ r = \frac{x^5}{x^3} = x^2 \]

Using sum formula: \[ S_n = \frac{x^3(1 - (x^2)^n)}{1 - x^2} \]

Simplifying: \[ S_n = \frac{x^3(1 - x^{2n})}{1 - x^2} \]

Final Answer

\[ S_n = \frac{x^3(1 - x^{2n})}{1 - x^2}, \quad (x \ne \pm 1) \]

Exam Significance

For Board Exams: Tests pattern recognition and correct application of G.P. formula.

For JEE/NEET: This structure appears in:

• Power series problems
• Algebraic summations
• Function-based sequences

Recognizing exponent patterns quickly leads to faster solving.

Theory Insight

Summation is linear: \[ \sum (a + b) = \sum a + \sum b \]

Key formulas:

Constant sum: \[ \sum_{k=1}^{n} c = cn \]

Geometric sum: \[ \sum_{k=1}^{n} r^k = \frac{r(r^n - 1)}{r - 1}, \quad r \neq 1 \]

This allows fast evaluation without expanding terms.

The exponential part grows rapidly, dominating the sum.

Solution Roadmap

Step 1: Split summation
Step 2: Evaluate constant sum
Step 3: Apply GP formula
Step 4: Combine results

Solution

\[ \sum_{k=1}^{11}(2+3^k) = \sum_{k=1}^{11}2 + \sum_{k=1}^{11}3^k \]

First part: \[ \sum_{k=1}^{11}2 = 2 \times 11 = 22 \]

Second part: \[ \sum_{k=1}^{11}3^k = \frac{3(3^{11} - 1)}{3 - 1} \]

\[ = \frac{3(3^{11} - 1)}{2} \]

Total sum: \[ 22 + \frac{3(3^{11} - 1)}{2} \]

\[ = 22 + \frac{3^{12} - 3}{2} \]

\[ \begin{aligned} &= \frac{44 + 3^{12} - 3}{2}\\ &= \frac{3^{12} + 41}{2} \end{aligned} \]

Since \(3^{12} = 531441\),

\[ \begin{aligned} \frac{531441 + 41}{2} &= \frac{531482}{2} \\&= 265741 \end{aligned} \]

Final Answer

\[ 265741 \]

Exam Significance

For Board Exams: Tests summation splitting and correct formula usage.

For JEE/NEET: Frequently appears in:

• Mixed summation problems
• Series decomposition techniques
• Fast calculation-based MCQs

Recognizing separable sums is a major time-saving strategy.

Theory Insight

A standard symmetric representation of three consecutive terms of a G.P. is: \[ \frac{a}{r},\; a,\; ar \]

This form simplifies both sum and product calculations:

Product: \[ \frac{a}{r} \cdot a \cdot ar = a^3 \]

This symmetry is a powerful shortcut in exams.

The middle term acts as a balancing center of the progression.

Solution Roadmap

Step 1: Assume terms as \(\frac{a}{r}, a, ar\)
Step 2: Use product to find \(a\)
Step 3: Substitute into sum equation
Step 4: Solve quadratic in \(r\)

Solution

Let the three terms be: \[ \frac{a}{r},\; a,\; ar \]

Product condition: \[ \frac{a}{r} \cdot a \cdot ar = a^3 = 1 \]

\[ a = 1 \]

Sum condition: \[ \frac{1}{r} + 1 + r = \frac{39}{10} \]

Multiply by \(r\): \[ 1 + r + r^2 = \frac{39r}{10} \]

Multiply by 10: \[ 10r^2 + 10r + 10 - 39r = 0 \]

\[ 10r^2 - 29r + 10 = 0 \]

Factorizing: \[ (2r - 5)(5r - 2) = 0 \]

\[ r = \frac{5}{2} \quad \text{or} \quad r = \frac{2}{5} \]

Corresponding terms:

If \(r = \frac{5}{2}\): \[ \frac{2}{5},\; 1,\; \frac{5}{2} \]

If \(r = \frac{2}{5}\): \[ \frac{5}{2},\; 1,\; \frac{2}{5} \]

Final Answer

\[ r = \frac{5}{2} \;\text{or}\; \frac{2}{5} \]

Terms: \[ \left(\frac{2}{5},\;1,\;\frac{5}{2}\right) \quad \text{or} \quad \left(\frac{5}{2},\;1,\;\frac{2}{5}\right) \]

Exam Significance

For Board Exams: Tests correct formation of G.P. and algebraic simplification.

For JEE/NEET: This pattern is highly important in:

• Symmetric substitution problems
• Quadratic formation from sequences
• Multi-condition sequence problems

Using \(\frac{a}{r}, a, ar\) instead of \(a, ar, ar^2\) significantly reduces computation.

Theory Insight

Sum of first \(n\) terms of a G.P.: \[ S_n = \frac{a(r^n - 1)}{r - 1}, \quad r \neq 1 \]

For exponential sequences like \(3, 3^2, 3^3,\ldots\), rewrite everything in powers of the same base.

Key idea: \[ \text{Convert equation into } r^n = \text{constant and compare powers} \]

Rapid growth shows why only a few terms are needed to reach large sums.

Solution Roadmap

Step 1: Identify \(a\) and \(r\)
Step 2: Apply sum formula
Step 3: Simplify equation
Step 4: Convert to powers and solve for \(n\)

Solution

Given: \[ a = 3,\quad r = 3,\quad S_n = 120 \]

Using sum formula: \[ 120 = \frac{3(3^n - 1)}{3 - 1} \]

\[ 120 = \frac{3(3^n - 1)}{2} \]

Multiply by 2: \[ 240 = 3(3^n - 1) \]

Divide by 3: \[ 80 = 3^n - 1 \]

\[ 3^n = 81 \]

\[ 81 = 3^4 \Rightarrow n = 4 \]

Final Answer

\[ n = 4 \]

Exam Significance

For Board Exams: Direct application of GP sum formula.

For JEE/NEET: This type appears frequently in:

• Exponential equation solving
• Logarithmic transformations
• Series-based equation problems

Fast conversion into powers (like \(3^n\)) is the key to speed.

Theory Insight

Sum of first \(n\) terms: \[ S_n = \frac{a(1 - r^n)}{1 - r}, \quad r \neq 1 \]

Important shortcut:

Sum of terms 4 to 6 = \[ S_6 - S_3 = r^3 \cdot S_3 \]

This avoids heavy algebra and is extremely useful in exams.

Later groups scale by powers of \(r\).

Solution Roadmap

Step 1: Use relation between grouped sums
Step 2: Find \(r\)
Step 3: Substitute into \(S_3\) to find \(a\)
Step 4: Write general sum formula

Solution

Given: \[ \begin{aligned} S_3 &= 16,\\ \text{Sum of next 3 terms} &= 128 \end{aligned} \]

Using shortcut: \[ \text{Sum of terms 4–6} = r^3 \cdot S_3 \]

\[ 128 = r^3 \cdot 16 \]

\[ r^3 = 8 \Rightarrow r = 2 \]

Now using: \[ S_3 = \frac{a(1 - r^3)}{1 - r} \]

\[ \begin{aligned} 16 &= \frac{a(1 - 8)}{1 - 2}\\ &= \frac{a(-7)}{-1} = 7a \end{aligned} \]

\[ a = \frac{16}{7} \]

General sum: \[ S_n = \frac{a(1 - r^n)}{1 - r} \]

\[ \begin{aligned} S_n &= \frac{16}{7} \cdot \frac{1 - 2^n}{1 - 2}\\ &= \frac{16}{7}(2^n - 1) \end{aligned} \]

Final Answer

\[ a = \frac{16}{7}, \quad r = 2 \]

\[ S_n = \frac{16}{7}(2^n - 1) \]

Exam Significance

For Board Exams: Focus on correct use of GP sum formulas.

For JEE/NEET: This is a high-value pattern:

• Grouped sum relationships
• Exponential scaling in sequences
• Shortcut-based solving

Recognizing \( \text{next block} = r^k \times \text{previous block} \) is a major speed booster.

Theory Insight

nth term of G.P.: \[ a_n = ar^{n-1} \]

Sum formula: \[ S_n = \frac{a(1 - r^n)}{1 - r}, \quad r \neq 1 \]

Key strategy: Express numbers in powers to simplify ratios.

Decreasing pattern shows \(0 < r < 1\).

Solution Roadmap

Step 1: Use \(a_7 = ar^6\) to find \(r\)
Step 2: Express in powers
Step 3: Apply sum formula
Step 4: Simplify carefully

Solution

Given: \[ a = 729,\quad a_7 = 64 \]

Using: \[ a_7 = ar^6 \]

\[ 64 = 729r^6 \]

\[ r^6 = \frac{64}{729} = \left(\frac{2}{3}\right)^6 \]

\[ r = \frac{2}{3} \]

Now, \[ \begin{aligned} S_7 &= \frac{729(1 - (2/3)^7)}{1 - 2/3} \\ &= 729 \cdot \frac{1 - \frac{128}{2187}}{\frac{1}{3}}\\ &= 729 \cdot 3 \left(1 - \frac{128}{2187}\right)\\ &= 2187 \left(\frac{2187 - 128}{2187}\right)\\ &= 2187 \cdot \frac{2059}{2187} \end{aligned} \]

\[ S_7 = 2059 \]

Final Answer

\[ S_7 = 2059 \]

Exam Significance

For Board Exams: Focus on correct substitution and simplification.

For JEE/NEET: Key learning:

• Power comparison \((729 = 3^6,\; 64 = 2^6)\)
• Handling fractional ratios efficiently
• Avoiding calculation errors in large numbers

Converting into powers early makes the solution almost mental.

Theory Insight

General G.P. terms: \[ a,\; ar,\; ar^2,\; ar^3,\; \ldots \]

Key relations:

\[ S_2 = a + ar = a(1 + r) \]

\[ a_n = ar^{n-1} \]

Always prefer direct expressions like \(a + ar\) instead of using sum formula for small \(n\).

Terms grow or alternate depending on sign of \(r\).

Solution Roadmap

Step 1: Use \(S_2 = a(1+r)\)
Step 2: Use relation between \(a_5\) and \(a_3\)
Step 3: Solve for \(r\)
Step 4: Substitute to find \(a\)

Solution

Given: \[ \begin{align} a + ar &= -4\\ \Rightarrow a(1 + r) &= -4 \tag{1} \end{align} \]

Also, \[ \begin{aligned} a_3 &= ar^2,\\ a_5 &= ar^4 \end{aligned} \]

Given: \[ ar^4 = 4ar^2 \]

Dividing by \(ar^2\) (\(a \neq 0\)): \[ \begin{aligned} r^2 &= 4\\ \Rightarrow r &= \pm 2 \end{aligned} \]

Case 1: \(r = 2\)

From (1): \[ \begin{aligned} a(1+2) &= -4 \\ \Rightarrow 3a &= -4 \\ \Rightarrow a &= -\frac{4}{3} \end{aligned} \]

G.P.: \[ -\frac{4}{3},\; -\frac{8}{3},\; -\frac{16}{3},\; \ldots \]

Case 2: \(r = -2\)

From (1): \[ \begin{aligned} a(1-2) &= -4 \\ \Rightarrow -a &= -4 \\ \Rightarrow a &= 4 \end{aligned} \]

G.P.: \[ 4,\; -8,\; 16,\; -32,\; \ldots \]

Final Answer

Two possible G.P.s:

\[ -\frac{4}{3},\; -\frac{8}{3},\; -\frac{16}{3},\; \ldots \]

\[ 4,\; -8,\; 16,\; -32,\; \ldots \]

Exam Significance

For Board Exams: Focus on forming correct equations from conditions.

For JEE/NEET: Important learnings:

• Prefer \(a(1+r)\) over formula for small sums
• Always consider both positive and negative roots
• Handle alternating G.P. carefully

Missing the \(r=-2\) case is a very common mistake.

Theory Insight

In a G.P.: \[ a_n = ar^{n-1} \]

Key property:

Three numbers \(x, y, z\) are in G.P. if: \[ y^2 = xz \quad \text{or} \quad \frac{y}{x} = \frac{z}{y} \]

Equal spacing in indices leads to equal powers of \(r\).

Equal gaps (4 → 10 → 16) ensure equal multiplication by \(r^6\).

Solution Roadmap

Step 1: Express terms using \(a_n = ar^{n-1}\)
Step 2: Form ratios
Step 3: Compare powers of \(r\)
Step 4: Conclude G.P.

Solution

Let first term \(a\), common ratio \(r\).

\[ \begin{aligned} x &= a_4 = ar^3,\\\\ y &= a_{10} = ar^9,\\\\ z &= a_{16} = ar^{15} \end{aligned} \]

Consider ratios:

\[ \begin{aligned} \frac{y}{x} &= \frac{ar^9}{ar^3} \\ &= r^{6} \end{aligned} \]

\[ \begin{aligned} \frac{z}{y} &= \frac{ar^{15}}{ar^9} \\ &= r^{6} \end{aligned} \]

Hence, \[ \frac{y}{x} = \frac{z}{y} \]

Therefore, \(x,\; y,\; z\) are in G.P.

Alternative Proof (Powerful Shortcut)

\[ \begin{aligned} y^2 &= (ar^9)^2 \\ &= a^2 r^{18} \end{aligned} \]

\[ \begin{aligned} xz &= (ar^3)(ar^{15}) \\ &= a^2 r^{18} \end{aligned} \]

\[ y^2 = xz \]

Hence, \(x,\; y,\; z\) are in G.P.

Final Result

\[ x,\; y,\; z \text{ are in G.P.} \]

Exam Significance

For Board Exams: Standard proof using ratio or product method.

For JEE/NEET: Key takeaway:

• Equal index gaps ⇒ equal powers of \(r\)
• Use \(y^2 = xz\) for fastest verification
• Avoid lengthy ratio comparisons in MCQs

Spotting index symmetry is a major speed advantage.

Theory Insight

Numbers like \(8, 88, 888,\ldots\) are constructed using powers of 10.

Key identity: \[ 11\ldots1 \; (k \text{ digits}) = \frac{10^k - 1}{9} \]

Hence, \[ 88\ldots8 \; (k \text{ digits}) = 8 \cdot \frac{10^k - 1}{9} \]

This converts the sequence into a combination of a geometric series and a simple sum.

Each term grows by appending a digit, equivalent to multiplying by 10 and adding 8.

Solution Roadmap

Step 1: Express general term using powers of 10
Step 2: Convert into summation form
Step 3: Split into GP and simple sum
Step 4: Simplify expression

Solution

The \(k^{th}\) term: \[ T_k = 8 \cdot \frac{10^k - 1}{9} \]

Sum of first \(n\) terms: \[ \begin{aligned} S_n &= \sum_{k=1}^{n} 8 \cdot \frac{10^k - 1}{9}\\ S_n &= \frac{8}{9} \sum_{k=1}^{n} (10^k - 1)\\ S_n &= \frac{8}{9} \left(\sum_{k=1}^{n} 10^k - \sum_{k=1}^{n} 1 \right) \end{aligned} \]

\[ \sum_{k=1}^{n} 1 = n \]

\[ \sum_{k=1}^{n} 10^k = \frac{10(10^n - 1)}{9} \]

Substituting: \[ S_n = \frac{8}{9} \left[\frac{10(10^n - 1)}{9} - n \right] \]

\[ S_n = \frac{80}{81}(10^n - 1) - \frac{8n}{9} \]

Final Answer

\[ S_n = \frac{80}{81}(10^n - 1) - \frac{8n}{9} \]

Exam Significance

For Board Exams: Tests your ability to convert patterns into algebraic form.

For JEE/NEET: This is a high-value concept:

• Number pattern → series transformation
• Combination of GP + linear sum
• Useful in digit-based problems

Recognizing \(\frac{10^n - 1}{9}\) pattern is a major speed advantage.

Theory Insight

When two sequences are multiplied term-by-term, always check for patterns.

If one sequence increases geometrically and the other decreases correspondingly, their product may form a new G.P.

Key idea: \[ (a r^{k}) \cdot (b r^{-k}) = ab \]

This often reduces the problem to a simpler geometric series.

One sequence grows while the other shrinks, creating a structured product pattern.

Solution Roadmap

Step 1: Multiply corresponding terms
Step 2: Identify pattern in products
Step 3: Recognize resulting G.P.
Step 4: Apply sum formula

Solution

Corresponding products:

\[ (2 \cdot 128),\; (4 \cdot 32),\; (8 \cdot 8),\; (16 \cdot 2),\; \left(32 \cdot \frac{1}{2}\right) \]

\[ = 256,\; 128,\; 64,\; 32,\; 16 \]

This forms a G.P. with: \[ a = 256,\quad r = \frac{1}{2} \]

Sum of 5 terms: \[ \begin{aligned} S &= \frac{256\left(1 - (1/2)^5\right)}{1 - 1/2}\\ &= 256 \cdot \frac{1 - \frac{1}{32}}{\frac{1}{2}}\\ &= 256 \cdot \frac{31}{32} \cdot 2\\ &= 16 \cdot 31 \\ &= 496 \end{aligned} \]

Final Answer

\[ 496 \]

Exam Significance

For Board Exams: Straightforward calculation with pattern recognition.

For JEE/NEET: Important insight:

• Product of sequences can form a G.P.
• Opposite ratios simplify expressions
• Always check for hidden structure before expanding

Recognizing the resulting G.P. saves significant time.

Theory Insight

General term of a G.P.: \[ T_k = ar^{k-1} \]

Product of two sequences:

\[ (ar^{k-1})(AR^{k-1}) = aA(rR)^{k-1} \]

This directly reveals a new G.P. with ratio equal to the product of individual ratios.

Each step multiplies by both \(r\) and \(R\), giving combined growth.

Solution Roadmap

Step 1: Write general terms of both G.P.s
Step 2: Multiply corresponding terms
Step 3: Simplify expression
Step 4: Identify new G.P. and its ratio

Solution

First sequence: \[ a,\; ar,\; ar^2,\;\ldots,\; ar^{n-1} \]

Second sequence: \[ A,\; AR,\; AR^2,\;\ldots,\; AR^{n-1} \]

Product of corresponding terms:

\[ aA,\; ar \cdot AR,\; ar^2 \cdot AR^2,\;\ldots \]

\[ = aA,\; aA(rR),\; aA(rR)^2,\;\ldots,\; aA(rR)^{n-1} \]

This is clearly a G.P. with:

First term: \[ aA \]

Common ratio: \[ rR \]

Final Result

The products form a G.P. with common ratio: \[ rR \]

Exam Significance

For Board Exams: Standard proof using general term.

For JEE/NEET: Important concept:

• Product of two G.P.s → new G.P.
• Ratios multiply: \(r_{\text{new}} = r \cdot R\)
• Useful in advanced sequence transformations

Recognizing this pattern avoids lengthy term-by-term verification.

Theory Insight

General G.P. with four terms: \[ a,\; ar,\; ar^2,\; ar^3 \]

Strategy:

Convert word conditions into algebraic equations and reduce everything in terms of \(a(r^2 - 1)\).

This allows quick substitution and avoids solving simultaneous equations directly.

Alternating growth indicates a negative ratio.

Solution Roadmap

Step 1: Assume terms \(a, ar, ar^2, ar^3\)
Step 2: Translate both conditions into equations
Step 3: Factor to extract common expression
Step 4: Substitute and solve for \(r\), then \(a\)

Solution

Let the G.P. be: \[ a,\; ar,\; ar^2,\; ar^3 \]

First condition: \[ ar^2 = a + 9 \]

\[ a(r^2 - 1) = 9 \quad (1) \]

Second condition: \[ ar = ar^3 + 18 \]

\[ ar - ar^3 = 18 \]

\[ ar(1 - r^2) = 18 \]

\[ -ar(r^2 - 1) = 18 \quad (2) \]

Using (1) in (2):

\[ \begin{aligned} -r \cdot 9 &= 18\\ \Rightarrow -9r &= 18\\ \Rightarrow r &= -2 \end{aligned} \]

Substitute in (1):

\[ \begin{aligned} a(4 - 1) &= 9\\ \Rightarrow 3a &= 9\\ \Rightarrow a &= 3 \end{aligned} \]

Hence, the G.P. is:

\[ 3,\; -6,\; 12,\; -24 \]

Final Answer

\[ 3,\; -6,\; 12,\; -24 \]

Exam Significance

For Board Exams: Focus on forming correct equations from conditions.

For JEE/NEET: Key learning:

• Reduce equations using common factor \(a(r^2 - 1)\)
• Expect negative ratio in alternating sequences
• Substitution method is faster than solving directly

Identifying reusable expressions significantly reduces computation time.

Theory Insight

General term of a G.P.: \[ a_n = Ar^{n-1} \]

Key idea:

When expressions involve powers of terms, convert everything into powers of \(A\) and \(r\), then combine exponents.

If total exponent becomes zero: \[ A^0 r^0 = 1 \]

Balanced exponent shifts cancel out completely.

Solution Roadmap

Step 1: Express \(a, b, c\) using \(Ar^{n-1}\)
Step 2: Raise each to given powers
Step 3: Combine exponents
Step 4: Show total exponent = 0

Solution

Let first term \(A\), common ratio \(r\).

\[ a = Ar^{p-1},\quad b = Ar^{q-1},\quad c = Ar^{r-1} \]

Consider: \[ a^{q-r} b^{r-p} c^{p-q} \]

Substitute: \[ = (Ar^{p-1})^{q-r}(Ar^{q-1})^{r-p}(Ar^{r-1})^{p-q} \]

\[ = A^{(q-r)+(r-p)+(p-q)} \cdot r^{(p-1)(q-r)+(q-1)(r-p)+(r-1)(p-q)} \]

First exponent: \[ (q-r)+(r-p)+(p-q) = 0 \]

Second exponent: \[ (p-1)(q-r)+(q-1)(r-p)+(r-1)(p-q) = 0 \]

Hence, \[ a^{q-r} b^{r-p} c^{p-q} = A^0 r^0 = 1 \]

Final Result

\[ a^{q-r} b^{r-p} c^{p-q} = 1 \]

Exam Significance

For Board Exams: Standard proof using exponent manipulation.

For JEE/NEET: Key takeaways:

• Convert everything to base form early
• Focus on exponent cancellation
• Symmetry in indices simplifies proofs

Recognizing zero-exponent patterns saves time in advanced problems.

Theory Insight

General term: \[ a_n = ar^{n-1} \]

Product of first \(n\) terms:

\[ P = a \cdot ar \cdot ar^2 \cdots ar^{n-1} \]

Key idea: Convert product into powers and use sum of exponents: \[ 1 + 2 + \cdots + (n-1) = \frac{n(n-1)}{2} \]

Exponents accumulate systematically across terms.

Solution Roadmap

Step 1: Write product \(P\)
Step 2: Convert into powers of \(a\) and \(r\)
Step 3: Square the expression
Step 4: Use relation \(b = ar^{n-1}\)

Solution

Given: \[ a_n = ar^{n-1} = b \]

Product of first \(n\) terms:

\[ \begin{aligned} P &= a \cdot ar \cdot ar^2 \cdots ar^{n-1}\\ &= a^n \cdot r^{1+2+\cdots+(n-1)}\\ &= a^n \cdot r^{\frac{n(n-1)}{2}} \end{aligned} \]

Squaring:

\[ P^2 = a^{2n} r^{n(n-1)} \]

From \(b = ar^{n-1}\):

\[ ab = a^2 r^{n-1} \]

Raise to power \(n\):

\[ \begin{aligned} (ab)^n &= (a^2 r^{n-1})^n \\&= a^{2n} r^{n(n-1)} \end{aligned} \]

Hence,

\[ P^2 = (ab)^n \]

Final Result

\[ P^2 = (ab)^n \]

Exam Significance

For Board Exams: Standard proof using product expansion.

For JEE/NEET: Key insights:

• Product of GP terms → exponent summation
• Use \(n(n-1)/2\) quickly
• Link nth term relation cleverly

Recognizing exponent symmetry is crucial for advanced proofs.

Theory Insight

Sum of first \(n\) terms: \[ S_n = \frac{a(1 - r^n)}{1 - r}, \quad r \neq 1 \]

Key idea:

The next block of \(n\) terms (from \(n+1\) to \(2n\)) is just the first block multiplied by \(r^n\).

Hence, \[ \text{Next } n \text{ terms} = r^n \cdot S_n \]

The second block is a scaled version of the first by factor \(r^n\).

Solution Roadmap

Step 1: Write \(S_n\)
Step 2: Express next \(n\) terms as a new G.P.
Step 3: Use scaling relation
Step 4: Take ratio

Solution

Sum of first \(n\) terms: \[ S_n = \frac{a(1 - r^n)}{1 - r} \]

The \((n+1)^{th}\) term: \[ a_{n+1} = ar^n \]

Sum of terms from \((n+1)^{th}\) to \((2n)^{th}\):

\[ S' = \frac{ar^n(1 - r^n)}{1 - r} \]

Taking ratio: \[ \frac{S_n}{S'} = \frac{\frac{a(1 - r^n)}{1 - r}}{\frac{ar^n(1 - r^n)}{1 - r}} \]

Canceling common factors: \[ \frac{S_n}{S'} = \frac{1}{r^n} \]

Final Result

\[ \frac{\text{Sum of first } n \text{ terms}}{\text{Sum of next } n \text{ terms}} = \frac{1}{r^n} \]

Exam Significance

For Board Exams: Standard proof using GP sum formula.

For JEE/NEET: Key insight:

• Block shifting → multiplication by \(r^n\)
• Avoid full formula by using scaling idea
• Useful in advanced summation problems

Recognizing block-scaling patterns saves significant time.

Theory Insight

For a G.P.: \[ a,\; ar,\; ar^2,\; ar^3 \]

Key idea:

Factor common expressions and reduce both sides to the same algebraic form.

Look for repeated structure like: \[ 1 + r^2 + r^4 \]

Symmetry across terms leads to identical expressions.

Solution Roadmap

Step 1: Express all terms in G.P. form
Step 2: Simplify both brackets separately
Step 3: Factor common expression
Step 4: Compare both sides

Solution

Let the G.P. be: \[ a,\; ar,\; ar^2,\; ar^3 \]

First bracket: \[ \begin{aligned} a^2 + b^2 + c^2 &= a^2 + (ar)^2 + (ar^2)^2\\ &= a^2(1 + r^2 + r^4) \end{aligned} \]

Second bracket: \[ \begin{aligned} b^2 + c^2 + d^2 &= (ar)^2 + (ar^2)^2 + (ar^3)^2\\ &= a^2 r^2 (1 + r^2 + r^4) \end{aligned} \]

Multiply: \[ (a^2 + b^2 + c^2)(b^2 + c^2 + d^2) = a^4 r^2 (1 + r^2 + r^4)^2 \]

Now RHS:

\[ ab + bc + cd = a(ar) + (ar)(ar^2) + (ar^2)(ar^3) \]

\[ = a^2 r (1 + r^2 + r^4) \]

Squaring: \[ (ab + bc + cd)^2 = a^4 r^2 (1 + r^2 + r^4)^2 \]

Hence,

\[ (a^2 + b^2 + c^2)(b^2 + c^2 + d^2) = (ab + bc + cd)^2 \]

Final Result

\[ (a^2 + b^2 + c^2)(b^2 + c^2 + d^2) = (ab + bc + cd)^2 \]

Exam Significance

For Board Exams: Standard identity-based proof.

For JEE/NEET: Key takeaways:

• Always convert to \(a, ar, ar^2, ar^3\)
• Factor common expressions early
• Symmetry simplifies complex identities

Spotting repeated structures like \(1+r^2+r^4\) is the key to speed.

Theory Insight

If \(k\) numbers are inserted between two numbers \(a\) and \(b\) in a G.P., then: \[ b = a r^{k+1} \]

Here \(k = 2\), so: \[ b = a r^3 \]

Key idea: Convert the problem into finding the common ratio using powers.

Each step multiplies by the same ratio \(r\).

Solution Roadmap

Step 1: Assume G.P. \(3,\; 3r,\; 3r^2,\; 3r^3\)
Step 2: Use last term condition
Step 3: Solve for \(r\)
Step 4: Find missing terms

Solution

Let the G.P. be: \[ 3,\; 3r,\; 3r^2,\; 3r^3 \]

Given last term: \[ 3r^3 = 81 \]

\[ \begin{aligned} r^3 &= \frac{81}{3} \\&= 27\\ \Rightarrow r &= 3 \end{aligned} \]

Now, \[ \begin{aligned} 3r &= 9,\\ 3r^2 &= 27 \end{aligned} \]

Hence, the G.P. is: \[ 3,\; 9,\; 27,\; 81 \]

Final Answer

Inserted numbers: \[ 9,\; 27 \]

Exam Significance

For Board Exams: Direct application of GP term formula.

For JEE/NEET: Key insight:

• Use \(b = ar^{k+1}\) shortcut
• Convert to powers for quick solving
• Avoid writing unnecessary intermediate steps

Recognizing insertion pattern speeds up solving significantly.

Theory Insight

Geometric mean of \(a\) and \(b\): \[ \sqrt{ab} \]

Key idea:

Convert everything into powers and use symmetry: \[ a^x = b^x \Rightarrow x = 0 \quad (\text{when } a \ne b) \]

Expressions of the form: \[ \frac{a^{n+1}+b^{n+1}}{a^n+b^n} \] often simplify using factorization patterns.

Symmetry between \(a\) and \(b\) is the key to simplification.

Solution Roadmap

Step 1: Equate given expression to \(\sqrt{ab}\)
Step 2: Multiply and expand
Step 3: Factor using symmetry
Step 4: Solve exponent equation

Solution

Given: \[ \frac{a^{n+1}+b^{n+1}}{a^n+b^n} = \sqrt{ab} \]

Multiply both sides: \[ a^{n+1}+b^{n+1} = \sqrt{ab}(a^n+b^n) \]

Write \(\sqrt{ab} = a^{1/2}b^{1/2}\):

\[ a^{n+1}+b^{n+1} = a^{n+\frac12}b^{\frac12} + a^{\frac12}b^{n+\frac12} \]

Rearranging:

\[ a^{n+\frac12}(a^{\frac12}-b^{\frac12}) + b^{n+\frac12}(b^{\frac12}-a^{\frac12}) = 0 \]

Factor:

\[ (a^{\frac12}-b^{\frac12})(a^{n+\frac12}-b^{n+\frac12}) = 0 \]

Since \(a \ne b\):

\[ a^{n+\frac12} = b^{n+\frac12} \]

\[ \begin{aligned} n + \frac{1}{2} &= 0\\ \Rightarrow n =& -\frac{1}{2} \end{aligned} \]

Final Answer

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

Exam Significance

For Board Exams: Tests algebraic manipulation and exponent rules.

For JEE/NEET: Key takeaways:

• Symmetric expressions → factorization shortcut
• Use exponent equality instead of expansion
• Recognize GM structure quickly

Spotting symmetry reduces multi-step problems to one key step.

Theory Insight

Geometric mean: \[ \text{G.M.} = \sqrt{ab} \]

Key idea:

For ratio problems, assume: \[ \frac{a}{b} = x \] instead of solving directly in \(a, b\).

This converts the problem into a single-variable equation.

Symmetry between \(a\) and \(b\) helps reduce complexity.

Solution Roadmap

Step 1: Assume ratio \(a:b = x:1\)
Step 2: Substitute into given condition
Step 3: Solve quadratic
Step 4: Write ratio

Solution

Let: \[ \frac{a}{b} = x \Rightarrow a = xb \]

Given: \[ a + b = 6\sqrt{ab} \]

Substitute: \[ xb + b = 6\sqrt{xb \cdot b} \]

\[ b(x+1) = 6b\sqrt{x} \]

Divide by \(b\): \[ x+1 = 6\sqrt{x} \]

Let \(\sqrt{x} = t\), so \(x = t^2\):

\[ t^2 + 1 = 6t \]

\[ t^2 - 6t + 1 = 0 \]

\[ t = 3 \pm 2\sqrt{2} \]

Hence, \[ \begin{aligned} \sqrt{x} &= 3 \pm 2\sqrt{2}\\ \Rightarrow x &= (3 \pm 2\sqrt{2})^2 \end{aligned} \]

But ratio can be written directly as: \[ a : b = (3+2\sqrt{2}) : (3-2\sqrt{2}) \]

Final Answer

\[ (3+2\sqrt{2}) : (3-2\sqrt{2}) \]

Exam Significance

For Board Exams: Tests algebraic manipulation and identities.

For JEE/NEET: Key takeaways:

• Use ratio substitution \(a = xb\)
• Convert to quadratic in \(\sqrt{x}\)
• Avoid expanding large expressions

This “ratio-first” method is significantly faster than direct solving.

Theory Insight

Arithmetic Mean: \[ A = \frac{x+y}{2} \]

Geometric Mean: \[ G = \sqrt{xy} \]

Key identity: \[ (x-y)^2 = (x+y)^2 - 4xy \]

This identity connects A.M. and G.M. directly.

Symmetry allows expressing numbers using sum and difference.

Solution Roadmap

Step 1: Use definitions of A.M. and G.M.
Step 2: Apply identity for \((x-y)^2\)
Step 3: Express \(x-y\)
Step 4: Solve using sum and difference

Solution

Let the numbers be \(x\) and \(y\).

\[ x+y = 2A,\quad xy = G^2 \]

Using identity: \[ (x-y)^2 = (x+y)^2 - 4xy \]

\[ \begin{aligned} (x-y)^2 &= (2A)^2 - 4G^2 \\&= 4(A^2 - G^2) \end{aligned} \]

\[ x-y = 2\sqrt{A^2 - G^2} \]

Since: \[ A^2 - G^2 = (A+G)(A-G) \]

\[ x-y = 2\sqrt{(A+G)(A-G)} \]

Now,

\[ \begin{aligned} x+y &= 2A,\\ x-y &= 2\sqrt{(A+G)(A-G)} \end{aligned} \]

Adding: \[ x = A + \sqrt{(A+G)(A-G)} \]

Subtracting: \[ y = A - \sqrt{(A+G)(A-G)} \]

Final Result

\[ x,\; y = A \pm \sqrt{(A+G)(A-G)} \]

Exam Significance

For Board Exams: Direct proof using identities.

For JEE/NEET: Key insights:

• Use \((x+y)\) and \((x-y)\) approach
• Connect A.M. and G.M. using identities
• Avoid solving quadratic explicitly

This is a standard transformation used in many advanced problems.

Theory Insight

Exponential growth follows a G.P.: \[ \text{Population after } t \text{ hours} = a \cdot r^t \]

Here:

\[ a = 30,\quad r = 2 \]

Important: Initial value is at \(t = 0\), so after \(t\) hours: \[ N(t) = 30 \cdot 2^t \]

Each hour multiplies the population by 2.

Solution Roadmap

Step 1: Identify G.P. parameters
Step 2: Use time-based formula \(N(t) = a \cdot r^t\)
Step 3: Substitute required values

Solution

Given: \[ a = 30,\quad r = 2 \]

Number after \(t\) hours: \[ N(t) = 30 \cdot 2^t \]

At end of \(2^{nd}\) hour: \[ N(2) = 30 \cdot 2^2 = 120 \]

At end of \(4^{th}\) hour: \[ N(4) = 30 \cdot 2^4 = 480 \]

At end of \(n^{th}\) hour: \[ N(n) = 30 \cdot 2^n \]

Final Answer

\[ 120,\quad 480,\quad 30 \cdot 2^n \]

Exam Significance

For Board Exams: Direct application of GP concept.

For JEE/NEET: Key insights:

• Time-based GP modeling (very common)
• Correct indexing: start from \(t=0\)
• Recognize exponential growth instantly

Misinterpreting time indexing is a common mistake—always define \(t=0\) clearly.

Theory Insight

Compound interest follows a geometric progression:

\[ A = P\left(1 + \frac{r}{100}\right)^n \]

Here each year multiplies the amount by: \[ 1 + \frac{r}{100} \]

So this is equivalent to a G.P. with:

\[ a = P,\quad r = \left(1 + \frac{r}{100}\right) \]

Each year grows the amount by a constant multiplier.

Solution Roadmap

Step 1: Identify \(P, r, n\)
Step 2: Substitute into compound interest formula
Step 3: Simplify and evaluate

Solution

Given: \[ \begin{aligned} P &= 500,\\ r &= 10\%,\\ n &= 10 \end{aligned} \]

\[ \begin{aligned} A &= 500\left(1 + \frac{10}{100}\right)^{10}\\ &= 500\left(\frac{11}{10}\right)^{10}\\ &= 500 \times (1.1)^{10}\\ &=(1.1)^{10} \\&\approx 2.59374 \end{aligned} \]

\[ A \approx 500 \times 2.59374 = 1296.87 \]

Final Answer

\[ \text{Rs }1296.87 \;(\text{approx.}) \]

Exam Significance

For Board Exams: Direct application of compound interest formula.

For JEE/NEET: Key insights:

• Compound interest = geometric growth
• Use \( (1+r)^n \) efficiently
• Approximation skills save time

Recognizing CI as a G.P. helps connect finance with sequences.

Theory Insight

If roots are \(a\) and \(b\), then:

\[ \text{A.M.} = \frac{a+b}{2}, \quad \text{G.M.} = \sqrt{ab} \]

Standard quadratic from roots:

\[ x^2 - (a+b)x + ab = 0 \]

Key idea: Convert A.M. → sum and G.M. → product directly.

A.M. controls sum, G.M. controls product.

Solution Roadmap

Step 1: Use A.M. to find \(a+b\)
Step 2: Use G.M. to find \(ab\)
Step 3: Substitute into standard quadratic form

Solution

Let roots be \(a\) and \(b\).

Given: \[ \begin{aligned} \frac{a+b}{2} &= 8 \\ \Rightarrow a+b &= 16 \end{aligned} \]

\[ \begin{aligned} \sqrt{ab} &= 5 \\ \Rightarrow ab &= 25 \end{aligned} \]

Required quadratic:

\[ x^2 - (a+b)x + ab = 0 \]

\[ x^2 - 16x + 25 = 0 \]

Final Answer

\[ x^2 - 16x + 25 = 0 \]

Exam Significance

For Board Exams: Direct application of A.M.–G.M. definitions.

For JEE/NEET: Key insights:

• A.M. → sum of roots
• G.M. → product of roots
• Immediate quadratic formation

This is a standard “convert mean → equation” pattern.