Chapter 8 — Sequences and Series

Theory Insight

A function satisfying \(f(x+y)=f(x)f(y)\) over natural numbers behaves like an exponential function. Such functions follow the standard result:

\[ f(x) = [f(1)]^x \]

This converts a functional equation into a geometric sequence problem, which is a very common transformation in algebra and sequences.

Solution Roadmap

• Convert functional equation → exponential form
• Identify sequence as a geometric progression (GP)
• Apply sum of GP formula
• Solve resulting exponential equation

Visual Understanding

Growth pattern of \(3^x\) (Geometric Progression with ratio 3)

Solution

Given \(f(x+y)=f(x)f(y)\) and \(f(1)=3\),

\[ f(x)=3^x \]

Now,

\[ \sum_{x=1}^{n} f(x)=\sum_{x=1}^{n} 3^x = 120 \]

Using sum of GP:

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

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

\[ 3^n - 1 = 80 \]

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

\[ n=4 \]

Final Answer: \(n=4\)

Exam Significance

• Direct application of functional equation → exponential mapping (JEE favorite)
• Conversion to GP is a standard trick in objective exams
• Tests speed in recognizing patterns rather than lengthy computation
• Frequently appears in boards as 3–5 mark conceptual problem

Theory Insight

A geometric progression (G.P.) follows exponential growth when \(r>1\). The sum of first \(n\) terms is given by:

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

Once \(n\) is found, the last term is obtained using:

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

Solution Roadmap

• Apply GP sum formula to find \(n\)
• Solve exponential equation
• Use nth term formula to get last term

Visual Understanding

Rapid exponential rise in GP with \(r=2\)

Solution

Given:

\[ S_n = 315,\quad a=5,\quad r=2 \]

Using GP sum formula:

\[ 315 = \frac{5(2^n - 1)}{2-1} \]

\[ 315 = 5(2^n - 1) \]

\[ 2^n - 1 = 63 \]

\[ 2^n = 64 = 2^6 \]

\[ n = 6 \]

Now, last term:

\[ a_n = ar^{n-1} = 5 \cdot 2^5 = 5 \cdot 32 = 160 \]

Final Answer: Number of terms = \(6\), Last term = \(160\)

Exam Significance

• Classic GP sum + nth term combined problem (very common in boards)
• Direct exponential solving (JEE Main level)
• Tests ability to move between sum and term formulas quickly
• Frequently appears in MCQs and short-answer formats

Theory Insight

In a G.P., the general term is:

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

When terms are spaced apart (like 3rd and 5th), they naturally form powers of \(r\). Such problems often reduce to a quadratic equation by substitution.

Solution Roadmap

• Write required terms using \(a_n = ar^{n-1}\)
• Form equation using given condition
• Reduce to quadratic via substitution
• Back-substitute to find \(r\)

Visual Understanding

Relationship: \(a_3 = r^2\), \(a_5 = r^4\)

Solution

Given:

\[ a = 1,\quad a_3 + a_5 = 90 \]

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

\[ a_3 = r^2,\quad a_5 = r^4 \]

\[ r^2 + r^4 = 90 \]

\[ r^2(1 + r^2) = 90 \]

Let \(r^2 = x\), then:

\[ x(1+x) = 90 \]

\[ x^2 + x - 90 = 0 \]

\[ (x+10)(x-9) = 0 \]

\[ x = 9 \quad (\text{since } x=r^2 \ge 0) \]

\[ r^2 = 9 \Rightarrow r = \pm 3 \]

Final Answer: \(r = 3\) or \(r = -3\)

Exam Significance

• Tests identification of non-consecutive GP terms
• Standard substitution trick \(r^2 = x\) (very common in JEE)
• Concept of both positive and negative ratio often asked in MCQs
• High probability question pattern in boards (3–4 marks)

Theory Insight

Three numbers in G.P. are conveniently represented as:

\[ \frac{a}{r},\; a,\; ar \]

If three numbers form an A.P., then:

\[ 2 \times (\text{middle term}) = \text{first term} + \text{third term} \]

This problem combines G.P. structure with A.P. condition, a very common hybrid concept.

Solution Roadmap

• Assume G.P. terms in standard symmetric form
• Apply sum condition
• Apply A.P. condition after transformation
• Solve system to find \(a\) and \(r\)

Visual Understanding

G.P. structure transformed into A.P. after subtraction

Solution

Let the numbers be:

\[ \frac{a}{r},\; a,\; ar \]

Given sum:

\[ \frac{a}{r} + a + ar = 56 \quad \text{(1)} \]

After subtraction, numbers become:

\[ \frac{a}{r} - 1,\; a - 7,\; ar - 21 \]

These are in A.P., so:

\[ 2(a - 7) = \left(\frac{a}{r} - 1\right) + (ar - 21) \]

\[ 2a - 14 = \frac{a}{r} + ar - 22 \]

\[ 2a + 8 = \frac{a}{r} + ar \quad \text{(2)} \]

From (1):

\[ \frac{a}{r} + ar = 56 - a \]

Substitute in (2):

\[ 2a + 8 = 56 - a \]

\[ 3a = 48 \Rightarrow a = 16 \]

Substitute \(a=16\) into (1):

\[ \frac{16}{r} + 16 + 16r = 56 \]

\[ \frac{16}{r} + 16r = 40 \]

\[ \frac{2}{r} + 2r = 5 \]

\[ 2r^2 - 5r + 2 = 0 \]

\[ (2r - 1)(r - 2) = 0 \]

\[ r = 2 \quad \text{or} \quad r = \frac{1}{2} \]

Therefore, numbers are:

\[ (8,\;16,\;32) \quad \text{or} \quad (32,\;16,\;8) \]

Final Answer: \(8, 16, 32\) (or \(32, 16, 8\))

Exam Significance

• Classic GP + AP hybrid question (very important for boards)
• Tests transformation thinking (sequence → modified sequence)
• Frequently appears in JEE Main subjective/numerical type
• Strengthens algebraic equation formation skills

Theory Insight

In a G.P. with even number of terms \(2n\), the sequence can be split into:

• Odd-position terms → form a G.P. with ratio \(r^2\)
• Even-position terms → also form a G.P. with ratio \(r^2\)

This splitting technique is a powerful tool in sequence problems and frequently used in competitive exams.

Solution Roadmap

• Represent total terms as \(2n\)
• Write sum of full G.P.
• Write sum of odd-position terms
• Apply given condition and simplify

Solution

Let total terms = \(2n\), first term = \(a\), common ratio = \(r\)

Sum of all terms:

\[ S_{2n} = a\frac{1 - r^{2n}}{1 - r}, \quad r \neq 1 \]

Odd-position terms: \(a, ar^2, ar^4, \dots\)

This is a G.P. with:

first term \(a\), ratio \(r^2\), number of terms \(n\)

\[ S_{\text{odd}} = a\frac{1 - r^{2n}}{1 - r^2} \]

Given:

\[ S_{2n} = 5 \times S_{\text{odd}} \]

\[ a\frac{1 - r^{2n}}{1 - r} = 5a\frac{1 - r^{2n}}{1 - r^2} \]

Cancel common factors \(a\) and \((1 - r^{2n})\):

\[ \frac{1}{1 - r} = \frac{5}{1 - r^2} \]

Using:

\[ 1 - r^2 = (1 - r)(1 + r) \]

\[ \frac{1}{1 - r} = \frac{5}{(1 - r)(1 + r)} \]

Cancel \((1 - r)\):

\[ 1 + r = 5 \]

\[ r = 4 \]

Final Answer: \(r = 4\)

Exam Significance

• Very important pattern: splitting GP into odd/even indexed terms
• Direct algebraic simplification using factorization
• Frequently asked in JEE Main and advanced objective questions
• Tests conceptual clarity over computation

Theory Insight

When multiple rational expressions are equal, we can equate them pairwise. Cross-multiplication then converts the expressions into algebraic equations.

A key identity for G.P. is:

\[ b^2 = ac \quad \text{and} \quad c^2 = bd \]

If such relations are obtained, it directly implies a geometric progression.

Solution Roadmap

• Equate first two ratios and cross-multiply
• Simplify to obtain relation between \(a, b, c\)
• Repeat for second and third ratios
• Combine results to prove G.P.

Solution

Given:

\[ \frac{a+bx}{a-bx}=\frac{b+cx}{b-cx}=\frac{c+dx}{c-dx}, \quad x \ne 0 \]

Equating first two:

\[ \frac{a+bx}{a-bx}=\frac{b+cx}{b-cx} \]

Cross-multiplying:

\[ (a+bx)(b-cx) = (b+cx)(a-bx) \]

Expanding:

\[ ab - acx + b^2x - bcx^2 = ab - b^2x + acx - bcx^2 \]

Cancel common terms:

\[ -acx + b^2x = -b^2x + acx \]

\[ 2b^2x = 2acx \]

Since \(x \ne 0\):

\[ b^2 = ac \quad \text{(1)} \]

Now equating second and third:

\[ \frac{b+cx}{b-cx}=\frac{c+dx}{c-dx} \]

Cross-multiplying:

\[ (b+cx)(c-dx) = (c+dx)(b-cx) \]

Simplifying similarly:

\[ c^2 = bd \quad \text{(2)} \]

From (1) and (2):

\[ \frac{b}{a} = \frac{c}{b} = \frac{d}{c} \]

Hence, \(a, b, c, d\) are in G.P.

Final Result: \(a, b, c, d\) form a geometric progression

Exam Significance

• Standard proof-based question in boards (4–5 marks)
• Tests algebraic manipulation via cross-multiplication
• Recognition of \(b^2=ac\) pattern is crucial for fast solving
• Frequently used concept in higher algebra and sequences

Theory Insight

In a G.P., expressions involving sum, product, and reciprocals often simplify using symmetry of exponents.

Key identities:

• Sum of exponents: \(0 + 1 + 2 + \cdots + (n-1) = \frac{n(n-1)}{2}\)
• Reciprocal G.P. has ratio \(\frac{1}{r}\)

Such problems rely on expressing everything in powers of \(a\) and \(r\), then simplifying systematically.

Solution Roadmap

• Write G.P. terms explicitly
• Compute \(S\), \(P\), and \(R\)
• Relate \(R\) with \(S\) using factorization
• Substitute and simplify to prove identity

Solution

Let the G.P. be:

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

Sum:

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

Product:

\[ \begin{aligned} P &= a \cdot ar \cdot ar^2 \cdots ar^{n-1}\\ &= a^n r^{0+1+2+\cdots+(n-1)}\\ &= a^n r^{\frac{n(n-1)}{2}} \end{aligned} \]

Reciprocals form G.P.:

\[ \frac{1}{a},\; \frac{1}{ar},\; \frac{1}{ar^2},\; \ldots,\; \frac{1}{ar^{n-1}} \]

Sum of reciprocals:

\[ R = \frac{1}{a}\cdot\frac{1 - r^{-n}}{1 - r^{-1}} \]

Simplify:

\[ R = \frac{1}{a} \cdot \frac{r^n - 1}{r^{n-1}(r - 1)} \]

Using expression of \(S\):

\[ \begin{aligned} S &= a\frac{r^n - 1}{r - 1} \Rightarrow \frac{r^n - 1}{r - 1} \\&= \frac{S}{a} \end{aligned} \]

Substitute in \(R\):

\[ \begin{aligned} R &= \frac{1}{a} \cdot \frac{S}{a r^{n-1}} \\&= \frac{S}{a^2 r^{n-1}} \end{aligned} \]

Now compute:

\[ R^n = \frac{S^n}{a^{2n} r^{n(n-1)}} \]

Also,

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

Therefore,

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

Hence proved: \(P^2 R^n = S^n\)

Exam Significance

• Important proof combining sum, product, and reciprocal GP
• Tests exponent handling and algebraic simplification
• Common in board exams (5 marks proof)
• Useful for advanced algebra and sequence manipulation problems

Theory Insight

If numbers are in G.P., they can always be expressed in exponential form:

\[ a,\; ar,\; ar^2,\; ar^3 \]

Raising terms to power \(n\) preserves the geometric structure, converting ratio \(r\) into \(r^n\).

Key idea: Factorization helps expose hidden G.P. structure.

Solution Roadmap

• Express terms using common ratio \(r\)
• Raise each term to power \(n\)
• Factor expressions to reveal common multiplier
• Identify constant ratio between consecutive terms

Solution

Since \(a, b, c, d\) are in G.P., let:

\[ b = ar,\quad c = ar^2,\quad d = ar^3 \]

Now consider:

\[ \begin{aligned} a^n + b^n &= a^n + (ar)^n \\&= a^n(1 + r^n) \end{aligned} \]

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

\[ \begin{aligned} c^n + d^n &= (ar^2)^n + (ar^3)^n \\&= a^n r^{2n} (1 + r^n) \end{aligned} \]

Thus, the three terms become:

\[ a^n(1 + r^n),\quad a^n r^n(1 + r^n),\quad a^n r^{2n}(1 + r^n) \]

Each term is obtained from the previous one by multiplying with \(r^n\).

Hence, they form a G.P. with common ratio \(r^n\).

Final Result: \((a^n + b^n), (b^n + c^n), (c^n + d^n)\) are in G.P.

Exam Significance

• Classic transformation: GP → powers → new GP
• Tests factorization and pattern recognition
• Very common in board proofs (4–5 marks)
• Helpful in JEE for identifying hidden ratios quickly

Theory Insight

Using Vieta’s formulas:

• Sum of roots = coefficient of \(x\) (with sign change)
• Product of roots = constant term

When roots form a G.P., assume them in exponential form:

\[ A,\; Ar,\; Ar^2,\; Ar^3 \]

This converts the problem into solving simple algebraic relations.

Solution Roadmap

• Express roots using G.P. form
• Apply Vieta’s formulas for both equations
• Form equations using sums
• Find \(r\) and \(A\)
• Compute \(p\) and \(q\) and evaluate ratio

Solution

Let the four numbers in G.P. be:

\[ \begin{aligned} a &= A,\\ b &= Ar,\\ c &= Ar^2,\\ d &= Ar^3 \end{aligned} \]

From \(x^2 - 3x + p = 0\):

\[ \begin{align} a + b &= 3 \\\Rightarrow A(1 + r) &= 3 \tag{1} \end{align} \]

From \(x^2 - 12x + q = 0\):

\[ \begin{align} c + d &= 12 \\ \Rightarrow Ar^2(1 + r) &= 12 \tag{2} \end{align} \]

Dividing (2) by (1):

\[ \begin{aligned} \frac{Ar^2(1 + r)}{A(1 + r)} &= \frac{12}{3}\\ \Rightarrow r^2 &= 4\\ \Rightarrow r &= 2 \end{aligned} \]

Substitute in (1):

\[ A(1 + 2) = 3 \Rightarrow A = 1 \]

Hence,

\[ \begin{aligned} a = 1,\quad b = 2,\\ c = 4,\quad d = 8 \end{aligned} \]

Now,

\[ \begin{aligned} p = ab = 2,\\ q = cd = 32 \end{aligned} \]

\[ \begin{aligned} (q + p):(q - p) &= (32 + 2):(32 - 2) \\&= 34:30 \\&= 17:15 \end{aligned} \]

Final Result: \((q + p):(q - p) = 17:15\)

Exam Significance

• High-frequency pattern: roots in G.P. + Vieta application
• Shortcut: divide equations to eliminate \(A(1+r)\)
• Very important for JEE Main MCQs
• Board exams: typical 4–5 mark structured question

Theory Insight

For two positive numbers:

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

A standard transformation in such problems is:

\[ \frac{a+b}{\sqrt{ab}} = \sqrt{\frac{a}{b}} + \sqrt{\frac{b}{a}} \]

This converts the problem into a quadratic form using substitution.

Solution Roadmap

• Use A.M./G.M. ratio
• Convert expression into symmetric form
• Substitute \(t = \sqrt{\frac{a}{b}}\)
• Solve quadratic and back-substitute

Solution

Given:

\[ \frac{\frac{a+b}{2}}{\sqrt{ab}} = \frac{m}{n} \]

\[ \Rightarrow \frac{a+b}{\sqrt{ab}} = \frac{2m}{n} \]

\[ \frac{a}{\sqrt{ab}} + \frac{b}{\sqrt{ab}} = \frac{2m}{n} \]

\[ \sqrt{\frac{a}{b}} + \sqrt{\frac{b}{a}} = \frac{2m}{n} \]

Let \(t = \sqrt{\frac{a}{b}}\), then:

\[ t + \frac{1}{t} = \frac{2m}{n} \]

Multiply by \(t\):

\[ t^2 - \frac{2m}{n}t + 1 = 0 \]

Solving:

\[ \begin{aligned} t &= \frac{\frac{2m}{n} \pm \sqrt{\left(\frac{2m}{n}\right)^2 - 4}}{2}\\ &= \frac{m \pm \sqrt{m^2 - n^2}}{n} \end{aligned} \]

Since \(t = \sqrt{\frac{a}{b}}\), we get:

\[ \begin{aligned} \frac{a}{b} &= t^2\\ &= \left(\frac{m + \sqrt{m^2 - n^2}}{n}\right)^2 \end{aligned} \]

Taking ratio form (eliminating square):

\[ a:b = (m+\sqrt{m^2-n^2}) : (m-\sqrt{m^2-n^2}) \]

Hence proved.

Exam Significance

• Classic A.M.–G.M. transformation problem
• Substitution \(t=\sqrt{a/b}\) is a key exam trick
• Appears frequently in JEE algebra MCQs
• Tests symmetry handling and quadratic solving speed

Theory Insight

Numbers formed by repeating digits can be expressed using powers of 10.

• \(5, 55, 555\) → built using \(10^k\)
• Decimals like \(0.6, 0.66\) → built using \(10^{-k}\)

Key idea: Convert each term into a geometric sum, then apply summation formulas.

Solution Roadmap

• Write general term using GP structure
• Split into two summations
• Apply GP sum formulas
• Simplify expression

(i) Series: \(5 + 55 + 555 + \cdots\)

The \(k\)-th term:

\[ \begin{aligned} T_k &= 5(1 + 10 + 10^2 + \cdots + 10^{k-1})\\ &= 5 \cdot \frac{10^k - 1}{9} \end{aligned} \]

Sum of first \(n\) terms:

\[ \begin{aligned} S_n &= \sum_{k=1}^{n} \frac{5}{9}(10^k - 1)\\ &= \frac{5}{9} \left(\sum_{k=1}^{n} 10^k - n \right) \end{aligned} \]

Using:

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

\[ S_n = \frac{5}{9} \left( \frac{10(10^n - 1)}{9} - n \right) \]

\[ S_n = \frac{5}{81}\left[10(10^n - 1) - 9n\right] \]

(ii) Series: \(0.6 + 0.66 + 0.666 + \cdots\)

The \(k\)-th term:

\[ T_k = 6(10^{-1} + 10^{-2} + \cdots + 10^{-k}) \]

\[ \begin{aligned} T_k &= 6 \cdot \frac{1 - 10^{-k}}{9}\\ &= \frac{2}{3}(1 - 10^{-k}) \end{aligned} \]

Sum of first \(n\) terms:

\[ \begin{aligned} S_n &= \sum_{k=1}^{n} \frac{2}{3}(1 - 10^{-k})\\ &= \frac{2}{3}\left(n - \sum_{k=1}^{n} 10^{-k}\right) \end{aligned} \]

Using:

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

\[ S_n = \frac{2}{3}\left(n - \frac{1 - 10^{-n}}{9}\right) \]

Final Answers:
(i) \(S_n = \frac{5}{81}[10(10^n - 1) - 9n]\)
(ii) \(S_n = \frac{2}{3}\left(n - \frac{1 - 10^{-n}}{9}\right)\)

Exam Significance

• Very important pattern: repeated digits → GP transformation
• Common in boards (long answer) and JEE (conceptual MCQ)
• Tests ability to break complex terms into standard forms
• Useful shortcut: always convert digits into powers of 10

Theory Insight

The given series is formed by multiplying consecutive even numbers:

\[ (2,4),\; (4,6),\; (6,8),\; \dots \]

Such patterns can be generalized using index-based representation:

\[ \text{Even numbers} = 2k \]

This allows conversion into a standard algebraic expression.

Solution Roadmap

• Identify pattern of consecutive even numbers
• Express general term using index \(k\)
• Simplify expression
• Substitute required value

Solution

The \(k^{th}\) term is:

\[ T_k = (2k)(2k + 2) \]

\[ T_k = 4k(k+1) \]

For \(k = 20\):

\[ T_{20} = 4 \cdot 20 \cdot 21 \]

\[ T_{20} = 80 \cdot 21 = 1680 \]

Final Answer: \(T_{20} = 1680\)

Exam Significance

• Tests pattern recognition (very common in JEE MCQs)
• Converts sequence → algebraic formula quickly
• Simple but scoring question in boards
• Useful trick: represent even numbers as \(2k\)

Theory Insight

This is a classic reducing balance interest problem.

• Interest is charged on the remaining unpaid amount each year
• Unpaid balances form an A.P.
• Total interest = rate × sum of unpaid balances

Solution Roadmap

• Find unpaid balance
• Determine number of instalments
• Form A.P. of unpaid balances
• Compute total interest
• Add interest to total cost

Solution

Unpaid balance:

\[ 12000 - 6000 = 6000 \]

Annual instalment = Rs \(500\), so number of years:

\[ n = \frac{6000}{500} = 12 \]

Unpaid balances each year form an A.P.:

\[ 6000,\; 5500,\; 5000,\; \ldots,\; 500 \]

Here:

\[ a = 6000,\quad l = 500,\quad n = 12 \]

Sum of unpaid balances:

\[ \begin{aligned} S &= \frac{n}{2}(a + l)\\ &= \frac{12}{2}(6000 + 500)\\ &= 6 \times 6500\\ &= 39000 \end{aligned} \]

Total interest:

\[ \begin{aligned} \text{Interest} &= 12\% \times 39000\\ &= 0.12 \times 39000\\ &= 4680 \end{aligned} \]

Total cost:

\[ 12000 + 4680 = 16680 \]

Final Answer: Total cost = Rs \(16{,}680\)

Exam Significance

• Very important application of A.P. in finance (boards favorite)
• Key idea: unpaid balances form A.P.
• Common in JEE as word problem with algebraic modelling
• Tests interpretation + sequence application together

Theory Insight

This is a reducing balance interest model, where:

• Interest is charged on the remaining unpaid amount each year
• Unpaid balances decrease uniformly → form an A.P.
• Total interest = rate × sum of unpaid balances

Solution Roadmap

• Find unpaid balance after initial payment
• Determine number of instalments
• Model unpaid balances as A.P.
• Compute total interest
• Add interest to original cost

Solution

Unpaid balance:

\[ 22000 - 4000 = 18000 \]

Instalment per year = Rs \(1000\), so:

\[ n = \frac{18000}{1000} = 18 \]

Unpaid balances form an A.P.:

\[ 18000,\; 17000,\; 16000,\; \ldots,\; 1000 \]

Here:

\[ a = 18000,\quad l = 1000,\quad n = 18 \]

Sum of unpaid balances:

\[ \begin{aligned} S &= \frac{n}{2}(a + l)\\ &= \frac{18}{2}(18000 + 1000)\\ &= 9 \times 19000\\ &= 171000 \end{aligned} \]

Total interest:

\[ \begin{aligned} \text{Interest} &= 10\% \times 171000\\ &= 0.10 \times 171000\\ &= 17100 \end{aligned} \]

Total cost:

\[ 22000 + 17100 = 39100 \]

Final Answer: Total cost = Rs \(39{,}100\)

Exam Significance

• Very common application of A.P. in real-life finance problems
• Key recognition: decreasing balance → A.P.
• Boards: frequently asked long-answer question
• JEE: tests modelling + quick summation skills

Theory Insight

This is a classic chain-reaction growth problem, where each stage multiplies by a fixed factor.

• Each person sends letters to 4 others → multiplication by 4
• Number of letters forms a geometric progression (G.P.)
• \(n^{th}\) term of GP: \(T_n = ar^{n-1}\)

Solution Roadmap

• Identify GP pattern
• Find number of letters in 8th set
• Multiply by cost per letter

Solution

Number of letters in each set:

\[ 4,\; 16,\; 64,\; 256,\; \ldots \]

This is a G.P. with:

\[ a = 4,\quad r = 4 \]

Number of letters in \(8^{th}\) set:

\[ T_8 = a r^{7} = 4 \cdot 4^{7} = 4^{8} = 65536 \]

Cost per letter = 50 paise = Rs \(0.5\)

Total cost:

\[ 65536 \times 0.5 = 32768 \]

Final Answer: Rs \(32{,}768\)

Exam Significance

• Very common GP application (chain growth / population model)
• Key shortcut: \(4 \cdot 4^{7} = 4^8\)
• Tests exponential thinking (important for JEE)
• Boards: standard application-based question

Theory Insight

Under simple interest:

• Interest earned each year is constant
• Total amount forms an A.P.

Important distinction:

• Amount in \(n^{th}\) year = interest earned during that year
• Total amount after \(n\) years = principal + total interest

Solution Roadmap

• Find yearly interest
• Interpret “15th year” correctly
• Compute total amount after 20 years

Solution

Given:

\[ P = 10000,\quad R = 5\% \]

Interest per year:

\[ \begin{aligned} \text{SI per year} &= \frac{10000 \times 5}{100} \\&= 500 \end{aligned} \]

Thus, yearly earnings form:

\[ 500,\; 500,\; 500,\; \ldots \]

Amount in the \(15^{th}\) year:

Since simple interest is constant,

\[ \text{Interest in 15th year} = 500 \]

Total amount after 20 years:

\[ \begin{aligned} A_{20} &= P + \frac{PRT}{100}\\ &= 10000 + \frac{10000 \times 5 \times 20}{100} \end{aligned} \]

\[ A_{20} = 10000 + 10000 = 20000 \]

Final Answer:
Amount earned in 15th year = Rs \(500\)
Total amount after 20 years = Rs \(20{,}000\)

Exam Significance

• Very common conceptual trap: “nth year” vs “after n years”
• Simple interest → constant yearly income (A.P. concept)
• Frequently asked in boards and competitive exams
• Tests interpretation more than calculation

Theory Insight

Depreciation follows compound decay, similar to compound interest.

• Each year value is multiplied by \((1 - \frac{R}{100})\)
• Successive values form a G.P.

General formula:

\[ V = P\left(1 - \frac{R}{100}\right)^n \]

Solution Roadmap

• Identify depreciation factor
• Apply compound decay formula
• Compute efficiently using powers

Solution

Given:

\[ P = 15625,\quad R = 20\%,\quad n = 5 \]

Depreciation factor:

\[ 1 - \frac{20}{100} = 0.8 = \frac{4}{5} \]

Value after 5 years:

\[ V = 15625 \times (0.8)^5 \]

Using fractional form:

\[ V = 15625 \times \left(\frac{4}{5}\right)^5 \]

\[ = 15625 \times \frac{1024}{3125} \]

\[ = 5 \times 1024 = 5120 \]

Final Answer: Rs \(5{,}120\)

Exam Significance

• Very common application of G.P. in depreciation problems
• Key trick: convert decimal to fraction \((0.8 = 4/5)\) for faster calculation
• Appears in boards and JEE as application-based question
• Tests understanding of compound decay vs simple change

Theory Insight

This is a work–time modelling problem using sequences.

• Total work = (workers) × (days)
• Original plan assumes constant workers → fixed work rate
• Actual workers form an A.P. due to daily reduction

Key idea: Total work remains constant in both situations.

Solution Roadmap

• Assume original time = \(x\)
• Form A.P. of workers per day
• Compute total work using sum of A.P.
• Equate both works and solve

Solution

Let original number of days = \(x\)

Total work:

\[ \text{Work} = 150x \]

Actual workers per day form an A.P.:

\[ 150,\; 146,\; 142,\; 138,\; \ldots \]

Here:

\[ a = 150,\quad d = -4,\quad n = x + 8 \]

Last term:

\[ \begin{aligned} l &= 150 + (x+8-1)(-4) \\&= 150 - 4(x+7) \end{aligned} \]

Total work done:

\[ \begin{aligned} S &= \frac{x+8}{2}[a + l]\\ &= \frac{x+8}{2}[150 + 150 - 4(x+7)]\\ &= \frac{x+8}{2}(300 - 4x - 28)\\ &= \frac{x+8}{2}(272 - 4x) \end{aligned} \]

Equating work:

\[ \begin{aligned} \frac{(x+8)(272 - 4x)}{2} &= 150x\\ (x+8)(272 - 4x) &= 300x \end{aligned} \]

Expanding:

\[ \begin{aligned} 272x + 2176 - 4x^2 - 32x &= 300x\\ -4x^2 + 240x + 2176 &= 300x\\ -4x^2 - 60x + 2176 &= 0\\ x^2 + 15x - 544 &= 0\\ (x + 32)(x - 17) &= 0\\ \Rightarrow x &= 17 \end{aligned} \]

Total time taken:

\[ \begin{aligned} x + 8 &= 17 + 8\\ &= 25 \end{aligned} \]

Final Answer: \(25\) days

Exam Significance

• Very important mixed concept: Work + A.P.
• Key idea: decreasing workforce → A.P.
• Frequently asked in boards (5 marks)
• JEE: tests modelling + algebra accuracy
• Common trap: wrong last term → incorrect equation