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🎓 Class 11 Mathematics CBSE Theory Ch 8 — Sequences and Series ⏱ ~15 min
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Targeting Class 11 level in Sequences Series, with Advanced difficulty.

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End-of-Chapter Exercises

Exercise 8.1 — Find terms / write formulas
1. \(a_n=n(n+2)\). Write first 5 terms.
3, 8, 15, 24, 35.
2. \(a_n=n/(n+1)\). Write first 5 terms.
1/2, 2/3, 3/4, 4/5, 5/6.
3. \(a_n=2^n\). Write first 5 terms.
2, 4, 8, 16, 32.
4. \(a_n=(2n-3)/6\). Write first 5 terms.
−1/6, 1/6, 3/6=1/2, 5/6, 7/6.
5. \(a_n=(-1)^{n-1}5^{n+1}\). Write first 5 terms.
+25, −125, +625, −3125, +15625 (signs alternate; magnitudes are 5²,5³,5⁴,…).
6. \(a_n=n(n^2+5)/4\). Write first 5 terms.
a₁=1·6/4=3/2; a₂=2·9/4=9/2; a₃=3·14/4=21/2; a₄=4·21/4=21; a₅=5·30/4=75/2.
7. \(a_n=4n-3\). Find \(a_{17}\) and \(a_{24}\).
a₁₇=4·17−3=65. a₂₄=4·24−3=93.
8. \(a_n=n^2/2^n\). Find \(a_7\).
a₇=49/128.
9. \(a_n=(-1)^{n-1}n^3\). Find \(a_9\).
a₉=(−1)⁸·9³=729.
10. \(a_n=n(n-2)/(n+3)\). Find \(a_{20}\).
a₂₀=20·18/23=360/23.
11. \(a_1=3,\ a_n=3a_{n-1}+2\) for \(n\ge 2\). Find first 5 terms.
a₁=3, a₂=11, a₃=35, a₄=107, a₅=323.
12. \(a_1=-1,\ a_n=a_{n-1}/n,\ n\ge 2\). First 5 terms.
a₁=−1, a₂=−1/2, a₃=−1/6, a₄=−1/24, a₅=−1/120 (i.e. \(-1/n!\)).
13. \(a_1=2,\ a_n=na_{n-1},\ n\ge 2\). First 5 terms and corresponding series.
a₁=2, a₂=4, a₃=12, a₄=48, a₅=240. Series: 2+4+12+48+240+…
14. Fibonacci with \(a_1=a_2=1\). Find \(a_{n+1}/a_n\) for \(n=1,2,3,4,5\).
Sequence: 1, 1, 2, 3, 5, 8. Ratios: 1/1, 2/1, 3/2, 5/3, 8/5 = 1, 2, 1.5, 1.667, 1.6 — converging to φ ≈ 1.618.
Exercise 8.2 — G.P. problems
1. Find the 20th and \(n\)th terms of the G.P. 5/2, 5/4, 5/8, … .
a=5/2, r=1/2. \(a_n=(5/2)\cdot(1/2)^{n-1}=5/2^n\). \(a_{20}=5/2^{20}=5/1048576\).
2. Find the 12th term of a G.P. whose 8th term is 192 and common ratio is 2.
a·2⁷=192, so a=192/128=3/2. a₁₂=(3/2)·2¹¹=3·2¹⁰=3072.
3. The 5th, 8th, 11th terms of a G.P. are p, q, s. Show q²=ps. (Same as Example 8.)
p=ar⁴, q=ar⁷, s=ar¹⁰. q²=a²r¹⁴=ps. ✓
4. The 4th term of a G.P. is square of its 2nd term and the first term is −3. Determine its 7th term.
a=−3, ar³=(ar)². So −3r³=9r², giving r=−3. a₇=ar⁶=−3·729=−2187.
5. Which term of the G.P. 2, 2√2, 4, 4√2, … is 128?
a=2, r=√2. \(2\cdot(\sqrt 2)^{n-1}=128\Rightarrow(\sqrt 2)^{n-1}=64=2^6=(\sqrt 2)^{12}\Rightarrow n-1=12\Rightarrow n=13\).
6. For what values of x are the numbers \(-2/7,\ x,\ -7/2\) in G.P.?
x²=(−2/7)(−7/2)=1, so x=±1.
7. Find the sum to 20 terms of the G.P. \(0.15+0.015+0.0015+\cdots\).
a=0.15, r=0.1. \(S_{20}=0.15(1-0.1^{20})/(1-0.1)=0.15(1-10^{-20})/0.9=(1/6)(1-10^{-20})\). Numerically ≈ 0.16666… = 1/6.
8. Sum to n terms of \(\sqrt 7,\sqrt{21},3\sqrt 7,\ldots\).
a=√7, r=√3. \(S_n=\sqrt 7\cdot[(\sqrt 3)^n-1]/(\sqrt 3-1)\). Multiply top and bottom by \(\sqrt 3+1\): \(S_n=\sqrt 7(\sqrt 3+1)[(\sqrt 3)^n-1]/2\).
9. Sum to n terms of \(1,-a,a^2,-a^3,\ldots\) (\(a\ne -1\)).
G.P. with first term 1, ratio −a. \(S_n=[1-(-a)^n]/(1-(-a))=[1-(-a)^n]/(1+a)\).
10. Sum to n terms of \(x^3,x^5,x^7,\ldots\) (\(x\ne\pm 1\)).
a=x³, r=x². \(S_n=x^3(x^{2n}-1)/(x^2-1)\).
12. The sum of the first three terms of a G.P. is 39/10 and their product is 1. Find the common ratio and the terms.
Let terms be a/r, a, ar. Product a³=1⇒a=1. Sum: 1/r+1+r=39/10, so 10r²−29r+10=0, giving r=2/5 or 5/2. Either G.P. works: (5/2, 1, 2/5) or (2/5, 1, 5/2).
Miscellaneous Exercise on Chapter 8
11. Find the sum to n terms of \(5+55+555+5555+\cdots\).
\(S=5(1+11+111+\cdots)=(5/9)(9+99+999+\cdots)=(5/9)[(10-1)+(100-1)+\cdots]\) \(=(5/9)[10(10^n-1)/9 - n]=(5/81)[10(10^n-1)-9n]\).
8. Find the sum of the products of the corresponding terms of the sequences 2, 4, 8, 16, 32 and 128, 32, 8, 2, 1/2.
Products: 256, 128, 64, 32, 16. This is a G.P. with a=256, r=1/2, n=5. \(S_5=256(1-1/32)/(1/2)=256\cdot(31/32)\cdot 2=496\).
15. The number of bacteria in a culture doubles every hour. If there were 30 originally, how many at the end of 2nd hour, 4th hour, n-th hour?
G.P. with a=30, r=2 (count is 30·2ⁿ at end of nth hour, taking 0th hour as start). End of 2nd hour: 30·2²=120. End of 4th: 30·16=480. End of nth hour: 30·2ⁿ.
17. If \(a, b, c\) are in A.P. and \(b, c, d\) are in G.P. and \(1/c, 1/d, 1/e\) are in A.P., prove that \(a, c, e\) are in G.P.
From AP: 2b = a+c, so b = (a+c)/2. From GP: c² = bd, so d = c²/b = 2c²/(a+c). From AP of reciprocals: 2/d = 1/c + 1/e = (e+c)/(ce), so e = c·d/(2c−d). Substitute and simplify to get c² = ae, hence a, c, e in G.P. \(\square\)
26. Let A = ³√(b·c·d) (any GM expression). Use the AM–GM inequality.
AM ≥ GM: for positive reals, (b+c+d)/3 ≥ ³√(bcd). Equality when b=c=d. The same idea generalises: arithmetic mean ≥ geometric mean ≥ harmonic mean.
Activity: Estimate π via a Series
L4 Analyse
Materials: Calculator.
Predict: Add: 4(1 − 1/3 + 1/5 − 1/7 + 1/9 − ...). Where does the partial sum head?
  1. Compute partial sums: 4·1 = 4; 4(1−1/3) = 8/3 ≈ 2.667; 4(1−1/3+1/5) ≈ 3.467; 4(1−1/3+1/5−1/7) ≈ 2.895.
  2. Continue: the partial sums oscillate, slowly converging.
  3. After 10000 terms, partial sum ≈ 3.1414926…
  4. This is the Madhava–Leibniz series for π. Discovered by Madhava of Sangamagrama (Kerala, c. 1340–1425) — three centuries before Leibniz!
  5. Convergence is very slow because the terms alternate and shrink only as 1/n.
The Madhava–Leibniz series \(\pi/4=\sum_k(-1)^k/(2k+1)\) is conditionally convergent. Faster series (Machin's, Ramanujan's) compute π to billions of digits. The slow convergence here illustrates why series acceleration is a major branch of numerical analysis.

Consolidation Competency-Based Questions

Scenario: A study of bacterial growth tracks colony size hourly: 100, 200, 400, 800, … cells.
Q1. The total cells after 10 hours (cumulative sum 100 + 200 + … + 100·2⁹):
L3 Apply
Answer: G.P. sum: \(100(2^{10}-1)/(2-1)=100\cdot 1023=102\,300\) cells.
Q2. (T/F) "Σ 1/2ⁿ from n=1 to ∞ converges to 1." Verify.
L5 Evaluate
True. a=1/2, r=1/2, |r|<1, S∞ = (1/2)/(1/2) = 1. Geometrically, this is the "Zeno paradox": you cover 1/2 + 1/4 + 1/8 + … of the way, and reach the destination.
Q3. Find the sum 1 + 11 + 111 + 1111 + … to n terms.
L4 Analyse
Solution: = (1/9)[(10−1)+(100−1)+(1000−1)+…+(10ⁿ−1)] = (1/9)[10(10ⁿ−1)/9 − n] = (1/81)[10(10ⁿ−1)−9n].
Q4. (Apply AM–GM) For positive a, b with a+b=10, find max value of ab.
L4 Analyse
Solution: AM = 5; GM = √(ab). AM ≥ GM ⇒ √(ab) ≤ 5, so ab ≤ 25. Maximum 25 at a=b=5.
Q5. Design: a chess-board fortune. Place 1 grain of rice on the 1st square, 2 on the 2nd, 4 on the 3rd, doubling each time, for all 64 squares. How many grains in total?
L6 Create
Solution: Σ from 0 to 63 of 2ᵏ = 2⁶⁴ − 1 = 18,446,744,073,709,551,615 grains. At ~25,000 grains per kg of rice, that's ~7.4×10¹¹ tonnes — about 1000× the world's annual rice production. The classic king-and-the-mathematician fable about exponential growth.

Consolidation Assertion–Reason

Assertion (A): The sum 1 + 1/3 + 1/9 + 1/27 + … = 3/2.
Reason (R): Infinite-GP formula \(S_\infty=a/(1-r)\) with a = 1, r = 1/3 gives 1/(2/3) = 3/2.
(a) Both true, R explains A.
(b) Both true, R doesn't explain A.
(c) A true, R false.
(d) A false, R true.
Answer: (a). Direct application.
Assertion (A): For positive reals a, b, c: (a+b+c)/3 ≥ ³√(abc).
Reason (R): AM ≥ GM holds for any number of positive reals.
(a) Both true, R explains A.
(b) Both true, R doesn't explain A.
(c) A true, R false.
(d) A false, R true.
Answer: (a). A is the n=3 case of R.

Chapter Summary

Key formulas at a glance
  • Sequence: ordered list \(a_1, a_2, a_3, \ldots\); a function on natural numbers.
  • Series: sum of a sequence: \(\sum a_k\).
  • Recurrence: Fibonacci \(a_n = a_{n-1} + a_{n-2}\), \(a_1 = a_2 = 1\).
  • G.P.: \(a_n = ar^{n-1}\); each term is \(r\) times the previous.
  • Sum of GP: \(S_n = a(r^n-1)/(r-1)\) for \(r\ne 1\).
  • Infinite GP (|r| < 1): \(S_\infty = a/(1-r)\).
  • Geometric mean: \(G = \sqrt{ab}\) for positive \(a, b\); the middle term of \(a, G, b\) in G.P.
  • AM ≥ GM: \((a+b)/2 \ge \sqrt{ab}\), equality iff \(a = b\).
  • Special sums: \(\sum k = n(n+1)/2\), \(\sum k^2 = n(n+1)(2n+1)/6\), \(\sum k^3 = [n(n+1)/2]^2\).

Historical Note

Evidence is found that Babylonians, some 4000 years ago, knew of arithmetic and geometric sequences. According to the Boethius (510), the Greek philosopher Pythagoras (570–500 B.C.) seems to have known about arithmetic, geometric and harmonic sequences. The arithmetic, geometric and harmonic means of two numbers were already known to early Indian mathematicians.

The Indian mathematician Aryabhata (476 CE) was the first to give the formula for the sum of squares and cubes of natural numbers in his famous work Aryabhatiyam, written around 499 CE. He also gave the formula for finding the sum to \(n\) terms of an arithmetic sequence. Brahmagupta (598), Mahavira (850) and Bhaskara (1114–1185) also considered the sum of squares and cubes. Another specific type of sequence having important applications in mathematics, called Fibonacci sequence, was discovered by Italian mathematician Leonardo Fibonacci (1170–1250). Seventeenth century witnessed the classification of series into specific forms. In 1671 James Gregory used the term infinite series in connection with infinite sequence. It was only through the rigorous development of algebraic and set theoretic tools that the concepts related to sequence and series could be formulated suitably.

Frequently Asked Questions

What is the chapter summary of Class 11 Maths Chapter 8?
A sequence is a function on the natural numbers; a series is the sum of a sequence. A geometric progression has aₙ = a·rⁿ⁻¹; sum Sₙ = a(rⁿ−1)/(r−1) for r ≠ 1; infinite sum a/(1−r) for |r| < 1. Geometric mean of a, b is √(ab); AM ≥ GM always.
Who first studied geometric progressions?
Indian mathematicians Aryabhata (476 CE) and Brahmagupta (598 CE) gave formulas for sums of arithmetic and geometric progressions. Mahavira (850) and Bhaskara II (12th century) extended these to sums of squares and cubes. Fibonacci (1202) introduced his eponymous sequence.
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