📈 Have You Ever Wondered?
How does an exponential viral tweet reach 1 billion views in 48 hours, or why does a bouncing ball losing 10% height on every bounce travel a finite t...
How does an exponential viral tweet reach 1 billion views in 48 hours, or why does a bouncing ball losing 10% height on every bounce travel a finite total distance before coming to rest? Geometric progressions model compounding physical phenomena.
यह अध्याय क्यों महत्वपूर्ण है
In Class 11 Mathematics, "Sequences and Series" provides an authoritative, curriculum-verified master resource aligned with the 2026–27 NCERT syllabus.
अध्ययन से पूर्व (आवश्यक ज्ञान)
- Arithmetic Progression (AP) from Class 10.
- Linear sequences.
- Sum of terms.
इस अध्याय के लक्ष्य
- Define Sequence, Series, and Progression.
- Analyze Geometric Progression (GP): General term $a_n = ar^{n-1}$.
- Derive sum of first $n$ terms of a GP: $S_n = \frac{a(r^n - 1)}{r - 1}$.
- Compute Geometric Mean (GM) between two numbers: $G = \sqrt{ab}$.
- State and apply the AM-GM Inequality: $AM \ge GM$ for positive real numbers.
अध्याय रूपरेखा एवं प्रगति
1
1. Geometric Progression (GP) Essen...
2
2. Sum of $n$ Terms of a GP
3
3. AM, GM & The Cardinal Inequality
सम्पूर्ण सैद्धांतिक एवं वैचारिक अध्ययन
1. Geometric Progression (GP) Essentials
A sequence where the ratio of any term to its preceding term is constant (Common Ratio $r$): $$\mathbf{a, ar, ar^2, ar^3, \dots, ar^{n-1}} \quad (a_n = ar^{n-1})$$
2. Sum of $n$ Terms of a GP
Sum of $n$ terms: $$\mathbf{S_n = \begin{cases} \frac{a(1 - r^n)}{1 - r}, & r < 1 \\ \frac{a(r^n - 1)}{r - 1}, & r > 1 \end{cases}}$$ For an infinite decaying GP ($|r| < 1$): $\mathbf{S_\infty = \frac{a}{1 - r}}$.
3. AM, GM & The Cardinal Inequality
For two positive real numbers $a$ and $b$:
• Arithmetic Mean (AM): $A = \frac{a + b}{2}$.
• Geometric Mean (GM): $G = \sqrt{ab}$.
• AM-GM Inequality: $$\mathbf{A \ge G} \quad \left(\frac{a+b}{2} \ge \sqrt{ab}\right)$$ with equality holding if and only if $a = b$.
स्व-मूल्यांकन अभ्यास (Check Your Understanding)
मूल वैचारिक स्पष्टता की जांच के लिए नैदानिक प्रश्न। पहले स्वयं हल करें, फिर उत्तर देखें।
1
Find the 10th and $n$th terms of the GP: $5, 25, 125, \dots$
उत्तर एवं व्याख्या देखें
उत्तर: $a = 5, r = 5$. $a_{10} = 5 \times 5^9 = 5^{10}$. General term $a_n = 5 \times 5^{n-1} = 5^n$.
a10 = 5^10, an = 5^n.
2
Find the sum of the first 8 terms of the GP: $1, \frac{2}{3}, \frac{4}{9}, \dots$
उत्तर एवं व्याख्या देखें
उत्तर: $a = 1, r = 2/3 < 1$. $S_8 = \frac{1(1 - (2/3)^8)}{1 - 2/3} = 3[1 - \frac{256}{6561}] = 3 \times \frac{6305}{6561} = \frac{6305}{2187}$.
6305 / 2187.
3
Insert two geometric means between 1 and 27.
उत्तर एवं व्याख्या देखें
उत्तर: Let the sequence be $1, G_1, G_2, 27$. Total 4 terms: $a_4 = ar^3 \implies 27 = 1 \cdot r^3 \implies r = 3$. Means: $G_1 = 1 \times 3 = 3$, $G_2 = 3 \times 3 = 9$.
Means are 3 and 9.
4
If AM and GM of two positive numbers are 10 and 8 respectively, find the numbers.
उत्तर एवं व्याख्या देखें
उत्तर: $A = \frac{a+b}{2} = 10 \implies a + b = 20$. $G = \sqrt{ab} = 8 \implies ab = 64$. $(a - b)^2 = (a+b)^2 - 4ab = 400 - 256 = 144 \implies a - b = 12$. Solving: $a = 16, b = 4$.
Numbers are 16 and 4.
5
Find the sum to infinity of the GP: $6, 1.2, 0.24, \dots$
उत्तर एवं व्याख्या देखें
उत्तर: $a = 6, r = \frac{1.2}{6} = 0.2 < 1$. $S_\infty = \frac{a}{1 - r} = \frac{6}{1 - 0.2} = \frac{6}{0.8} = 7.5$.
7.5.
