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Ch 8Sequences and Seriesअनुक्रम तथा श्रेणीSequences aur Series

Sequences and series, the geometric progression (G.P.) — its nth term, the sum of n terms and the geometric mean — and the relation A.M. ≥ G.M. — with every NCERT exercise solved.अनुक्रम और श्रेणी, गुणोत्तर श्रेढ़ी (G.P.) — इसका n वाँ पद, n पदों का योग और गुणोत्तर माध्य — तथा संबंध A.M. ≥ G.M. — NCERT के सभी प्रश्नों के हल के साथ।Sequence aur series, geometric progression (G.P.) — iska nth term, n terms ka sum aur geometric mean — aur relation A.M. ≥ G.M. — NCERT ke saare questions ke hal ke saath.

Didn’t understand a paragraph? Tap it (or its अ button) → हिंदी / Hinglish / Simple English.कोई पैराग्राफ़ समझ नहीं आया? उस पर (या अ बटन पर) टैप करें → English / Hinglish।Koi paragraph samajh nahi aaya? Us par (ya अ button par) tap kijiye → हिंदी / English.

👆 Words with a open a diagram or graph — tap any label to read what it means.👆 वाले शब्द पर टैप करें — चित्र या ग्राफ़ खुलेगा (हर लेबल पर टैप = उसका अर्थ)।👆 wale word par tap kijiye — diagram ya graph khulega (har label par tap = uska matlab).
📘 NCERT chapter (English PDF)📗 NCERT हिंदी PDF∑ Formulas✎ NCERT solutions🎯 Quiz

Diagrams in this chapterइस अध्याय के चित्रIs chapter ke diagrams

G.P. vs A.P.: multiplying grows much faster than addingG.P. बनाम A.P.: गुणा करने से वृद्धि जोड़ने से बहुत तेज़G.P. vs A.P.: multiplying grows much faster than addingSection 8.4A.M. ≥ G.M. in a semicircleअर्धवृत्त में A.M. ≥ G.M.A.M. ≥ G.M. in a semicircleSection 8.5
In this chapter (NCERT sections)इस अध्याय में (NCERT खंड)Is chapter mein (NCERT sections)
  1. 8.2Sequencesअनुक्रमSequences
  2. 8.3Seriesश्रेणीSeries
  3. 8.4Geometric Progression (G.P.)गुणोत्तर श्रेढ़ी (G.P.)Geometric Progression (G.P.)
  4. 8.5Relationship Between A.M. and G.M.A.M. और G.M. में संबंधA.M. aur G.M. ka relation
  5. ★Yaad rakho — key formulasयाद रखो — मुख्य सूत्रYaad rakho — key formulas
  6. ⚠Common mistakesआम गलतियाँCommon mistakes
  7. ✎NCERT exercises — solutionsNCERT प्रश्नावली — हलNCERT exercises — solutions
  8. 🎯Quizक्विज़Quiz

8.2Sequencesअनुक्रमSequences

SequenceNumbers arranged in a definite order by some rule: a_1, a_2, a_3, \dots, a_n, \dots. a_n is the nth term (general term). A sequence is a function whose domain is \mathbf{N} (or \{1, 2, \dots, k\}). It is finite or infinite.अनुक्रमकिसी नियम से निश्चित क्रम में रखी संख्याएँ: a_1, a_2, a_3, \dots, a_n, \dots। a_n n वाँ पद (व्यापक पद) है। अनुक्रम एक फलन है जिसका प्रांत \mathbf{N} (या \{1, 2, \dots, k\}) है। यह परिमित या अपरिमित होता है।SequenceKisi rule se fixed order mein rakhe numbers: a_1, a_2, a_3, \dots, a_n, \dots. a_n nth term (general term) hai. Sequence ek function hai jiska domain \mathbf{N} (ya \{1, 2, \dots, k\}) hai. Ye finite ya infinite hota hai.

Some sequences are given by a recurrence: e.g. the Fibonacci sequence a_1 = a_2 = 1, a_n = a_{n-1} + a_{n-2} gives 1, 1, 2, 3, 5, 8, …कुछ अनुक्रम पुनरावृत्ति संबंध से दिए जाते हैं: जैसे फिबोनाकी अनुक्रम a_1 = a_2 = 1, a_n = a_{n-1} + a_{n-2} से 1, 1, 2, 3, 5, 8, …Kuch sequences recurrence se diye jaate hain: jaise Fibonacci sequence a_1 = a_2 = 1, a_n = a_{n-1} + a_{n-2} se 1, 1, 2, 3, 5, 8, …

Example 1Write the first three terms of (i) a_n = 2n + 5 (ii) a_n = \frac{n - 3}{4}.उदाहरण 1पहले तीन पद लिखिए: (i) a_n = 2n + 5 (ii) a_n = \frac{n - 3}{4}।Example 1Pehle teen terms likhiye: (i) a_n = 2n + 5 (ii) a_n = \frac{n - 3}{4}.
Solution — step by step (1)हल — चरण दर चरण (1)Hal — step by step (1)
  1. (i)\ 7, 9, 11 \qquad (ii)\ -\frac{1}{2}, -\frac{1}{4}, 0

8.3Seriesश्रेणीSeries

The sum a_1 + a_2 + a_3 + \dots of the terms of a sequence is a series, written briefly as \sum_{k=1}^{n} a_k.अनुक्रम के पदों का योग a_1 + a_2 + a_3 + \dots श्रेणी है, संक्षेप में \sum_{k=1}^{n} a_k।Sequence ke terms ka sum a_1 + a_2 + a_3 + \dots series hai, short mein \sum_{k=1}^{n} a_k.

8.4Geometric Progression (G.P.)गुणोत्तर श्रेढ़ी (G.P.)Geometric Progression (G.P.)

In a G.P. each term after the first is the previous term multiplied by a fixed non-zero number r, the common ratio: \frac{a_{k+1}}{a_k} = r. Examples: 2, 4, 8, 16, … (r = 2); \frac{1}{9}, -\frac{1}{27}, \frac{1}{81}, \dots (r = -\frac{1}{3}). Compare it with an A.P. in the chart.G.P. में पहले पद के बाद हर पद पिछले पद को एक निश्चित अशून्य संख्या r (सार्व अनुपात) से गुणा करके मिलता है: \frac{a_{k+1}}{a_k} = r। उदाहरण: 2, 4, 8, 16, … (r = 2); \frac{1}{9}, -\frac{1}{27}, \frac{1}{81}, \dots (r = -\frac{1}{3})। चार्ट में इसकी A.P. से तुलना कीजिए।G.P. mein pehle term ke baad har term pichhle term ko ek fixed non-zero number r (common ratio) se guna karke milta hai: \frac{a_{k+1}}{a_k} = r. Examples: 2, 4, 8, 16, … (r = 2); \frac{1}{9}, -\frac{1}{27}, \frac{1}{81}, \dots (r = -\frac{1}{3}). Chart mein iski A.P. se tulna kijiye.

G.P. vs A.P.: multiplying grows much faster than addingG.P. बनाम A.P.: गुणा करने से वृद्धि जोड़ने से बहुत तेज़G.P. vs A.P.: multiplying grows much faster than addingSection 8.4 · 2 tappable partsअनुभाग 8.4 · 2 टैप करने लायक हिस्सेSection 8.4 · 2 tappable partsOpen
a_n = ar^{n-1}
nth term (general term)n वाँ पद (व्यापक पद)nth term (general term)
S_n = \frac{a(r^n - 1)}{r - 1} = \frac{a(1 - r^n)}{1 - r}
sum of n terms; if r = 1, S_n = nan पदों का योग; यदि r = 1 तो S_n = nan terms ka sum; agar r = 1 to S_n = na
r \ne 1
G = \sqrt{ab}
geometric mean of positive a, bधनात्मक a, b का गुणोत्तर माध्यpositive a, b ka geometric mean

To insert n numbers G_1, \dots, G_n between a and b so that a, G_1, \dots, G_n, b is a G.P., use b = ar^{n+1}, i.e. r = \left(\frac{b}{a}\right)^{\frac{1}{n+1}}.a और b के बीच n संख्याएँ G_1, \dots, G_n ऐसे डालने के लिए कि a, G_1, \dots, G_n, b G.P. हो, b = ar^{n+1} अर्थात् r = \left(\frac{b}{a}\right)^{\frac{1}{n+1}} लीजिए।a aur b ke beech n numbers G_1, \dots, G_n aise daalne ke liye ki a, G_1, \dots, G_n, b G.P. ho, b = ar^{n+1} yaani r = \left(\frac{b}{a}\right)^{\frac{1}{n+1}} lijiye.

Example 4Find the 10th and nth terms of the G.P. 5, 25, 125, …उदाहरण 4G.P. 5, 25, 125, … का 10वाँ और n वाँ पद ज्ञात कीजिए।Example 4G.P. 5, 25, 125, … ka 10th aur nth term nikaliye.
Solution — step by step (1)हल — चरण दर चरण (1)Hal — step by step (1)
  1. a = 5,\ r = 5:\ a_{10} = 5 \cdot 5^9 = 5^{10},\quad a_n = 5 \cdot 5^{n-1} = 5^n
Example 5Which term of the G.P. 2, 8, 32, … is 131072?उदाहरण 5G.P. 2, 8, 32, … का कौन-सा पद 131072 है?Example 5G.P. 2, 8, 32, … ka kaun sa term 131072 hai?
Solution — step by step (1)हल — चरण दर चरण (1)Hal — step by step (1)
  1. 2 \cdot 4^{n-1} = 131072 \Rightarrow 4^{n-1} = 65536 = 4^8 \Rightarrow n = 9
Example 6In a G.P. the 3rd term is 24 and the 6th term is 192. Find the 10th term.उदाहरण 6किसी G.P. का तीसरा पद 24 और छठा पद 192 है। 10वाँ पद ज्ञात कीजिए।Example 6Kisi G.P. ka 3rd term 24 aur 6th term 192 hai. 10th term nikaliye.
Solution — step by step (1)हल — चरण दर चरण (1)Hal — step by step (1)
  1. \frac{ar^5}{ar^2} = \frac{192}{24} \Rightarrow r^3 = 8 \Rightarrow r = 2,\ a = 6 \Rightarrow a_{10} = 6 \cdot 2^9 = 3072
Example 8How many terms of the G.P. 3, \frac{3}{2}, \frac{3}{4}, \dots give the sum \frac{3069}{512}?उदाहरण 8G.P. 3, \frac{3}{2}, \frac{3}{4}, \dots के कितने पदों का योग \frac{3069}{512} है?Example 8G.P. 3, \frac{3}{2}, \frac{3}{4}, \dots ke kitne terms ka sum \frac{3069}{512} hai?
Solution — step by step (2)हल — चरण दर चरण (2)Hal — step by step (2)
  1. S_n = \frac{3\left(1 - \frac{1}{2^n}\right)}{1 - \frac{1}{2}} = 6\left(1 - \frac{1}{2^n}\right) = \frac{3069}{512}
  2. 1 - \frac{1}{2^n} = \frac{3069}{3072} \Rightarrow 2^n = 1024 \Rightarrow n = 10
Example 9The sum of the first three terms of a G.P. is \frac{13}{12} and their product is −1. Find r and the terms.उदाहरण 9किसी G.P. के पहले तीन पदों का योग \frac{13}{12} और गुणनफल −1 है। r और पद ज्ञात कीजिए।Example 9Kisi G.P. ke pehle teen terms ka sum \frac{13}{12} aur product −1 hai. r aur terms nikaliye.
Solution — step by step (1)हल — चरण दर चरण (1)Hal — step by step (1)
  1. Take the terms as \frac{a}{r}, a, ar — the product is then just a^3.पदों को \frac{a}{r}, a, ar लीजिए — तब गुणनफल केवल a^3 है।Terms ko \frac{a}{r}, a, ar lijiye — tab product sirf a^3 hai.
    a^3 = -1 \Rightarrow a = -1;\quad -\frac{1}{r} - 1 - r = \frac{13}{12} \Rightarrow 12r^2 + 25r + 12 = 0 \Rightarrow r = -\frac{3}{4} \text{ or } -\frac{4}{3}

✔ Terms \frac{4}{3}, -1, \frac{3}{4} or \frac{3}{4}, -1, \frac{4}{3}.पद \frac{4}{3}, -1, \frac{3}{4} या \frac{3}{4}, -1, \frac{4}{3}।Terms \frac{4}{3}, -1, \frac{3}{4} ya \frac{3}{4}, -1, \frac{4}{3}.

Example 11A person has 2 parents, 4 grandparents, 8 great grandparents, … How many ancestors in the 10 generations before him?उदाहरण 11किसी व्यक्ति के 2 माता-पिता, 4 दादा-दादी/नाना-नानी, 8 परदादा… हैं। उससे पहले की 10 पीढ़ियों में कितने पूर्वज?Example 11Kisi insaan ke 2 parents, 4 grandparents, 8 great grandparents… hain. Usse pehle ki 10 generations mein kitne ancestors?
Solution — step by step (1)हल — चरण दर चरण (1)Hal — step by step (1)
  1. S_{10} = \frac{2(2^{10} - 1)}{2 - 1} = 2046

8.5Relationship Between A.M. and G.M.A.M. और G.M. में संबंधA.M. aur G.M. ka relation

A - G = \frac{a + b}{2} - \sqrt{ab} = \frac{(\sqrt{a} - \sqrt{b})^2}{2} \ge 0
so A ≥ G (equal only if a = b)अतः A ≥ G (बराबर केवल जब a = b)isliye A ≥ G (barabar sirf jab a = b)
A.M. ≥ G.M. in a semicircleअर्धवृत्त में A.M. ≥ G.M.A.M. ≥ G.M. in a semicircleSection 8.5 · 5 tappable partsअनुभाग 8.5 · 5 टैप करने लायक हिस्सेSection 8.5 · 5 tappable partsOpen
Example 13The A.M. and G.M. of two positive numbers are 10 and 8. Find the numbers.उदाहरण 13दो धनात्मक संख्याओं के A.M. और G.M. 10 और 8 हैं। संख्याएँ ज्ञात कीजिए।Example 13Do positive numbers ke A.M. aur G.M. 10 aur 8 hain. Numbers nikaliye.
Solution — step by step (1)हल — चरण दर चरण (1)Hal — step by step (1)
  1. a + b = 20,\ ab = 64 \Rightarrow (a - b)^2 = 400 - 256 = 144 \Rightarrow a - b = 12

✔ The numbers are 16 and 4.संख्याएँ 16 और 4 हैं।Numbers 16 aur 4 hain.

Rationalised bookArithmetic progressions are not a separate section in the current book (you know them from Class 10), but some Miscellaneous questions use A.P. sums: S_n = \frac{n}{2}[2a + (n - 1)d].संशोधित पुस्तकवर्तमान पुस्तक में समांतर श्रेढ़ी अलग खंड नहीं है (कक्षा 10 से आप जानते हैं), पर कुछ विविध प्रश्नों में A.P. योग लगता है: S_n = \frac{n}{2}[2a + (n - 1)d]।Rationalised bookCurrent book mein A.P. alag section nahi hai (Class 10 se aap jaante hain), par kuch Misc questions mein A.P. sum lagta hai: S_n = \frac{n}{2}[2a + (n - 1)d].

★ Yaad rakho — key formulas★ याद रखो — मुख्य सूत्र★ Yaad rakho — key formulas

  1. G.P.: a_n = ar^{n-1}, S_n = \frac{a(r^n - 1)}{r - 1} (r \ne 1).G.P.: a_n = ar^{n-1}, S_n = \frac{a(r^n - 1)}{r - 1} (r \ne 1)।G.P.: a_n = ar^{n-1}, S_n = \frac{a(r^n - 1)}{r - 1} (r \ne 1).
  2. Three terms in G.P.: take \frac{a}{r}, a, ar (product a^3).G.P. में तीन पद: \frac{a}{r}, a, ar लीजिए (गुणनफल a^3)।G.P. mein teen terms: \frac{a}{r}, a, ar lijiye (product a^3).
  3. a, b, c in G.P. \iff b^2 = ac.a, b, c G.P. में \iff b^2 = ac।a, b, c G.P. mein \iff b^2 = ac.
  4. G.M. = \sqrt{ab}; A.M. \ge G.M. for positive a, b.G.M. = \sqrt{ab}; धनात्मक a, b के लिए A.M. \ge G.M.।G.M. = \sqrt{ab}; positive a, b ke liye A.M. \ge G.M.
  5. 8 + 88 + 888 + \dots = \frac{8}{9}(9 + 99 + 999 + \dots) = \frac{8}{9}\left[(10 + 100 + \dots) - n\right].8 + 88 + 888 + \dots = \frac{8}{9}(9 + 99 + 999 + \dots) = \frac{8}{9}\left[(10 + 100 + \dots) - n\right]।8 + 88 + 888 + \dots = \frac{8}{9}(9 + 99 + 999 + \dots) = \frac{8}{9}\left[(10 + 100 + \dots) - n\right].

⚠ Common mistakes⚠ आम गलतियाँ⚠ Common mistakes

  • Using ar^n for the nth term — it is ar^{n-1}.n वें पद के लिए ar^n लेना — यह ar^{n-1} है।nth term ke liye ar^n lena — ye ar^{n-1} hai.
  • Applying S_n = \frac{a(r^n - 1)}{r - 1} when r = 1 (division by zero); then S_n = na.r = 1 पर S_n = \frac{a(r^n - 1)}{r - 1} लगाना (शून्य से भाग); तब S_n = na।r = 1 par S_n = \frac{a(r^n - 1)}{r - 1} lagana (zero se bhaag); tab S_n = na.
  • Taking only the positive root when r^2 = 4: r can be −2 too — check both.r^2 = 4 पर केवल धन मूल लेना: r = −2 भी हो सकता है — दोनों जाँचिए।r^2 = 4 par sirf positive root lena: r = −2 bhi ho sakta hai — dono check kijiye.
  • Writing G.M. = \frac{a + b}{2} — that is the A.M.; G.M. = \sqrt{ab}.G.M. = \frac{a + b}{2} लिखना — वह A.M. है; G.M. = \sqrt{ab}।G.M. = \frac{a + b}{2} likhna — wo A.M. hai; G.M. = \sqrt{ab}.
  • Mixing up the common ratio of a decreasing G.P.: for 3, 3/2, 3/4, … r = 1/2, not 2.घटती G.P. का सार्व अनुपात उलटना: 3, 3/2, 3/4, … में r = 1/2, न कि 2।Ghatti G.P. ka common ratio ulatna: 3, 3/2, 3/4, … mein r = 1/2, na ki 2.

✎ NCERT exercises — solutions✎ NCERT प्रश्नावली — हल✎ NCERT exercises — solutions

All 64 questions of the NCERT exercises. Tap a question to see the solution — try it yourself first!NCERT प्रश्नावली के सभी 64 प्रश्न। हल देखने के लिए प्रश्न पर टैप करें — पहले स्वयं हल करके देखें!NCERT exercises ke saare 64 questions. Hal dekhne ke liye question par tap kijiye — pehle khud try kijiye!

Exercise 8.1प्रश्नावली 8.1Exercise 8.1 · 14 Q

Exercises 11–13: first five terms and the series.प्रश्न 11–13: पहले पाँच पद और श्रेणी।Exercises 11–13: first five terms and the series.

Q1. a_n = n(n + 2)

✔ 3, 8, 15, 24, 35

Q2. a_n = \frac{n}{n + 1}

✔ \frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \frac{4}{5}, \frac{5}{6}

Q3. a_n = 2^n

✔ 2, 4, 8, 16, 32

Q4. a_n = \frac{2n - 3}{6}

✔ -\frac{1}{6}, \frac{1}{6}, \frac{1}{2}, \frac{5}{6}, \frac{7}{6}

Q5. a_n = (-1)^{n-1} 5^{n+1}

✔ 25, −125, 625, −3125, 15625

Q6. a_n = n\,\frac{n^2 + 5}{4}
  1. \frac{1 \cdot 6}{4}, \frac{2 \cdot 9}{4}, \frac{3 \cdot 14}{4}, \frac{4 \cdot 21}{4}, \frac{5 \cdot 30}{4}

✔ \frac{3}{2}, \frac{9}{2}, \frac{21}{2}, 21, \frac{75}{2}

Q7. a_n = 4n - 3; a_{17}, a_{24}
  1. a_{17} = 68 - 3 = 65,\quad a_{24} = 96 - 3 = 93
Q8. a_n = \frac{n^2}{2^n}; a_7
  1. a_7 = \frac{49}{128}
Q9. a_n = (-1)^{n-1} n^3; a_9
  1. a_9 = (-1)^8 \cdot 729 = 729
Q10. a_n = \frac{n(n - 2)}{n + 3}; a_{20}
  1. a_{20} = \frac{20 \times 18}{23} = \frac{360}{23}
Q11. a_1 = 3, a_n = 3a_{n-1} + 2 for n > 1a_1 = 3, a_n = 3a_{n-1} + 2, n > 1a_1 = 3, a_n = 3a_{n-1} + 2 for n > 1
  1. 3,\ 3(3) + 2 = 11,\ 35,\ 107,\ 323

✔ Series: 3 + 11 + 35 + 107 + 323 + …श्रेणी: 3 + 11 + 35 + 107 + 323 + …Series: 3 + 11 + 35 + 107 + 323 + …

Q12. a_1 = -1, a_n = \frac{a_{n-1}}{n}, n \ge 2
  1. -1,\ -\frac{1}{2},\ -\frac{1}{6},\ -\frac{1}{24},\ -\frac{1}{120}

✔ Series: -1 - \frac{1}{2} - \frac{1}{6} - \frac{1}{24} - \frac{1}{120} - \dotsश्रेणी: -1 - \frac{1}{2} - \frac{1}{6} - \frac{1}{24} - \frac{1}{120} - \dotsSeries: -1 - \frac{1}{2} - \frac{1}{6} - \frac{1}{24} - \frac{1}{120} - \dots

Q13. a_1 = a_2 = 2, a_n = a_{n-1} - 1, n > 2
  1. 2, 2, 1, 0, -1

✔ Series: 2 + 2 + 1 + 0 + (−1) + …श्रेणी: 2 + 2 + 1 + 0 + (−1) + …Series: 2 + 2 + 1 + 0 + (−1) + …

Q14. Fibonacci: 1 = a_1 = a_2, a_n = a_{n-1} + a_{n-2}. Find \frac{a_{n+1}}{a_n} for n = 1, …, 5.फिबोनाकी: 1 = a_1 = a_2, a_n = a_{n-1} + a_{n-2}। n = 1, …, 5 के लिए \frac{a_{n+1}}{a_n} ज्ञात कीजिए।Fibonacci: 1 = a_1 = a_2, a_n = a_{n-1} + a_{n-2}. Find \frac{a_{n+1}}{a_n} for n = 1, …, 5.
  1. a: 1, 1, 2, 3, 5, 8 \Rightarrow \frac{1}{1}, \frac{2}{1}, \frac{3}{2}, \frac{5}{3}, \frac{8}{5}

✔ 1, 2, \frac{3}{2}, \frac{5}{3}, \frac{8}{5}

Exercise 8.2प्रश्नावली 8.2Exercise 8.2 · 32 Q

Exercises 7–10: sum to the given number of terms.प्रश्न 7–10: दिए गए पदों तक योग।Exercises 7–10: sum to the given number of terms.

Q1. Find the 20th and nth terms of the G.P. \frac{5}{2}, \frac{5}{4}, \frac{5}{8}, \dotsG.P. \frac{5}{2}, \frac{5}{4}, \frac{5}{8}, \dots का 20वाँ और n वाँ पद ज्ञात कीजिए।Find the 20th and nth terms of the G.P. \frac{5}{2}, \frac{5}{4}, \frac{5}{8}, \dots
  1. a = \frac{5}{2},\ r = \frac{1}{2}:\ a_{20} = \frac{5}{2}\left(\frac{1}{2}\right)^{19} = \frac{5}{2^{20}},\quad a_n = \frac{5}{2^n}
Q2. Find the 12th term of a G.P. whose 8th term is 192 and r = 2.उस G.P. का 12वाँ पद ज्ञात कीजिए जिसका 8वाँ पद 192 और r = 2 है।Find the 12th term of a G.P. whose 8th term is 192 and r = 2.
  1. a_{12} = a_8 r^4 = 192 \times 16 = 3072
Q3. The 5th, 8th and 11th terms of a G.P. are p, q, s. Show that q^2 = ps.किसी G.P. के 5वें, 8वें और 11वें पद p, q, s हैं। दर्शाइए कि q^2 = ps।The 5th, 8th and 11th terms of a G.P. are p, q, s. Show that q^2 = ps.
  1. p = ar^4,\ q = ar^7,\ s = ar^{10} \Rightarrow ps = a^2r^{14} = (ar^7)^2 = q^2
Q4. The 4th term of a G.P. is the square of its 2nd term and the first term is −3. Find the 7th term.किसी G.P. का चौथा पद उसके दूसरे पद का वर्ग है और पहला पद −3 है। 7वाँ पद ज्ञात कीजिए।The 4th term of a G.P. is the square of its 2nd term and the first term is −3. Find the 7th term.
  1. ar^3 = (ar)^2 \Rightarrow r = a = -3 \Rightarrow a_7 = -3(-3)^6 = -2187
Q5. Which term of (a) 2, 2√2, 4, … is 128? (b) √3, 3, 3√3, … is 729? (c) \frac{1}{3}, \frac{1}{9}, \frac{1}{27}, \dots is \frac{1}{19683}?कौन-सा पद: (a) 2, 2√2, 4, … में 128? (b) √3, 3, 3√3, … में 729? (c) \frac{1}{3}, \frac{1}{9}, \frac{1}{27}, \dots में \frac{1}{19683}?Which term of (a) 2, 2√2, 4, … is 128? (b) √3, 3, 3√3, … is 729? (c) \frac{1}{3}, \frac{1}{9}, \frac{1}{27}, \dots is \frac{1}{19683}?
  1. (a)\ 2(\sqrt{2})^{n-1} = 128 \Rightarrow 2^{\frac{n-1}{2}} = 2^6 \Rightarrow n = 13
  2. (b)\ (\sqrt{3})^n = 729 = 3^6 \Rightarrow n = 12 \qquad (c)\ \left(\frac{1}{3}\right)^n = 3^{-9} \Rightarrow n = 9
Q6. For what x are -\frac{2}{7}, x, -\frac{7}{2} in G.P.?x के किन मानों के लिए -\frac{2}{7}, x, -\frac{7}{2} G.P. में हैं?For what x are -\frac{2}{7}, x, -\frac{7}{2} in G.P.?
  1. x^2 = \left(-\frac{2}{7}\right)\left(-\frac{7}{2}\right) = 1 \Rightarrow x = \pm 1
Q7. 0.15, 0.015, 0.0015, … 20 terms0.15, 0.015, 0.0015, … 20 पद0.15, 0.015, 0.0015, … 20 terms
  1. S_{20} = \frac{0.15(1 - 0.1^{20})}{1 - 0.1} = \frac{1}{6}\left(1 - (0.1)^{20}\right)
Q8. √7, √21, 3√7, … n terms√7, √21, 3√7, … n पद√7, √21, 3√7, … n terms
  1. r = \sqrt{3}:\ S_n = \frac{\sqrt{7}\left((\sqrt{3})^n - 1\right)}{\sqrt{3} - 1} = \frac{\sqrt{7}(\sqrt{3} + 1)}{2}\left(3^{\frac{n}{2}} - 1\right)
Q9. 1, −a, a², −a³, … n terms (a ≠ −1)1, −a, a², −a³, … n पद (a ≠ −1)1, −a, a², −a³, … n terms (a ≠ −1)
  1. r = -a:\ S_n = \frac{1 - (-a)^n}{1 + a}
Q10. x³, x⁵, x⁷, … n terms (x ≠ ±1)x³, x⁵, x⁷, … n पद (x ≠ ±1)x³, x⁵, x⁷, … n terms (x ≠ ±1)
  1. r = x^2:\ S_n = \frac{x^3(1 - x^{2n})}{1 - x^2}
Q11. Evaluate \sum_{k=1}^{11}(2 + 3^k).\sum_{k=1}^{11}(2 + 3^k) का मान ज्ञात कीजिए।Evaluate \sum_{k=1}^{11}(2 + 3^k).
  1. = 22 + \frac{3(3^{11} - 1)}{2} = 22 + \frac{3 \times 177146}{2} = 22 + 265719 = 265741
Q12. The sum of the first three terms of a G.P. is \frac{39}{10} and their product is 1. Find r and the terms.किसी G.P. के पहले तीन पदों का योग \frac{39}{10} और गुणनफल 1 है। r और पद ज्ञात कीजिए।The sum of the first three terms of a G.P. is \frac{39}{10} and their product is 1. Find r and the terms.
  1. \frac{a}{r} \cdot a \cdot ar = a^3 = 1 \Rightarrow a = 1;\ \frac{1}{r} + 1 + r = \frac{39}{10} \Rightarrow 10r^2 - 29r + 10 = 0
  2. r = \frac{29 \pm 21}{20} = \frac{5}{2} \text{ or } \frac{2}{5}

✔ Terms \frac{2}{5}, 1, \frac{5}{2} or \frac{5}{2}, 1, \frac{2}{5}.पद \frac{2}{5}, 1, \frac{5}{2} या \frac{5}{2}, 1, \frac{2}{5}।Terms \frac{2}{5}, 1, \frac{5}{2} ya \frac{5}{2}, 1, \frac{2}{5}.

Q13. How many terms of 3, 3², 3³, … give the sum 120?3, 3², 3³, … के कितने पदों का योग 120 है?How many terms of 3, 3², 3³, … give the sum 120?
  1. \frac{3(3^n - 1)}{2} = 120 \Rightarrow 3^n = 81 \Rightarrow n = 4
Q14. The sum of the first three terms of a G.P. is 16 and of the next three is 128. Find a, r and S_n.किसी G.P. के पहले तीन पदों का योग 16 और अगले तीन का 128 है। a, r और S_n ज्ञात कीजिए।The sum of the first three terms of a G.P. is 16 and of the next three is 128. Find a, r and S_n.
  1. a(1 + r + r^2) = 16,\ ar^3(1 + r + r^2) = 128 \Rightarrow r^3 = 8 \Rightarrow r = 2,\ a = \frac{16}{7}
  2. S_n = \frac{16}{7}(2^n - 1)
Q15. A G.P. has a = 729 and 7th term 64. Find S_7.एक G.P. में a = 729 और 7वाँ पद 64 है। S_7 ज्ञात कीजिए।A G.P. has a = 729 and 7th term 64. Find S_7.
  1. 729r^6 = 64 \Rightarrow r = \pm\frac{2}{3}
  2. r = \frac{2}{3}: S_7 = \frac{729\left(1 - \frac{128}{2187}\right)}{\frac{1}{3}} = 2059;\quad r = -\frac{2}{3}: S_7 = \frac{729\left(1 + \frac{128}{2187}\right)}{\frac{5}{3}} = 463
Q16. Find a G.P. whose first two terms add to −4 and whose 5th term is 4 times the 3rd.ऐसी G.P. ज्ञात कीजिए जिसके पहले दो पदों का योग −4 और 5वाँ पद तीसरे का 4 गुना हो।Find a G.P. whose first two terms add to −4 and whose 5th term is 4 times the 3rd.
  1. ar^4 = 4ar^2 \Rightarrow r = \pm 2;\quad a(1 + r) = -4
  2. r = 2: a = -\frac{4}{3};\quad r = -2: a = 4

✔ -\frac{4}{3}, -\frac{8}{3}, -\frac{16}{3}, \dots or 4, −8, 16, −32, …-\frac{4}{3}, -\frac{8}{3}, -\frac{16}{3}, \dots या 4, −8, 16, −32, …-\frac{4}{3}, -\frac{8}{3}, -\frac{16}{3}, \dots ya 4, −8, 16, −32, …

Q17. The 4th, 10th and 16th terms of a G.P. are x, y, z. Prove that x, y, z are in G.P.किसी G.P. के 4थे, 10वें और 16वें पद x, y, z हैं। सिद्ध कीजिए कि x, y, z G.P. में हैं।The 4th, 10th and 16th terms of a G.P. are x, y, z. Prove that x, y, z are in G.P.
  1. x = ar^3,\ y = ar^9,\ z = ar^{15} \Rightarrow xz = a^2r^{18} = y^2
Q18. Sum to n terms of 8, 88, 888, 8888, …8, 88, 888, 8888, … का n पदों तक योगSum to n terms of 8, 88, 888, 8888, …
  1. = \frac{8}{9}(9 + 99 + 999 + \dots) = \frac{8}{9}\left[(10 + 10^2 + \dots + 10^n) - n\right]
  2. = \frac{8}{9}\left[\frac{10(10^n - 1)}{9} - n\right] = \frac{80}{81}(10^n - 1) - \frac{8n}{9}
Q19. Sum of the products of corresponding terms of 2, 4, 8, 16, 32 and 128, 32, 8, 2, ½.2, 4, 8, 16, 32 और 128, 32, 8, 2, ½ के संगत पदों के गुणनफलों का योग।Sum of the products of corresponding terms of 2, 4, 8, 16, 32 and 128, 32, 8, 2, ½.
  1. 256 + 128 + 64 + 32 + 16 = 496
Q20. Show that the products of corresponding terms of a, ar, …, arⁿ⁻¹ and A, AR, …, ARⁿ⁻¹ form a G.P. and find its ratio.दर्शाइए कि a, ar, …, arⁿ⁻¹ और A, AR, …, ARⁿ⁻¹ के संगत पदों के गुणनफल G.P. बनाते हैं और उसका अनुपात ज्ञात कीजिए।Show that the products of corresponding terms of a, ar, …, arⁿ⁻¹ and A, AR, …, ARⁿ⁻¹ form a G.P. and find its ratio.
  1. aA,\ (aA)(rR),\ (aA)(rR)^2, \dots \Rightarrow \text{ratio} = rR
Q21. Find four numbers in G.P. where the 3rd exceeds the 1st by 9 and the 2nd exceeds the 4th by 18.G.P. में चार संख्याएँ जिनमें तीसरी पहली से 9 अधिक और दूसरी चौथी से 18 अधिक हो।Find four numbers in G.P. where the 3rd exceeds the 1st by 9 and the 2nd exceeds the 4th by 18.
  1. a(r^2 - 1) = 9,\ ar(1 - r^2) = 18 \Rightarrow -r = 2 \Rightarrow r = -2,\ a = 3

✔ 3, −6, 12, −24

Q22. If the pth, qth and rth terms of a G.P. are a, b, c, prove that a^{q-r} b^{r-p} c^{p-q} = 1.यदि किसी G.P. के p वें, q वें और r वें पद a, b, c हैं, तो सिद्ध कीजिए a^{q-r} b^{r-p} c^{p-q} = 1।If the pth, qth and rth terms of a G.P. are a, b, c, prove that a^{q-r} b^{r-p} c^{p-q} = 1.
  1. a = AR^{p-1},\ b = AR^{q-1},\ c = AR^{r-1}
  2. \text{product} = A^{(q-r)+(r-p)+(p-q)} R^{(p-1)(q-r)+(q-1)(r-p)+(r-1)(p-q)} = A^0 R^0 = 1
Q23. The first and nth terms of a G.P. are a and b, and P is the product of n terms. Prove P^2 = (ab)^n.किसी G.P. का पहला और n वाँ पद a और b हैं, और P, n पदों का गुणनफल है। सिद्ध कीजिए P^2 = (ab)^n।The first and nth terms of a G.P. are a and b, and P is the product of n terms. Prove P^2 = (ab)^n.
  1. P = a^n r^{0 + 1 + \dots + (n-1)} = a^n r^{\frac{n(n-1)}{2}} \Rightarrow P^2 = a^{2n} r^{n(n-1)} = \left(a \cdot ar^{n-1}\right)^n = (ab)^n
Q24. Show that the ratio of the sum of the first n terms of a G.P. to the sum of terms (n + 1)th to (2n)th is \frac{1}{r^n}.दर्शाइए कि किसी G.P. के पहले n पदों के योग और (n + 1) वें से (2n) वें पदों के योग का अनुपात \frac{1}{r^n} है।Show that the ratio of the sum of the first n terms of a G.P. to the sum of terms (n + 1)th to (2n)th is \frac{1}{r^n}.
  1. \frac{\frac{a(r^n - 1)}{r - 1}}{\frac{ar^n(r^n - 1)}{r - 1}} = \frac{1}{r^n}
Q25. If a, b, c, d are in G.P., show that (a^2 + b^2 + c^2)(b^2 + c^2 + d^2) = (ab + bc + cd)^2.यदि a, b, c, d G.P. में हैं, तो दर्शाइए (a^2 + b^2 + c^2)(b^2 + c^2 + d^2) = (ab + bc + cd)^2।If a, b, c, d are in G.P., show that (a^2 + b^2 + c^2)(b^2 + c^2 + d^2) = (ab + bc + cd)^2.
  1. b = ar, c = ar^2, d = ar^3:\ \text{LHS} = a^2(1 + r^2 + r^4) \cdot a^2r^2(1 + r^2 + r^4)
  2. \text{RHS} = \left(a^2r(1 + r^2 + r^4)\right)^2 = a^4r^2(1 + r^2 + r^4)^2 = \text{LHS}
Q26. Insert two numbers between 3 and 81 to make a G.P.3 और 81 के बीच दो संख्याएँ डालिए ताकि G.P. बने।Insert two numbers between 3 and 81 to make a G.P.
  1. 81 = 3r^3 \Rightarrow r = 3

✔ 9 and 279 और 279 aur 27

Q27. Find n so that \frac{a^{n+1} + b^{n+1}}{a^n + b^n} is the G.M. of a and b.n ज्ञात कीजिए ताकि \frac{a^{n+1} + b^{n+1}}{a^n + b^n}, a और b का G.M. हो।Find n so that \frac{a^{n+1} + b^{n+1}}{a^n + b^n} is the G.M. of a and b.
  1. a^{n+1} + b^{n+1} = \sqrt{ab}(a^n + b^n) \Rightarrow a^{n+\frac{1}{2}}\left(\sqrt{a} - \sqrt{b}\right) = b^{n+\frac{1}{2}}\left(\sqrt{a} - \sqrt{b}\right)
  2. \left(\frac{a}{b}\right)^{n+\frac{1}{2}} = 1 = \left(\frac{a}{b}\right)^0 \Rightarrow n = -\frac{1}{2}
Q28. The sum of two numbers is 6 times their G.M. Show that the numbers are in the ratio (3 + 2\sqrt{2}) : (3 - 2\sqrt{2}).दो संख्याओं का योग उनके G.M. का 6 गुना है। दर्शाइए कि संख्याएँ (3 + 2\sqrt{2}) : (3 - 2\sqrt{2}) के अनुपात में हैं।The sum of two numbers is 6 times their G.M. Show that the numbers are in the ratio (3 + 2\sqrt{2}) : (3 - 2\sqrt{2}).
  1. a + b = 6\sqrt{ab};\ t = \sqrt{\frac{a}{b}}:\ t^2 - 6t + 1 = 0 \Rightarrow t = 3 + 2\sqrt{2}
  2. \frac{a}{b} = (3 + 2\sqrt{2})^2 = \frac{3 + 2\sqrt{2}}{3 - 2\sqrt{2}} \quad (\text{since } (3 + 2\sqrt{2})(3 - 2\sqrt{2}) = 1)
Q29. If A and G are the A.M. and G.M. of two positive numbers, prove the numbers are A \pm \sqrt{(A + G)(A - G)}.यदि A और G दो धनात्मक संख्याओं के A.M. और G.M. हैं, तो सिद्ध कीजिए कि संख्याएँ A \pm \sqrt{(A + G)(A - G)} हैं।If A and G are the A.M. and G.M. of two positive numbers, prove the numbers are A \pm \sqrt{(A + G)(A - G)}.
  1. a + b = 2A,\ ab = G^2 \Rightarrow a, b \text{ are roots of } x^2 - 2Ax + G^2 = 0a + b = 2A,\ ab = G^2 \Rightarrow a, b \text{ इसके मूल हैं: } x^2 - 2Ax + G^2 = 0a + b = 2A,\ ab = G^2 \Rightarrow a, b \text{ iske roots hain: } x^2 - 2Ax + G^2 = 0
  2. x = A \pm \sqrt{A^2 - G^2} = A \pm \sqrt{(A + G)(A - G)}
Q30. Bacteria double every hour; 30 at the start. How many at the end of the 2nd hour, 4th hour, nth hour?जीवाणु हर घंटे दोगुने होते हैं; शुरू में 30। दूसरे, चौथे और n वें घंटे के अंत में कितने?Bacteria double every hour; 30 at the start. How many at the end of the 2nd hour, 4th hour, nth hour?
  1. 30 \cdot 2^2 = 120,\quad 30 \cdot 2^4 = 480,\quad 30 \cdot 2^n
Q31. What will ₹500 amount to in 10 years at 10% compounded annually?10% वार्षिक चक्रवृद्धि पर ₹500, 10 वर्ष में कितना होगा?What will ₹500 amount to in 10 years at 10% compounded annually?
  1. 500(1.1)^{10} \approx 500 \times 2.5937 \approx ₹1296.87
Q32. The A.M. and G.M. of the roots of a quadratic equation are 8 and 5. Find the equation.किसी द्विघात समीकरण के मूलों के A.M. और G.M. 8 और 5 हैं। समीकरण ज्ञात कीजिए।The A.M. and G.M. of the roots of a quadratic equation are 8 and 5. Find the equation.
  1. \text{sum} = 16,\ \text{product} = 25 \Rightarrow x^2 - 16x + 25 = 0\text{योग} = 16,\ \text{गुणनफल} = 25 \Rightarrow x^2 - 16x + 25 = 0\text{sum} = 16,\ \text{product} = 25 \Rightarrow x^2 - 16x + 25 = 0

Miscellaneous Exercise on Chapter 8अध्याय 8 पर विविध प्रश्नावलीMiscellaneous Exercise on Chapter 8 · 18 Q

Q1. f(x + y) = f(x)f(y) for x, y ∈ N, f(1) = 3 and \sum_{x=1}^{n} f(x) = 120. Find n.x, y ∈ N के लिए f(x + y) = f(x)f(y), f(1) = 3 और \sum_{x=1}^{n} f(x) = 120। n ज्ञात कीजिए।f(x + y) = f(x)f(y) for x, y ∈ N, f(1) = 3 and \sum_{x=1}^{n} f(x) = 120. Find n.
  1. f(2) = 9,\ f(3) = 27, \dots \Rightarrow f(x) = 3^x;\ \frac{3(3^n - 1)}{2} = 120 \Rightarrow 3^n = 81 \Rightarrow n = 4
Q2. The sum of some terms of a G.P. is 315; a = 5, r = 2. Find the last term and the number of terms.किसी G.P. के कुछ पदों का योग 315 है; a = 5, r = 2। अंतिम पद और पदों की संख्या ज्ञात कीजिए।The sum of some terms of a G.P. is 315; a = 5, r = 2. Find the last term and the number of terms.
  1. 5(2^n - 1) = 315 \Rightarrow 2^n = 64 \Rightarrow n = 6;\ \text{last} = 5 \cdot 2^5 = 160
Q3. The first term of a G.P. is 1 and the 3rd + 5th terms = 90. Find r.किसी G.P. का पहला पद 1 है और तीसरे + पाँचवें पद = 90। r ज्ञात कीजिए।The first term of a G.P. is 1 and the 3rd + 5th terms = 90. Find r.
  1. r^2 + r^4 = 90 \Rightarrow (r^2 + 10)(r^2 - 9) = 0 \Rightarrow r^2 = 9 \Rightarrow r = \pm 3
Q4. Three numbers in G.P. add to 56. Subtracting 1, 7, 21 gives an A.P. Find the numbers.G.P. में तीन संख्याओं का योग 56 है। इनमें से 1, 7, 21 घटाने पर A.P. बनती है। संख्याएँ ज्ञात कीजिए।Three numbers in G.P. add to 56. Subtracting 1, 7, 21 gives an A.P. Find the numbers.
  1. 2(ar - 7) = (a - 1) + (ar^2 - 21) \Rightarrow a(r - 1)^2 = 8;\quad a(1 + r + r^2) = 56
  2. \frac{1 + r + r^2}{(r - 1)^2} = 7 \Rightarrow 2r^2 - 5r + 2 = 0 \Rightarrow r = 2 \ (a = 8) \text{ or } r = \frac{1}{2} \ (a = 32)

✔ 8, 16, 32 (or 32, 16, 8)8, 16, 32 (या 32, 16, 8)8, 16, 32 (ya 32, 16, 8)

Q5. A G.P. has an even number of terms. The sum of all terms is 5 times the sum of the terms in odd places. Find r.एक G.P. में पदों की संख्या सम है। सभी पदों का योग विषम स्थानों के पदों के योग का 5 गुना है। r ज्ञात कीजिए।A G.P. has an even number of terms. The sum of all terms is 5 times the sum of the terms in odd places. Find r.
  1. \text{odd places: } a, ar^2, \dots \text{ (ratio } r^2) \Rightarrow \frac{S}{S_{odd}} = \frac{\frac{a(r^{2n} - 1)}{r - 1}}{\frac{a(r^{2n} - 1)}{r^2 - 1}} = r + 1 = 5 \Rightarrow r = 4\text{विषम स्थान: } a, ar^2, \dots \text{ (अनुपात } r^2) \Rightarrow \frac{S}{S_{odd}} = \frac{\frac{a(r^{2n} - 1)}{r - 1}}{\frac{a(r^{2n} - 1)}{r^2 - 1}} = r + 1 = 5 \Rightarrow r = 4\text{odd places: } a, ar^2, \dots \text{ (ratio } r^2) \Rightarrow \frac{S}{S_{odd}} = \frac{\frac{a(r^{2n} - 1)}{r - 1}}{\frac{a(r^{2n} - 1)}{r^2 - 1}} = r + 1 = 5 \Rightarrow r = 4
Q6. If \frac{a + bx}{a - bx} = \frac{b + cx}{b - cx} = \frac{c + dx}{c - dx} (x ≠ 0), show that a, b, c, d are in G.P.यदि \frac{a + bx}{a - bx} = \frac{b + cx}{b - cx} = \frac{c + dx}{c - dx} (x ≠ 0), तो दर्शाइए कि a, b, c, d G.P. में हैं।If \frac{a + bx}{a - bx} = \frac{b + cx}{b - cx} = \frac{c + dx}{c - dx} (x ≠ 0), show that a, b, c, d are in G.P.
  1. Componendo–dividendo on each equal ratio.हर बराबर अनुपात पर योगांतरानुपात लगाइए।Har barabar ratio par componendo–dividendo lagaiye.
    \frac{a}{bx} = \frac{b}{cx} = \frac{c}{dx} \Rightarrow \frac{b}{a} = \frac{c}{b} = \frac{d}{c}
Q7. S, P, R are the sum, product and sum of reciprocals of n terms of a G.P. Prove P^2R^n = S^n.S, P, R किसी G.P. के n पदों का योग, गुणनफल और व्युत्क्रमों का योग हैं। सिद्ध कीजिए P^2R^n = S^n।S, P, R are the sum, product and sum of reciprocals of n terms of a G.P. Prove P^2R^n = S^n.
  1. S = \frac{a(r^n - 1)}{r - 1},\ P = a^n r^{\frac{n(n-1)}{2}},\ R = \frac{r^n - 1}{a r^{n-1}(r - 1)}
  2. P^2R^n = a^{2n} r^{n(n-1)} \cdot \frac{(r^n - 1)^n}{a^n r^{n(n-1)}(r - 1)^n} = \left[\frac{a(r^n - 1)}{r - 1}\right]^n = S^n
Q8. If a, b, c, d are in G.P., prove (a^n + b^n), (b^n + c^n), (c^n + d^n) are in G.P.यदि a, b, c, d G.P. में हैं, तो सिद्ध कीजिए (a^n + b^n), (b^n + c^n), (c^n + d^n) G.P. में हैं।If a, b, c, d are in G.P., prove (a^n + b^n), (b^n + c^n), (c^n + d^n) are in G.P.
  1. a^n + b^n = a^n(1 + r^n),\ b^n + c^n = a^nr^n(1 + r^n),\ c^n + d^n = a^nr^{2n}(1 + r^n)
  2. Each is r^n times the previous, so they are in G.P.हर एक पिछले का r^n गुना है, अतः G.P. में।Har ek pichhle ka r^n guna hai, isliye G.P. mein.
Q9. a, b are roots of x^2 - 3x + p = 0 and c, d of x^2 - 12x + q = 0, where a, b, c, d form a G.P. Prove (q + p) : (q - p) = 17 : 15.a, b समीकरण x^2 - 3x + p = 0 के और c, d समीकरण x^2 - 12x + q = 0 के मूल हैं, और a, b, c, d G.P. में हैं। सिद्ध कीजिए (q + p) : (q - p) = 17 : 15।a, b are roots of x^2 - 3x + p = 0 and c, d of x^2 - 12x + q = 0, where a, b, c, d form a G.P. Prove (q + p) : (q - p) = 17 : 15.
  1. a(1 + r) = 3,\ ar^2(1 + r) = 12 \Rightarrow r^2 = 4
  2. p = ab = a^2r,\ q = cd = a^2r^5 \Rightarrow \frac{q + p}{q - p} = \frac{r^4 + 1}{r^4 - 1} = \frac{17}{15}
Q10. The ratio of the A.M. and G.M. of two positive numbers a and b is m : n. Show that a : b = \left(m + \sqrt{m^2 - n^2}\right) : \left(m - \sqrt{m^2 - n^2}\right).दो धनात्मक संख्याओं a, b के A.M. और G.M. का अनुपात m : n है। दर्शाइए a : b = \left(m + \sqrt{m^2 - n^2}\right) : \left(m - \sqrt{m^2 - n^2}\right)।The ratio of the A.M. and G.M. of two positive numbers a and b is m : n. Show that a : b = \left(m + \sqrt{m^2 - n^2}\right) : \left(m - \sqrt{m^2 - n^2}\right).
  1. Let A = km, G = kn. By Ex 8.2 Q29 the numbers are A \pm \sqrt{A^2 - G^2}.मान लीजिए A = km, G = kn। प्रश्नावली 8.2 प्रश्न 29 से संख्याएँ A \pm \sqrt{A^2 - G^2} हैं।Maan lijiye A = km, G = kn. Exercise 8.2 Q29 se numbers A \pm \sqrt{A^2 - G^2} hain.
    a : b = k\left(m + \sqrt{m^2 - n^2}\right) : k\left(m - \sqrt{m^2 - n^2}\right)
Q11. Sum to n terms: (i) 5 + 55 + 555 + … (ii) .6 + .66 + .666 + …n पदों तक योग: (i) 5 + 55 + 555 + … (ii) .6 + .66 + .666 + …Sum to n terms: (i) 5 + 55 + 555 + … (ii) .6 + .66 + .666 + …
  1. (i)\ \frac{5}{9}\left[\frac{10(10^n - 1)}{9} - n\right] = \frac{50}{81}(10^n - 1) - \frac{5n}{9}
  2. (ii)\ \frac{6}{9}\left[(1 - 0.1) + (1 - 0.01) + \dots\right] = \frac{2}{3}\left[n - \frac{1}{9}\left(1 - 10^{-n}\right)\right] = \frac{2n}{3} - \frac{2}{27}\left(1 - 10^{-n}\right)
Q12. Find the 20th term of 2 × 4 + 4 × 6 + 6 × 8 + …2 × 4 + 4 × 6 + 6 × 8 + … का 20वाँ पद ज्ञात कीजिए।Find the 20th term of 2 × 4 + 4 × 6 + 6 × 8 + …
  1. a_n = 2n(2n + 2) \Rightarrow a_{20} = 40 \times 42 = 1680
Q13. A tractor costs ₹12000; ₹6000 cash, the rest in annual instalments of ₹500 plus 12% interest on the unpaid amount. Total cost?एक ट्रैक्टर ₹12000 का; ₹6000 नकद, शेष ₹500 की वार्षिक किस्तों में तथा बकाया पर 12% ब्याज। कुल लागत?A tractor costs ₹12000; ₹6000 cash, the rest in annual instalments of ₹500 plus 12% interest on the unpaid amount. Total cost?
  1. \text{interest} = 0.12(6000 + 5500 + \dots + 500) = 0.12 \times \frac{12}{2}(6000 + 500) = 0.12 \times 39000 = 4680

✔ ₹12000 + ₹4680 = ₹16680

Q14. A scooter costs ₹22000; ₹4000 cash, the rest in instalments of ₹1000 plus 10% interest on the unpaid amount. Total cost?एक स्कूटर ₹22000 का; ₹4000 नकद, शेष ₹1000 की किस्तों में तथा बकाया पर 10% ब्याज। कुल लागत?A scooter costs ₹22000; ₹4000 cash, the rest in instalments of ₹1000 plus 10% interest on the unpaid amount. Total cost?
  1. \text{interest} = 0.1(18000 + 17000 + \dots + 1000) = 0.1 \times \frac{18}{2}(18000 + 1000) = 17100

✔ ₹22000 + ₹17100 = ₹39100

Q15. A chain letter: each person sends 4 letters; postage 50 paise per letter. Amount spent when the 8th set is mailed?चेन पत्र: हर व्यक्ति 4 पत्र भेजता है; डाक खर्च 50 पैसे प्रति पत्र। 8वाँ सेट भेजे जाने तक कुल खर्च?A chain letter: each person sends 4 letters; postage 50 paise per letter. Amount spent when the 8th set is mailed?
  1. 4 + 4^2 + \dots + 4^8 = \frac{4(4^8 - 1)}{3} = 87380 \text{ letters} \Rightarrow 87380 \times 0.50 = ₹43690
Q16. ₹10000 at 5% simple interest annually. Amount in the 15th year, and total after 20 years?₹10000 पर 5% वार्षिक साधारण ब्याज। 15वें वर्ष में राशि, और 20 वर्ष बाद कुल राशि?₹10000 at 5% simple interest annually. Amount in the 15th year, and total after 20 years?
  1. \text{interest per year} = 500:\ 15\text{th year: } 10000 + 14 \times 500 = 17000;\ \text{after 20 years: } 10000 + 20 \times 500 = 20000\text{प्रति वर्ष ब्याज} = 500:\ 15\text{वाँ वर्ष: } 10000 + 14 \times 500 = 17000;\ \text{20 वर्ष बाद: } 10000 + 20 \times 500 = 20000\text{har saal byaaj} = 500:\ 15\text{va saal: } 10000 + 14 \times 500 = 17000;\ \text{20 saal baad: } 10000 + 20 \times 500 = 20000

✔ ₹17000; ₹20000

Q17. A machine costing ₹15625 depreciates by 20% each year. Value after 5 years?₹15625 की मशीन का मूल्य हर वर्ष 20% घटता है। 5 वर्ष बाद मूल्य?A machine costing ₹15625 depreciates by 20% each year. Value after 5 years?
  1. 15625 \times (0.8)^5 = 15625 \times 0.32768 = ₹5120
Q18. 150 workers start a job; 4 drop out on day 2, 4 more on day 3, and so on. It took 8 more days. In how many days was the work done?150 मज़दूर काम शुरू करते हैं; दूसरे दिन 4 हट जाते हैं, तीसरे दिन 4 और, इत्यादि। काम में 8 दिन अधिक लगे। काम कितने दिनों में पूरा हुआ?150 workers start a job; 4 drop out on day 2, 4 more on day 3, and so on. It took 8 more days. In how many days was the work done?
  1. \frac{n}{2}\left[2(150) - 4(n - 1)\right] = 150(n - 8)
  2. n(152 - 2n) = 150n - 1200 \Rightarrow n^2 - n - 600 = 0 \Rightarrow n = 25

✔ 25 days25 दिन25 din

🎯 Quiz — 12 questions🎯 क्विज़ — 12 प्रश्न🎯 Quiz — 12 questions

Tap an option. The answer and a short reason appear at once.किसी विकल्प पर टैप करें। उत्तर और छोटा कारण तुरंत दिखेगा।Kisi option par tap kijiye. Answer aur chhota reason turant dikhega.

1. The 10th term of the G.P. 5, 25, 125, … isG.P. 5, 25, 125, … का 10वाँ पद हैG.P. 5, 25, 125, … ka 10th term hai

2. The G.M. of 4 and 16 is4 और 16 का G.M. है4 aur 16 ka G.M. hai

3. Sum of 1 + 2 + 4 + … + 2⁹ is1 + 2 + 4 + … + 2⁹ का योग है1 + 2 + 4 + … + 2⁹ ka sum hai

4. If −2/7, x, −7/2 are in G.P., then x =यदि −2/7, x, −7/2 G.P. में हैं, तो x =Agar −2/7, x, −7/2 G.P. mein hain, to x =

5. For positive a ≠ b, which is always true?धनात्मक a ≠ b के लिए क्या हमेशा सत्य है?Positive a ≠ b ke liye kya hamesha sach hai?

6. The 4th term of a G.P. with a = 3, r = −2 isa = 3, r = −2 वाली G.P. का चौथा पद हैa = 3, r = −2 wali G.P. ka 4th term hai

7. Which is a G.P.?कौन-सी G.P. है?Kaun si G.P. hai?

8. The 3rd term of a G.P. is 24 and the 6th is 192. Then r =G.P. का तीसरा पद 24 और छठा 192 है। तो r =G.P. ka 3rd term 24 aur 6th 192 hai. To r =

9. aₙ = n(n + 2). Then a₄ =aₙ = n(n + 2)। तो a₄ =aₙ = n(n + 2). To a₄ =

10. A.M. and G.M. of two positive numbers are 10 and 8. The numbers areदो धनात्मक संख्याओं के A.M. और G.M. 10 और 8 हैं। संख्याएँ हैंDo positive numbers ke A.M. aur G.M. 10 aur 8 hain. Numbers hain

11. Three terms in G.P. are best taken asG.P. में तीन पद सबसे अच्छे लिए जाते हैंG.P. mein teen terms sabse achhe liye jaate hain

12. How many terms of 3, 3², 3³, … make 120?3, 3², 3³, … के कितने पदों का योग 120?3, 3², 3³, … ke kitne terms ka sum 120?

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These are Simple Learning Hub's own study notes and solutions, written to follow the NCERT Class 11 Mathematics textbook (Reprint 2026-27) section by section; read them with the official NCERT chapter. Diagrams are original SLH drawings.ये Simple Learning Hub के अपने स्टडी नोट्स हैं, जो NCERT कक्षा 11 गणित की किताब (2026-27) के हर अनुभाग के साथ लिखे गए हैं; इन्हें NCERT के आधिकारिक अध्याय के साथ पढ़िए। डायग्राम SLH के अपने बनाए चित्र हैं।Ye Simple Learning Hub ke apne study notes hain, jo NCERT Class 11 Maths ki kitaab (2026-27) ke har section ke saath likhe gaye hain; inhe NCERT ke official chapter ke saath padhiye. Diagrams SLH ki apni drawings hain.