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$Class 10 Math Unit 6 Sequence and Series$

Class 10 Mathematics | Unit 6 Sequence and Series

बीजगणित (Algebra)

Unit 6: Sequence and Series (अनुक्रम र श्रेणी) Formulae & Solutions

Solutions by Important Edu Notes Team

1. Key Concepts (मुख्य अवधारणाहरू)

Term	Definition
Sequence (अनुक्रम)	A set of numbers arranged in a definite order according to some rule. निश्चित नियम अनुसार मिलाएर राखिएका सङ्ख्याहरूको समूहलाई अनुक्रम भनिन्छ ।
Series (श्रेणी)	The sum of the terms of a sequence is called a series. अनुक्रमका पदहरूलाई जोड चिन्हले जोडेर बन्ने अभिव्यञ्जनालाई श्रेणी भनिन्छ ।
Arithmetic Sequence (अङ्कगणितीय अनुक्रम)	A sequence in which the difference between any two consecutive terms is constant. कुनै अनुक्रमको कुनै पनि पद र त्यसको अघिल्लो पदबिचको अन्तर सधैँ समान हुन्छ भने त्यसलाई अङ्कगणितीय अनुक्रम भनिन्छ ।
Common Difference (समान अन्तर)	The constant difference between a term and its preceding term in an arithmetic sequence. अङ्कगणितीय अनुक्रममा कुनै पद र त्यसको अघिल्लो पदबिचको फरकलाई समान अन्तर भनिन्छ ।
Arithmetic Mean (अङ्कगणितीय मध्यमान)	The terms lying between the first and last terms of an arithmetic sequence. अङ्कगणितीय अनुक्रमको पहिलो र अन्तिम पदबिचमा पर्ने पदहरूलाई अङ्कगणितीय मध्यमान भनिन्छ ।

2. Important Formulas (महत्वपूर्ण सूत्रहरू)

A. General Notation (साधारण सङ्केतहरू)

First term (पहिलो पद): $a$
Common difference (समान अन्तर): $d$
Number of terms (पद सङ्ख्या): $n$
nth term / Last term ($n$ औँ पद): $t_n$ or $l$
Sum of n terms ($n$ पदहरूको योगफल): $S_n$
Arithmetic Mean (अङ्कगणितीय मध्यमान): $AM$

B. Formulas for Arithmetic Sequence

Description	Formula
Common Difference (समान अन्तर)	$$d = t_2 – t_1$$
General Term / nth Term ($n$ औँ पद)	$$t_n = a + (n – 1)d$$
Arithmetic Mean (Single)	$$AM = \frac{a + b}{2}$$
Sum of first $n$ terms	$$S_n = \frac{n}{2} [2a + (n – 1)d]$$ $$S_n = \frac{n}{2} (a + l)$$
‘n’ Arithmetic Means	$$d = \frac{b – a}{n + 1}$$ Means: $m_1 = a + d, m_2 = a + 2d, \dots$

3. Exercise 6.1 Solution (Arithmetic Sequence)

Q1.a: What are arithmetic means? (समानान्तरीय मध्यमा भनेको के हो ?)

English: The terms lying between the first term and the last term of an arithmetic progression (AP) are called arithmetic means.

Nepali: समानान्तरीय श्रेणीको पहिलो पद र अन्तिम पदको बिचमा पर्ने पद वा पदहरूलाई समानान्तरीय मध्यमा भनिन्छ ।

Q1.b: If a, m, b are in arithmetic sequence, express m in terms of a and b. (यदि a, m, b समानान्तरीय अनुक्रममा भए m लाई a र b का रूपमा लेख्नुहोस् ।)

Since $a, m, b$ are in an arithmetic sequence, the common difference must be the same.

$$m – a = b – m$$
$$2m = a + b$$
$$m = \frac{a + b}{2}$$

$\therefore m = \frac{a + b}{2}$

Q1.c: Find the mid value of the numbers 12 and 18. (दुई सङ्ख्या 12 र 18 को मध्यमान कति हुन्छ, लेख्नुहोस् ।)

Here, First number ($a$) = 12, Last number ($b$) = 18

We know that,

$$m = \frac{a + b}{2}$$
$$m = \frac{12 + 18}{2}$$
$$m = \frac{30}{2}$$
$$m = 15$$

The mid value is 15.

Q2.a: Find the arithmetic mean between 6 and 10. (समानान्तरीय मध्यमा पत्ता लगाउनुहोस् : 6 र 10)

Here, $a = 6$ and $b = 10$.

$$m = \frac{a + b}{2}$$
$$= \frac{6 + 10}{2}$$
$$= \frac{16}{2}$$
$$= 8$$

Arithmetic Mean = 8

Q2.b: Find the arithmetic mean between -2 and 2. (समानान्तरीय मध्यमा पत्ता लगाउनुहोस् : -2 र 2)

Here, $a = -2$ and $b = 2$.

$$m = \frac{a + b}{2}$$
$$= \frac{-2 + 2}{2}$$
$$= \frac{0}{2}$$
$$= 0$$

Arithmetic Mean = 0

Q2.c: Find the arithmetic mean between 4 and 8. (समानान्तरीय मध्यमा पत्ता लगाउनुहोस् : 4 र 8)

Here, $a = 4$ and $b = 8$.

$$m = \frac{a + b}{2}$$
$$= \frac{4 + 8}{2}$$
$$= \frac{12}{2}$$
$$= 6$$

Arithmetic Mean = 6

Q2.d: Find the arithmetic mean between $(a+b)$ and $(a-b)$. (समानान्तरीय मध्यमा पत्ता लगाउनुहोस् : $(a+b)$ र $(a-b)$)

Here, $A = a + b$, $B = a – b$

$$m = \frac{A + B}{2}$$
$$= \frac{(a + b) + (a – b)}{2}$$
$$= \frac{2a}{2}$$
$$= a$$

Arithmetic Mean = a

Q3.a: Find 4 arithmetic means between 5 and 20. (5 र 20 का बिचमा 4 ओटा समानान्तरीय मध्यमा पत्ता लगाउनुहोस् ।)

Here, $a = 5, b = 20, n = 4$

$$d = \frac{b – a}{n + 1}$$
$$d = \frac{20 – 5}{4 + 1}$$
$$d = \frac{15}{5}$$
$$d = 3$$

Now, the means are:

$$m_1 = a + d = 5 + 3 = 8$$
$$m_2 = a + 2d = 5 + 6 = 11$$
$$m_3 = a + 3d = 5 + 9 = 14$$
$$m_4 = a + 4d = 5 + 12 = 17$$

Means are: 8, 11, 14, 17

Q3.b: Find 6 arithmetic means between 70 and 14. (70 र 14 का बिचमा 6 ओटा समानान्तरीय मध्यमा पत्ता लगाउनुहोस् ।)

Here, $a = 70, b = 14, n = 6$

$$d = \frac{b – a}{n + 1}$$
$$d = \frac{14 – 70}{6 + 1}$$
$$d = \frac{-56}{7}$$
$$d = -8$$

The means are:

$$m_1 = 70 + (-8) = 62$$
$$m_2 = 70 – 16 = 54$$
$$m_3 = 70 – 24 = 46$$
$$m_4 = 70 – 32 = 38$$
$$m_5 = 70 – 40 = 30$$
$$m_6 = 70 – 48 = 22$$

Means are: 62, 54, 46, 38, 30, 22

Q3.c: Find 6 arithmetic means between 5 and -9. (5 र -9 का बिचमा 6 ओटा समानान्तरीय मध्यमा पत्ता लगाउनुहोस् ।)

Here, $a = 5, b = -9, n = 6$

$$d = \frac{b – a}{n + 1}$$
$$d = \frac{-9 – 5}{6 + 1}$$
$$d = \frac{-14}{7}$$
$$d = -2$$

Means:

$$m_1 = 5 + (-2) = 3$$
$$m_2 = 5 + (-4) = 1$$
$$m_3 = 5 + (-6) = -1$$
$$m_4 = 5 + (-8) = -3$$
$$m_5 = 5 + (-10) = -5$$
$$m_6 = 5 + (-12) = -7$$

Means are: 3, 1, -1, -3, -5, -7

Q4.a: The following sequence is an arithmetic sequence. From the sequence, find the value of x: $5, x, 9, …$ (तल दिइएका अनुक्रम समानान्तर अनुक्रम हुन् । उक्त अनुक्रमबाट x को मान पत्ता लगाउनुहोस् : $5, x, 9, …$)

Since 5, x, 9 are in AP, x is the mean of 5 and 9.

$$x = \frac{5 + 9}{2}$$
$$x = \frac{14}{2}$$
$$x = 7$$

x = 7

Q4.b: The following sequence is an arithmetic sequence. From the sequence, find the value of x: $x+1, x+5, 3x+1, …$ (तल दिइएका अनुक्रम समानान्तर अनुक्रम हुन् । उक्त अनुक्रमबाट x को मान पत्ता लगाउनुहोस् : $x+1, x+5, 3x+1, …$)

Using Common Difference property ($t_2 – t_1 = t_3 – t_2$):

$$(x+5) – (x+1) = (3x+1) – (x+5)$$
$$x + 5 – x – 1 = 3x + 1 – x – 5$$
$$4 = 2x – 4$$
$$4 + 4 = 2x$$
$$8 = 2x$$
$$x = 4$$

x = 4

Q4.c: The following sequence is an arithmetic sequence. From the sequence, find the value of x: $x+2, 3x, 4x+1, …$ (तल दिइएका अनुक्रम समानान्तर अनुक्रम हुन् । उक्त अनुक्रमबाट x को मान पत्ता लगाउनुहोस् : $x+2, 3x, 4x+1, …$)

Using Mean property ($2 \times t_2 = t_1 + t_3$):

$$2(3x) = (x+2) + (4x+1)$$
$$6x = 5x + 3$$
$$6x – 5x = 3$$
$$x = 3$$

x = 3

Q5.a: Find the sum of the following arithmetic series: $7 + 11 + 15 + 19 …$, to 20 terms. (दिइएका समानान्तरीय श्रेणीको योगफल पत्ता लगाउनुहोस् : $7+11+15+19…$, 20 ओटा पद)

$a = 7, \quad d = 11 – 7 = 4, \quad n = 20$

$$S_{20} = \frac{20}{2}[2(7) + (20-1)4]$$
$$= 10[14 + 19 \times 4]$$
$$= 10[14 + 76]$$
$$= 10[90]$$
$$= 900$$

Sum = 900

Q5.b: Find the sum of: $4 – 1 – 6 – 11 – 16 – …$, to 7 terms. (दिइएका समानान्तरीय श्रेणीको योगफल पत्ता लगाउनुहोस् : $4-1-6-11-16-…$, 7 ओटा पद)

$a = 4, \quad d = -1 – 4 = -5, \quad n = 7$

$$S_7 = \frac{7}{2}[2(4) + (7-1)(-5)]$$
$$= \frac{7}{2}[8 + 6(-5)]$$
$$= \frac{7}{2}[8 – 30]$$
$$= \frac{7}{2}[-22]$$
$$= 7 \times (-11)$$
$$= -77$$

Sum = -77

Q5.c: Find the sum of: $\frac{1}{2} + \frac{3}{2} + \frac{5}{2} + …$, to 16 terms. (दिइएका समानान्तरीय श्रेणीको योगफल पत्ता लगाउनुहोस् : $\frac{1}{2}+\frac{3}{2}+\frac{5}{2}+…,$ 16 ओटा पद)

$a = 0.5, \quad d = 1, \quad n = 16$

$$S_{16} = \frac{16}{2}[2(0.5) + (16-1)1]$$
$$= 8[1 + 15]$$
$$= 8[16]$$
$$= 128$$

Sum = 128

Q5.d: Find the sum of: $5 + 10 + 15 + … + 65$. (दिइएका समानान्तरीय श्रेणीको योगफल पत्ता लगाउनुहोस् : $5+10+15+…+65$)

$a = 5, \quad d = 5, \quad l = 65$

First, find n:

$$l = a + (n-1)d$$
$$65 = 5 + (n-1)5$$
$$60 = 5(n-1)$$
$$12 = n-1$$
$$n = 13$$

Now sum:

$$S_{13} = \frac{13}{2}(a + l)$$
$$= \frac{13}{2}(5 + 65)$$
$$= \frac{13}{2}(70)$$
$$= 13 \times 35$$
$$= 455$$

Sum = 455

Q5.e: Find the sum of: $-64 – 48 – 32 – … + 32$. (दिइएका समानान्तरीय श्रेणीको योगफल पत्ता लगाउनुहोस् : $-64-48-32-…+32$)

$a = -64, \quad d = 16, \quad l = 32$

Find n:

$$32 = -64 + (n-1)16$$
$$96 = 16(n-1)$$
$$6 = n-1$$
$$n = 7$$

Sum:

$$S_7 = \frac{7}{2}(-64 + 32)$$
$$= \frac{7}{2}(-32)$$
$$= 7 \times (-16)$$
$$= -112$$

Sum = -112

Q5.f: Find the sum of the first 10 odd numbers. (पहिलो 10 ओटा बिजोर सङ्ख्याको योगफल)

Sequence: 1, 3, 5…

$$S_n = n^2$$
$$S_{10} = 10^2$$
$$= 100$$

Sum = 100

Q5.g: Find the sum of the first 100 natural numbers. (पहिलो 100 सम्मका प्राकृतिक सङ्ख्याको योगफल)

$$S_{100} = \frac{100}{2}(1 + 100)$$
$$= 50(101)$$
$$= 5050$$

Sum = 5050

Q5.h: Find the sum of natural numbers from 50 to 100. (50 देखि 100 सम्मका प्राकृतिक सङ्ख्याको योगफल)

$a=50, l=100$. $n = 100-50+1 = 51$.

$$S_{51} = \frac{51}{2}(50 + 100)$$
$$= \frac{51}{2}(150)$$
$$= 51 \times 75$$
$$= 3825$$

Sum = 3825

Q6.a: First term = 1, last term = 50 and sum of all terms = 204. Find the common difference. (पहिलो पद = 1, अन्तिम पद = 50 र जम्मा पदको योगफल = 204 भए समान अन्तर पत्ता लगाउनुहोस् ।)

$$204 = \frac{n}{2}(1 + 50)$$
$$408 = 51n$$
$$n = 8$$

Now find d:

$$l = a + (n-1)d$$
$$50 = 1 + (8-1)d$$
$$49 = 7d$$
$$d = 7$$

Common Difference (d) = 7

Q6.c: First term = 17, last term = -99/8 and sum of all terms = 407/16. Find the common difference. (पहिलो पद = 17, अन्तिम पद = -99/8 र जम्मा पदको योगफल = 407/16 भए समान अन्तर पत्ता लगाउनुहोस् ।)

$$\frac{407}{16} = \frac{n}{2}(17 – \frac{99}{8})$$
$$\frac{407}{8} = n(\frac{136 – 99}{8})$$
$$\frac{407}{8} = n(\frac{37}{8})$$
$$407 = 37n$$
$$n = 11$$

Now find d:

$$-\frac{99}{8} = 17 + (11-1)d$$
$$-\frac{99}{8} – 17 = 10d$$
$$\frac{-99 – 136}{8} = 10d$$
$$\frac{-235}{8} = 10d$$
$$d = -\frac{23.5}{8}$$
$$d = -\frac{47}{16}$$

d = -47/16

Q7.a: Common difference = -3, number of terms = 10 and sum of all terms = 325. Find its first term. (समान अन्तर = -3, पदको सङ्ख्या = 10 र जम्मा पदको योगफल = 325 भए पहिलो पद पत्ता लगाउनुहोस् ।)

$$325 = \frac{10}{2}[2a + (10-1)(-3)]$$
$$325 = 5[2a – 27]$$
$$65 = 2a – 27$$
$$92 = 2a$$
$$a = 46$$

a = 46

Q7.b: Common difference = 9, number of terms = 9 and sum of all terms = 108. Find its first term. (समान अन्तर = 9, पदको सङ्ख्या = 9 र जम्मा पदको योगफल = 108 भए पहिलो पद पत्ता लगाउनुहोस् ।)

$$108 = \frac{9}{2}[2a + (9-1)9]$$
$$216 = 9[2a + 72]$$
$$24 = 2a + 72$$
$$2a = 24 – 72$$
$$2a = -48$$
$$a = -24$$

a = -24

Q7.c: Common difference = 3, number of terms = 10 and sum of all terms = 155. Find its first term. (समान अन्तर = 3, पदको सङ्ख्या = 10 र जम्मा पदको योगफल = 155 भए पहिलो पद पत्ता लगाउनुहोस् ।)

$$155 = \frac{10}{2}[2a + (10-1)3]$$
$$155 = 5[2a + 27]$$
$$31 = 2a + 27$$
$$2a = 4$$
$$a = 2$$

a = 2

Q8.a: How many terms of the series $4 + 10 + 16 + 22 + …$ has a sum 374? (समानान्तरीय श्रेणी $4+10+16+22+…$ मा कतिओटा सङ्ख्याको जम्मा योगफल 374 हुन्छ, पत्ता लगाउनुहोस् ।)

$a=4, d=6, S=374$

$$374 = \frac{n}{2}[2(4) + (n-1)6]$$
$$748 = n[8 + 6n – 6]$$
$$748 = n[6n + 2]$$
$$748 = 6n^2 + 2n$$
$$3n^2 + n – 374 = 0$$

Solving quadratic equation:

$$n = \frac{-1 \pm \sqrt{1 – 4(3)(-374)}}{6}$$
$$n = \frac{-1 \pm 67}{6}$$
$$n = 11 \quad (\text{Taking positive})$$

11 terms

Q8.b: If the first term of an arithmetic series is 36 and its common difference is 9, how many terms will have a sum 540? (समानान्तरीय श्रेणीमा पहिलो पद 36 र समान अन्तर 9 छ । कतिओटा सङ्ख्याको जम्मा योगफल 540 हुन्छ, पत्ता लगाउनुहोस् ।)

$$540 = \frac{n}{2}[2(36) + (n-1)9]$$
$$1080 = n[72 + 9n – 9]$$
$$1080 = n[9n + 63]$$
$$1080 = 9n^2 + 63n$$
$$120 = n^2 + 7n$$
$$n^2 + 7n – 120 = 0$$
$$(n + 15)(n – 8) = 0$$
$$n = 8$$

8 terms

Q9.a: Find the sum of the first 16 terms of an arithmetic series, if its third term is -15 and eighth term is 10. (समानान्तरीय श्रेणीको तेस्रो पद 15 र आठौं पद 10 भए पहिलो 16 ओटा पदको योगफल पत्ता लगाउनुहोस् ।)

$$a + 2d = -15 \quad (i)$$
$$a + 7d = 10 \quad (ii)$$
$$(ii) – (i) \Rightarrow 5d = 25 \Rightarrow d = 5$$
$$a + 10 = -15 \Rightarrow a = -25$$

Sum of 16 terms:

$$S_{16} = \frac{16}{2}[2(-25) + 15(5)]$$
$$= 8[-50 + 75]$$
$$= 8[25]$$
$$= 200$$

Sum = 200

Q9.b: Find the sum of the first 20 terms of an arithmetic series, if its fifth term is 10 and eleventh term is 22. (समानान्तरीय श्रेणीको पाँचौँ पद 10 र एघारौं पद 22 भए पहिलो 20 ओटा पदको योगफल पत्ता लगाउनुहोस् ।)

$$a + 4d = 10$$
$$a + 10d = 22$$
$$6d = 12 \Rightarrow d = 2$$
$$a + 8 = 10 \Rightarrow a = 2$$

Sum of 20 terms:

$$S_{20} = \frac{20}{2}[2(2) + 19(2)]$$
$$= 10[4 + 38]$$
$$= 10[42]$$
$$= 420$$

Sum = 420

Q10.a: If the sum of the first 6 terms of an arithmetic series is 75 and the sum of its first 12 terms is 390, find the sum of the first 20 terms. (समानान्तरीय श्रेणीको पहिलो छ पदको योगफल 75 र पहिलो 12 पदको योगफल 390 भए पहिलो 20 पदको योगफल पत्ता लगाउनुहोस् ।)

$$S_6 = 3[2a + 5d] = 75 \Rightarrow 2a + 5d = 25$$
$$S_{12} = 6[2a + 11d] = 390 \Rightarrow 2a + 11d = 65$$
$$Subtracting: 6d = 40 \Rightarrow d = 20/3$$
$$2a = 25 – 100/3 = -25/3 \Rightarrow a = -25/6$$

Find $S_{20}$:

$$S_{20} = 10[2(-25/6) + 19(20/3)]$$
$$= 10[-25/3 + 380/3]$$
$$= 10[355/3]$$
$$= \frac{3550}{3}$$

Sum = 3550/3

Q10.b: If the sum of the first 7 terms of an arithmetic series is 21 and the sum of its first 12 terms is 126, then find the series. (समानान्तरीय श्रेणीको पहिलो सात पदको योगफल 21 र पहिलो 12 पदको योगफल 126 भए उक्त श्रेणी पत्ता लगाउनुहोस् ।)

$$S_7 = 7[a + 3d] = 21 \Rightarrow a + 3d = 3$$
$$S_{12} = 6[2a + 11d] = 126 \Rightarrow 2a + 11d = 21$$
$$Substitute a = 3-3d: 2(3-3d) + 11d = 21$$
$$6 – 6d + 11d = 21$$
$$5d = 15 \Rightarrow d = 3$$
$$a = 3 – 9 = -6$$

Series: -6, -3, 0, 3…

Q11.a: A worker in a garment factory produced 1000 Nepali cap (Dhaka Topi) at the first year and increased the production by 100 every year. Can he produce 15,000 caps in 10 years? (नेपाली ढाका उद्योगमा कार्यरत एक जना कामदारले पहिलो वर्षमा 1,000 ओटा टोपी बनाए । प्रत्येक वर्ष उनले 100 का दरले टोपीको सङ्ख्या बढाउँदै गए । के 10 वर्षपछि उनले टोपीको सङ्ख्या 15,000 पुर्याउन सक्लान् त ? गणना गरी लेख्नुहोस् ।)

$a=1000, d=100, n=10$. Target=15000.

$$S_{10} = 5[2(1000) + 9(100)]$$
$$= 5[2000 + 900]$$
$$= 5[2900]$$
$$= 14500$$

14500 < 15000

No, he will be short by 500 caps.

Q11.b: The monthly salary of a person appointed for a new job is Rs. 60,000 initially and it increases by Rs. 24,000 every year. In how many years he will earn the total amount of Rs. 1 crore? (The interest is not included in his earning) (एक जना कर्मचारी मासिक तलब रु. 60,000 कमाउने गरी नोकरीमा लागे । प्रत्येक वर्ष उनको तलबमा समान रु. 24,000 का दरले बढोत्तरी पाउँछन् । उसले जम्मा आम्दानी रु. 1 करोड पुऱ्याउन चाहन्छन् । त्यसका लागि उसले कति वर्षसम्म नोकरी गर्नुपर्ला ? गणना गरी कारणसहित उल्लेख गर्नुहोस् ।)

$a = 60,000 \times 12 = 720,000$

$d = 24,000$

Target Sum ($S_n$) = 10,000,000

$$10000000 = \frac{n}{2}[1440000 + (n-1)24000]$$
$$20000 = n[1440 + 24n – 24]$$
$$20000 = 1416n + 24n^2$$
$$n^2 + 59n – 833.33 = 0$$

Solving quadratic, $n \approx 11.77$.

He needs to work for approximately 12 years.

4. Exercise 6.2 Solution (Geometric Sequence)

Q1.a: What is meant by geometric mean? (गुणोत्तर मध्यमा भनेको के हो ?)

English: The term lying between the first term and the last term of a geometric progression (GP) is called the geometric mean.

Nepali: गुणोत्तर श्रेणीको पहिलो पद र अन्तिम पदको बिचमा पर्ने पदलाई गुणोत्तर मध्यमा भनिन्छ ।

Q1.b: If the positive numbers a, m and b are in geometric sequence, express m in terms of a and b. (यदि धनात्मक सङ्ख्याहरू a, m, b गुणोत्तर अनुक्रममा भए m लाई a र b का रूपमा लेख्नुहोस् ।)

$$\frac{m}{a} = \frac{b}{m}$$
$$m^2 = ab$$
$$m = \sqrt{ab}$$

$\therefore m = \sqrt{ab}$

Q1.c: What is the geometric mean between 3 and 27? (3 र 27 को बिचमा पर्ने गुणोत्तर मध्यमा कति हुन्छ ?)

$$m = \sqrt{3 \times 27}$$
$$m = \sqrt{81}$$
$$m = 9$$

Geometric Mean = 9

Q2.a: Find the geometric mean between -4 and -64. (-4 र -64 बिचको गुणोत्तर मध्यमा पत्ता लगाउनुहोस् ।)

$$m = \sqrt{-4 \times -64}$$
$$m = \sqrt{256}$$
$$m = -16 \quad (\text{Since terms are negative})$$

Geometric Mean = -16

Q2.b: Find the geometric mean between $\frac{1}{5}$ and 125. ($\frac{1}{5}$ र 125 बिचको गुणोत्तर मध्यमा पत्ता लगाउनुहोस् ।)

$$m = \sqrt{\frac{1}{5} \times 125}$$
$$m = \sqrt{25}$$
$$m = 5$$

Geometric Mean = 5

Q2.c: Find the geometric mean between 7 and 343. (7 र 343 बिचको गुणोत्तर मध्यमा पत्ता लगाउनुहोस् ।)

$$m = \sqrt{7 \times 343}$$
$$m = \sqrt{2401}$$
$$m = 49$$

Geometric Mean = 49

Q3.a: Find 4 geometric means between 6 and 192. (6 र 192 का बिचमा 4 ओटा गुणोत्तर मध्यमा पत्ता लगाउनुहोस् ।)

$a = 6, b = 192, n = 4 \Rightarrow N = 6$

$$192 = 6 r^5$$
$$32 = r^5$$
$$2^5 = r^5 \Rightarrow r = 2$$

Means:

$$m_1 = 6 \times 2 = 12$$
$$m_2 = 12 \times 2 = 24$$
$$m_3 = 24 \times 2 = 48$$
$$m_4 = 48 \times 2 = 96$$

Means are: 12, 24, 48, 96

Q3.b: Find 3 geometric means between 5 and 405. (5 र 405 का बिचमा 3 ओटा गुणोत्तर मध्यमा पत्ता लगाउनुहोस् ।)

$$405 = 5 r^4$$
$$81 = r^4$$
$$3^4 = r^4 \Rightarrow r = 3$$

Means: $15, 45, 135$

Means are: 15, 45, 135

Q3.c: Find 3 geometric means between $\frac{9}{4}$ and $\frac{4}{9}$. ($\frac{9}{4}$ र $\frac{4}{9}$ का बिचमा 3 ओटा गुणोत्तर मध्यमा पत्ता लगाउनुहोस् ।)

$$\frac{4}{9} = \frac{9}{4} r^4$$
$$r^4 = \frac{16}{81}$$
$$r^4 = (\frac{2}{3})^4 \Rightarrow r = \frac{2}{3}$$

Means: $\frac{9}{4}(\frac{2}{3}) = \frac{3}{2}, \quad 1, \quad \frac{2}{3}$

Means are: 3/2, 1, 2/3

Q4.a: Find the value of x from the geometric series: 4, x, 9. (दिइएको गुणोत्तर अनुक्रमबाट x को मान पत्ता लगाउनुहोस् : 4, x, 9)

$$x = \sqrt{4 \times 9}$$
$$x = \sqrt{36}$$
$$x = 6$$

x = 6

Q4.b: Find the value of x from the geometric series: x, 4, 8. (दिइएको गुणोत्तर अनुक्रमबाट x को मान पत्ता लगाउनुहोस् : x, 4, 8)

$$\frac{4}{x} = \frac{8}{4}$$
$$\frac{4}{x} = 2$$
$$4 = 2x$$
$$x = 2$$

x = 2

Q4.c: Find the value of x from the geometric series: 5, 25, x+1. (दिइएको गुणोत्तर अनुक्रमबाट x को मान पत्ता लगाउनुहोस् : 5, 25, x+1)

$$r = \frac{25}{5} = 5$$
$$x+1 = 25 \times 5$$
$$x+1 = 125$$
$$x = 124$$

x = 124

Q5.a: Find the sum of the geometric series: $2 + 4 + 8 + 16 …$ to 6 terms. (दिइएका गुणोत्तर श्रेणीको योगफल पत्ता लगाउनुहोस् : $2+4+8+16…$, 6 ओटा पद)

$a=2, r=2, n=6$

$$S_6 = \frac{2(2^6 – 1)}{2 – 1}$$
$$= 2(64 – 1)$$
$$= 2(63)$$
$$= 126$$

Sum = 126

Q5.b: Find the sum of the geometric series: $\frac{1}{9} + \frac{1}{3} + 1 + …$ to 5 terms. (दिइएका गुणोत्तर श्रेणीको योगफल पत्ता लगाउनुहोस् : $\frac{1}{9}+\frac{1}{3}+1…$, 5 ओटा पद)

$a=1/9, r=3, n=5$

$$S_5 = \frac{\frac{1}{9}(3^5 – 1)}{3 – 1}$$
$$= \frac{1}{9} \times \frac{242}{2}$$
$$= \frac{121}{9}$$

Sum = 121/9

Q5.c: Find the sum of the geometric series: $-\frac{1}{4} + \frac{1}{2} – 1 + …$ to 6 terms. (दिइएका गुणोत्तर श्रेणीको योगफल पत्ता लगाउनुहोस् : $-\frac{1}{4}+\frac{1}{2}-1…$, 6 ओटा पद)

$a=-1/4, r=-2, n=6$

$$S_6 = \frac{-\frac{1}{4}(1 – (-2)^6)}{1 – (-2)}$$
$$= \frac{-\frac{1}{4}(1 – 64)}{3}$$
$$= \frac{63}{12}$$
$$= \frac{21}{4}$$

Sum = 21/4

Q5.d: Find the sum of the geometric series: $16 + 8 + 4 + … + \frac{1}{16}$. (दिइएका गुणोत्तर श्रेणीको योगफल पत्ता लगाउनुहोस् : $16+8+4+…+\frac{1}{16}$)

$a=16, r=0.5, l=1/16$

$$S_n = \frac{16 – (1/16)(1/2)}{0.5}$$
$$= \frac{16 – 1/32}{0.5}$$
$$= 2(32 – 1/16)$$
$$= 2(511/16)$$
$$= \frac{511}{8}$$

Sum = 511/8 or 63 7/8

Q6.a: First term = 2, last term = 486 and sum of all terms = 728. Find the common ratio. (पहिलो पद = 2, अन्तिम पद = 486 र जम्मा पदको योगफल = 728 भए समान अनुपात पत्ता लगाउनुहोस् ।)

$$728 = \frac{486r – 2}{r – 1}$$
$$728r – 728 = 486r – 2$$
$$242r = 726$$
$$r = 3$$

Common Ratio (r) = 3

Q6.b: First term = 5, last term = 1215 and sum of all terms = 1820. Find the common ratio. (पहिलो पद = 5, अन्तिम पद = 1215 र जम्मा पदको योगफल = 1820 भए समान अनुपात पत्ता लगाउनुहोस् ।)

$$1820 = \frac{1215r – 5}{r – 1}$$
$$1820r – 1820 = 1215r – 5$$
$$605r = 1815$$
$$r = 3$$

Common Ratio (r) = 3

Q6.c: First term = 3, last term = 768 and sum of all terms = 1533. Find the common ratio. (पहिलो पद = 3, अन्तिम पद = 768 र जम्मा पदको योगफल = 1533 भए समान अनुपात पत्ता लगाउनुहोस् ।)

$$1533 = \frac{768r – 3}{r – 1}$$
$$1533r – 1533 = 768r – 3$$
$$765r = 1530$$
$$r = 2$$

Common Ratio (r) = 2

Q7.a: Find the sum of the first 10 terms of a geometric series, if its second term is 4 and the seventh term is 128. (गुणोत्तर श्रेणीको दोस्रो पद 4 र सातौं पद 128 भए पहिलो 10 ओटा पदको योगफल पत्ता लगाउनुहोस् ।)

$$r^5 = \frac{128}{4} = 32 \Rightarrow r = 2$$
$$a(2) = 4 \Rightarrow a = 2$$
$$S_{10} = \frac{2(2^{10} – 1)}{1}$$
$$= 2(1023)$$
$$= 2046$$

Sum = 2046

Q7.b: Find the sum of the first 7 terms of a geometric series if its second term is 3 and the fifth term is 81. (गुणोत्तर श्रेणीको दोस्रो पद 3 र पाँचौँ पद 81 भए पहिलो 7 ओटा पदको योगफल पत्ता लगाउनुहोस् ।)

$$r^3 = \frac{81}{3} = 27 \Rightarrow r = 3$$
$$a(3) = 3 \Rightarrow a = 1$$
$$S_7 = \frac{1(3^7 – 1)}{2}$$
$$= \frac{2186}{2}$$
$$= 1093$$

Sum = 1093

Q8.a: How many terms of a geometric series $32 + 48 + 72 + …$ have a sum 665? (गुणोत्तर श्रेणी $32+48+72+…$ मा कतिओटा पदको योगफल 665 हुन्छ, पत्ता लगाउनुहोस् ।)

$a=32, r=1.5$

$$665 = \frac{32(1.5^n – 1)}{0.5}$$
$$332.5 = 32(1.5^n – 1)$$
$$10.390625 = 1.5^n – 1$$
$$11.390625 = 1.5^n$$
$$n = 6$$

n = 6

Q8.b: How many terms of a geometric series $6 – 12 + 24 – 48 + …$ will have a sum -2046? (गुणोत्तर श्रेणी $6-12+24-48+…$ मा कतिओटा पदको योगफल -2046 हुन्छ, पत्ता लगाउनुहोस् ।)

$a=6, r=-2$

$$-2046 = \frac{6(1 – (-2)^n)}{3}$$
$$-1023 = 1 – (-2)^n$$
$$1024 = (-2)^n$$
$$n = 10$$

n = 10

Q9: Sarita borrowed Rs. 43680… (सरिताले उसकी साथी गरिमासँग 6 ओटा किस्ताबन्दीमा तिर्ने गरी रु. 43680 सापटी लिइन्… फरक कति रहेछ, पत्ता लगाउनुहोस् ।)

$S_6 = 43680, n=6, r=3$

$$43680 = \frac{a(3^6 – 1)}{2}$$
$$87360 = 728a$$
$$a = 120$$

First Installment = 120

Last Installment = $120(3^5) = 29160$

Difference = $29160 – 120$

Difference = 29040

Disclaimer: The solutions provided here are prepared by the Important Edu Notes Team and are based on the CDC curriculum.

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Unit 6: Sequence and Series