Answer :
- Let's analyze the mathematical problem step by step to derive the required values.
(a) समानान्तर श्रेणी र गुणोत्तर श्रेणीको कुनै एक फरक लेब्नुहोस्।
- Arithmetic Series (समानान्तर श्रेणी): In an arithmetic series, the difference between consecutive terms is constant. For example, if the first term is [tex]\(a\)[/tex] and the common difference is [tex]\(d\)[/tex], the terms are [tex]\(a, a + d, a + 2d, a + 3d,\)[/tex] and so on.
- Geometric Series (गुणोत्तर श्रेणी): In a geometric series, the ratio between consecutive terms is constant. For example, if the first term is [tex]\(a\)[/tex] and the common ratio is [tex]\(r\)[/tex], the terms are [tex]\(a, ar, ar^2, ar^3,\)[/tex] and so on.
(b) समानान्तर श्रेणीको दोस्रो पद पत्ता लगाउनुहोस्।
Given, the sum of three terms in the arithmetic series is 24. Let the terms be [tex]\(a - d\)[/tex], [tex]\(a\)[/tex], and [tex]\(a + d\)[/tex].
The sum of these terms is:
[tex]\[ (a - d) + a + (a + d) = 3a = 24 \][/tex]
Thus,
[tex]\[ a = \frac{24}{3} = 8 \][/tex]
So, the second term of the arithmetic series is [tex]\(8\)[/tex].
(c) धनात्मक समान अन्तर पत्ता लगाउनुहोस्।
Given that adding 1, 6, and 18 to the terms gives a geometric series, the terms transform to [tex]\(a - d + 1\)[/tex], [tex]\(a + 6\)[/tex], and [tex]\(a + d + 18\)[/tex].
These transformed terms are in a geometric series. Therefore, the ratio between consecutive terms must be the same:
[tex]\[ \frac{a + 6}{a - d + 1} = \frac{a + d + 18}{a + 6} \][/tex]
Substitute [tex]\(a = 8\)[/tex]:
[tex]\[ \frac{8 + 6}{8 - d + 1} = \frac{8 + d + 18}{8 + 6} \][/tex]
[tex]\[ \frac{14}{9 - d} = \frac{26 + d}{14} \][/tex]
Cross-multiplying:
[tex]\[ 14 \times 14 = (9 - d)(26 + d) \Rightarrow 196 = 234 + 17d - d^2 \][/tex]
Rearranging the equation:
[tex]\[ d^2 - 17d + 38 = 0 \][/tex]
Solving this quadratic equation using the quadratic formula [tex]\(d = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex] where [tex]\(a = 1\)[/tex], [tex]\(b = -17\)[/tex], and [tex]\(c = 38\)[/tex]:
[tex]\[ d = \frac{17 \pm \sqrt{17^2 - 4 \times 1 \times 38}}{2 \times 1} \][/tex]
[tex]\[ d = \frac{17 \pm \sqrt{289 - 152}}{2} \][/tex]
[tex]\[ d = \frac{17 \pm \sqrt{137}}{2} \][/tex]
Since [tex]\(d\)[/tex] must be positive and making an approximate evaluation:
[tex]\[ \sqrt{137} \approx 11.7 \][/tex]
[tex]\[ d \approx \frac{17 + 11.7}{2} = \frac{28.7}{2} = 14.35 \][/tex]
Thus, the common difference [tex]\(d\)[/tex] positive value is approximately [tex]\(14.35\)[/tex].
(d) सो समानान्तर श्रेणीको कति औ पद 104 हुन्छ ?
The [tex]\(n\)[/tex]th term of an arithmetic series is given by:
[tex]\[ a_n = a + (n-1)d \][/tex]
We need to find [tex]\(n\)[/tex] for which [tex]\(a_n = 104\)[/tex]:
[tex]\[ 104 = 8 + (n-1) \times 14.35 \][/tex]
[tex]\[ 104 - 8 = (n-1) \times 14.35 \][/tex]
[tex]\[ 96 = (n-1) \times 14.35 \][/tex]
[tex]\[ n - 1 = \frac{96}{14.35} \approx 6.69 \][/tex]
Therefore, the [tex]\(n\)[/tex]th term is approximately [tex]\(7\)[/tex].
(e) समानान्तर श्रेणीको पहिला 5 पदहरू र गुणोत्तर श्रेणीका पहिला 4 पदहरूको योगफल तुलना गर्नुहोस्।
Arithmetic series sum of the first 5 terms:
The sum of the first [tex]\(n\)[/tex] terms of an arithmetic series [tex]\(S_n\)[/tex] is given by:
[tex]\[ S_n = \frac{n}{2} \left( 2a + (n-1)d \right) \][/tex]
For the first 5 terms:
[tex]\[ S_5 = \frac{5}{2} \left( 2 \times 8 + 4 \times 14.35 \right) \][/tex]
[tex]\[ S_5 = \frac{5}{2} \left( 16 + 57.4 \right) \][/tex]
[tex]\[ S_5 = \frac{5}{2} \times 73.4 = 5 \times 36.7 = 183.5 \][/tex]
Geometric series sum of the first 4 terms:
The sum of the first [tex]\(n\)[/tex] terms of a geometric series [tex]\(S_n\)[/tex] is given by:
[tex]\[ S_n = a \frac{(r^n - 1)}{r - 1} \][/tex]
The first term of the geometric series is [tex]\(term1 = a - d + 1 = 8 - 14.35 + 1\)[/tex]:
[tex]\[ term1 = -5.35 \][/tex]
The common ratio is:
[tex]\[ term2 / term1 = 14 / -5.35 =-2.61 \ (approximately) \][/tex]
Thus,
\[
S_4 = -5.35 \frac{((-2.61)^4- 1)}{-2.61 - 1}
= -5.35 \left(\frac{46.46 - 1}{-3.61}\right)
= -5.35 \(-12.56)
=\approx 374.436\
Thus Arithmetic methods uses
\ = sum Arithmetic = 183.5 and
\= of G.Sum= 374.436
(a) समानान्तर श्रेणी र गुणोत्तर श्रेणीको कुनै एक फरक लेब्नुहोस्।
- Arithmetic Series (समानान्तर श्रेणी): In an arithmetic series, the difference between consecutive terms is constant. For example, if the first term is [tex]\(a\)[/tex] and the common difference is [tex]\(d\)[/tex], the terms are [tex]\(a, a + d, a + 2d, a + 3d,\)[/tex] and so on.
- Geometric Series (गुणोत्तर श्रेणी): In a geometric series, the ratio between consecutive terms is constant. For example, if the first term is [tex]\(a\)[/tex] and the common ratio is [tex]\(r\)[/tex], the terms are [tex]\(a, ar, ar^2, ar^3,\)[/tex] and so on.
(b) समानान्तर श्रेणीको दोस्रो पद पत्ता लगाउनुहोस्।
Given, the sum of three terms in the arithmetic series is 24. Let the terms be [tex]\(a - d\)[/tex], [tex]\(a\)[/tex], and [tex]\(a + d\)[/tex].
The sum of these terms is:
[tex]\[ (a - d) + a + (a + d) = 3a = 24 \][/tex]
Thus,
[tex]\[ a = \frac{24}{3} = 8 \][/tex]
So, the second term of the arithmetic series is [tex]\(8\)[/tex].
(c) धनात्मक समान अन्तर पत्ता लगाउनुहोस्।
Given that adding 1, 6, and 18 to the terms gives a geometric series, the terms transform to [tex]\(a - d + 1\)[/tex], [tex]\(a + 6\)[/tex], and [tex]\(a + d + 18\)[/tex].
These transformed terms are in a geometric series. Therefore, the ratio between consecutive terms must be the same:
[tex]\[ \frac{a + 6}{a - d + 1} = \frac{a + d + 18}{a + 6} \][/tex]
Substitute [tex]\(a = 8\)[/tex]:
[tex]\[ \frac{8 + 6}{8 - d + 1} = \frac{8 + d + 18}{8 + 6} \][/tex]
[tex]\[ \frac{14}{9 - d} = \frac{26 + d}{14} \][/tex]
Cross-multiplying:
[tex]\[ 14 \times 14 = (9 - d)(26 + d) \Rightarrow 196 = 234 + 17d - d^2 \][/tex]
Rearranging the equation:
[tex]\[ d^2 - 17d + 38 = 0 \][/tex]
Solving this quadratic equation using the quadratic formula [tex]\(d = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex] where [tex]\(a = 1\)[/tex], [tex]\(b = -17\)[/tex], and [tex]\(c = 38\)[/tex]:
[tex]\[ d = \frac{17 \pm \sqrt{17^2 - 4 \times 1 \times 38}}{2 \times 1} \][/tex]
[tex]\[ d = \frac{17 \pm \sqrt{289 - 152}}{2} \][/tex]
[tex]\[ d = \frac{17 \pm \sqrt{137}}{2} \][/tex]
Since [tex]\(d\)[/tex] must be positive and making an approximate evaluation:
[tex]\[ \sqrt{137} \approx 11.7 \][/tex]
[tex]\[ d \approx \frac{17 + 11.7}{2} = \frac{28.7}{2} = 14.35 \][/tex]
Thus, the common difference [tex]\(d\)[/tex] positive value is approximately [tex]\(14.35\)[/tex].
(d) सो समानान्तर श्रेणीको कति औ पद 104 हुन्छ ?
The [tex]\(n\)[/tex]th term of an arithmetic series is given by:
[tex]\[ a_n = a + (n-1)d \][/tex]
We need to find [tex]\(n\)[/tex] for which [tex]\(a_n = 104\)[/tex]:
[tex]\[ 104 = 8 + (n-1) \times 14.35 \][/tex]
[tex]\[ 104 - 8 = (n-1) \times 14.35 \][/tex]
[tex]\[ 96 = (n-1) \times 14.35 \][/tex]
[tex]\[ n - 1 = \frac{96}{14.35} \approx 6.69 \][/tex]
Therefore, the [tex]\(n\)[/tex]th term is approximately [tex]\(7\)[/tex].
(e) समानान्तर श्रेणीको पहिला 5 पदहरू र गुणोत्तर श्रेणीका पहिला 4 पदहरूको योगफल तुलना गर्नुहोस्।
Arithmetic series sum of the first 5 terms:
The sum of the first [tex]\(n\)[/tex] terms of an arithmetic series [tex]\(S_n\)[/tex] is given by:
[tex]\[ S_n = \frac{n}{2} \left( 2a + (n-1)d \right) \][/tex]
For the first 5 terms:
[tex]\[ S_5 = \frac{5}{2} \left( 2 \times 8 + 4 \times 14.35 \right) \][/tex]
[tex]\[ S_5 = \frac{5}{2} \left( 16 + 57.4 \right) \][/tex]
[tex]\[ S_5 = \frac{5}{2} \times 73.4 = 5 \times 36.7 = 183.5 \][/tex]
Geometric series sum of the first 4 terms:
The sum of the first [tex]\(n\)[/tex] terms of a geometric series [tex]\(S_n\)[/tex] is given by:
[tex]\[ S_n = a \frac{(r^n - 1)}{r - 1} \][/tex]
The first term of the geometric series is [tex]\(term1 = a - d + 1 = 8 - 14.35 + 1\)[/tex]:
[tex]\[ term1 = -5.35 \][/tex]
The common ratio is:
[tex]\[ term2 / term1 = 14 / -5.35 =-2.61 \ (approximately) \][/tex]
Thus,
\[
S_4 = -5.35 \frac{((-2.61)^4- 1)}{-2.61 - 1}
= -5.35 \left(\frac{46.46 - 1}{-3.61}\right)
= -5.35 \(-12.56)
=\approx 374.436\
Thus Arithmetic methods uses
\ = sum Arithmetic = 183.5 and
\= of G.Sum= 374.436
