यदि कुनै गुणोत्तर श्रेणीको तेस्रो पद र छैटौं पद क्रमशः 12 र 96 भए सो श्रेण पहिलो पद र पहिलो चारओटा पदहरूको योगफलको समानान्तरीय मध्यमा हुन्छ ? पत्तां लगाउनुहोस्।
If the third term and sixth term of a geometric series are 12 and respectively, what is the arithmetic mean between first term and the su of first four terms of the series? Find it.
फलन एउटा वास्तविक फलन हो । is a real valued function.
(a) र पत्ता लगाउनुहोस् ।

Find and .
(b) $र$ पत्ता लगाउनुहोस् ।

Find and
© के फ़लन विन्दु मा निरन्तर छ ? कारण दिनुहोस् ।

Is the function continuous at the point ? Give reason.
22. क्रामर विधिको प्रयोग गरी हल गर्नुहोस् (Solve using Cramer’s rule):

समीकरण ले प्रतिनिधित्व गरेका एकं जोडा रेखाहरूबिचक कोण पत्ता लग़ाउनुहोस्।
Find the angle between a pair of lines represented by the equatior .

Ask by Fernandez Murphy. in Nepal

Mar 26,2025

Question

यदि कुनै गुणोत्तर श्रेणीको तेस्रो पद र छैटौं पद क्रमशः 12 र 96 भए सो श्रेण पहिलो पद र पहिलो चारओटा पदहरूको योगफलको समानान्तरीय मध्यमा हुन्छ ? पत्तां लगाउनुहोस्।
If the third term and sixth term of a geometric series are 12 and respectively, what is the arithmetic mean between first term and the su of first four terms of the series? Find it.
फलन एउटा वास्तविक फलन हो । is a real valued function.
(a) र पत्ता लगाउनुहोस् ।

Find and .
(b) $र$ पत्ता लगाउनुहोस् ।

Find and
© के फ़लन विन्दु मा निरन्तर छ ? कारण दिनुहोस् ।

Is the function continuous at the point ? Give reason.
22. क्रामर विधिको प्रयोग गरी हल गर्नुहोस् (Solve using Cramer’s rule):

समीकरण ले प्रतिनिधित्व गरेका एकं जोडा रेखाहरूबिचक कोण पत्ता लग़ाउनुहोस्।
Find the angle between a pair of lines represented by the equatior .

Ask by Fernandez Murphy. in Nepal

Mar 26,2025

UpStudy AI · Accepted Answer

Let's break down the problem step by step.

### Part 1: Geometric Series

Given:
- Third term $$ a r^2 = 12 $$
- Sixth term $$ a r^5 = 96 $$

From these, we can form two equations:
1. $$ a r^2 = 12 $$ (1)
2. $$ a r^5 = 96 $$ (2)

To find $$ a $$ and $$ r $$, we can divide equation (2) by equation (1):

$$
\frac{a r^5}{a r^2} = \frac{96}{12}
$$

This simplifies to:

$$
r^3 = 8 \implies r = 2
$$

Now substituting $$ r $$ back into equation (1):

$$
a (2^2) = 12 \implies 4a = 12 \implies a = 3
$$

Now, we can find the sum of the first four terms:

$$
S_4 = a \frac{r^4 - 1}{r - 1} = 3 \frac{2^4 - 1}{2 - 1} = 3 \cdot 15 = 45
$$

The arithmetic mean between the first term $$ a $$ and the sum of the first four terms $$ S_4 $$ is:

$$
\text{AM} = \frac{a + S_4}{2} = \frac{3 + 45}{2} = 24
$$

### Part 2: Function Evaluation

Given the function $$ f(x) = 3x - 5 $$:

#### (a) Calculate $$ f(3.9), f(3.99), f(4.01), f(4.001) $$:

1. $$ f(3.9) = 3(3.9) - 5 = 11.7 - 5 = 6.7 $$
2. $$ f(3.99) = 3(3.99) - 5 = 11.97 - 5 = 6.97 $$
3. $$ f(4.01) = 3(4.01) - 5 = 12.03 - 5 = 7.03 $$
4. $$ f(4.001) = 3(4.001) - 5 = 12.003 - 5 = 7.003 $$

#### (b) Calculate the limits:

1. $$ \lim_{x \to 4^{-}} f(x) = f(4) = 3(4) - 5 = 12 - 5 = 7 $$
2. $$ \lim_{x \to 4^{+}} f(x) = f(4) = 3(4) - 5 = 12 - 5 = 7 $$

#### (c) Continuity at $$ x = 4 $$:

The function $$ f(x) $$ is continuous at $$ x = 4 $$ because:

- $$ \lim_{x \to 4^{-}} f(x) = \lim_{x \to 4^{+}} f(x) = f(4) = 7 $$

### Part 3: Cramer's Rule

Given the equations:

$$
4x + 3y = -18 \quad (1)
$$
$$
2x - 5y = 4 \quad (2)
$$

Using Cramer's rule, we need to find the determinants:

1. Coefficient matrix $$ D $$:
$$
D = \begin{vmatrix}
4 & 3 \
2 & -5
\end{vmatrix} = (4)(-5) - (3)(2) = -20 - 6 = -26
$$

2. Determinant $$ D_x $$ (replace first column with constants):
$$
D_x = \begin{vmatrix}
-18 & 3 \
4 & -5
\end{vmatrix} = (-18)(-5) - (3)(4) = 90 - 12 = 78
$$

3. Determinant $$ D_y $$ (replace second column with constants):
$$
D_y = \begin{vmatrix}
4 & -18 \
2 & 4
\end{vmatrix} = (4)(4) - (-18)(2) = 16 + 36 = 52
$$

Now, we can find $$ x $$ and $$ y $$:

$$
x = \frac{D_x}{D} = \frac{78}{-26} = -3
$$
$$
y = \frac{D_y}{D} = \frac{52}{-26} = -2
$$

### Part 4: Angle Between Lines

The equation $$ 2x^2 + 7xy + 3y^2 = 0 $$ represents two lines. The angle $$ \theta $$ between the lines can be found using the formula:

$$
\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|
$$

Where $$ m_1 $$ and $$ m_2 $$ are the slopes of the lines. The slopes can be found from the quadratic equation:

The slopes are given by:

$$
m_1, m_2 = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
$$

Here, $$ a = 2, b = 7, c = 3 $$:

Calculating the discriminant:

$$
b^2 - 4ac = 7^2 - 4(2)(3) = 49 - 24 = 25
$$

Thus, the slopes are:

$$
m_1, m_2 = \frac{-7 \pm 5}{4}
$$

Calculating the slopes:

1. $$ m_1 = \frac{-2}{4} = -0.5 $$
2. $$ m_2 = \frac{-12}{4} = -3 $$

Now, we can find the angle:

$$
\tan \theta = \left| \frac{-0.5 - (-3)}{1 + (-0.5)(-3)} \right| = \left| \frac{2.5}{1 + 1.5} \right| = \left| \frac{2.5}{2.5} \right| = 1
$$

Thus, $$ \theta = 45^\circ $$.

### Summary of Results:
1. Arithmetic Mean: $$ 24 $$
2. Function evaluations: $$ f(3.9) = 6.7, f(3.99) = 6.97, f(4.01) = 7.03, f(4.001) = 7.003 $$
3. Limits: $$ \lim_{x \to 4^{-}} f(x)
**Geometric Series:** - First term $$ a = 3 $$ - Sum of first four terms $$ S_4 = 45 $$ - Arithmetic mean between first term and $$ S_4 $$ is $$ 24 $$ **Function Evaluations:** - $$ f(3.9) = 6.7 $$ - $$ f(3.99) = 6.97 $$ - $$ f(4.01) = 7.03 $$ - $$ f(4.001) = 7.003 $$ **Limits:** - $$ \lim_{x \to 4^{-}} f(x) = 7 $$ - $$ \lim_{x \to 4^{+}} f(x) = 7 $$ **Continuity at $$ x = 4 $$:** - The function $$ f(x) = 3x - 5 $$ is continuous at $$ x = 4 $$ because both left and right limits equal $$ f(4) = 7 $$. **Cramer's Rule Solution:** - $$ x = -3 $$ - $$ y = -2 $$ **Angle Between Lines:** - The angle between the lines represented by $$ 2x^2 + 7xy + 3y^2 = 0 $$ is $$ 45^\circ $$.

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