6. (a) Find the value of x and y if 10, x, y, 25 are in AP.

(b) If-13, p, q, r, 7 are in AP, find the values of p, q and r.

7. (a) 1+n If /, m and n are in AP, show that m =.

(b) If the arithmetic mean between two number is 50 and the second number is 60, find the first number.

(c) Two numbers are in the ratio of 2:3. If their AM is 20, find the numbers.

Let's solve each of these problems one by one.

### 6. (a) Find the value of \( x \) and \( y \) if 10, \( x \), \( y \), 25 are in AP.

For any four numbers in AP, the difference between consecutive terms is constant.

Let's denote the common difference by \( d \).

- The second term \( x \) can be expressed as:
  \[
  x = 10 + d
  \]
- The third term \( y \) can be expressed as:
  \[
  y = x + d = 10 + 2d
  \]
- The fourth term is:
  \[
  25 = y + d = 10 + 3d
  \]

From \( 25 = 10 + 3d \), solve for \( d \):
\[
3d = 15 \Rightarrow d = 5
\]

Now, substitute \( d = 5 \) into the equations for \( x \) and \( y \):
\[
x = 10 + 5 = 15
\]
\[
y = 10 + 2(5) = 20
\]

So, the values of \( x \) and \( y \) are \( x = 15 \) and \( y = 20 \).

### 6. (b) If \(-13\), \( p \), \( q \), \( r \), 7 are in AP, find the values of \( p \), \( q \), and \( r \).

Again, the common difference \( d \) between consecutive terms is constant.

- The second term \( p \) can be expressed as:
  \[
  p = -13 + d
  \]
- The third term \( q \) can be expressed as:
  \[
  q = p + d = -13 + 2d
  \]
- The fourth term \( r \) can be expressed as:
  \[
  r = q + d = -13 + 3d
  \]
- The fifth term is:
  \[
  7 = r + d = -13 + 4d
  \]

From \( 7 = -13 + 4d \), solve for \( d \):
\[
4d = 20 \Rightarrow d = 5
\]

Now, substitute \( d = 5 \) into the equations for \( p \), \( q \), and \( r \):
\[
p = -13 + 5 = -8
\]
\[
q = -13 + 2(5) = -3
\]
\[
r = -13 + 3(5) = 2
\]

So, the values of \( p \), \( q \), and \( r \) are \( p = -8 \), \( q = -3 \), and \( r = 2 \).

### 7. (a) If \( l \), \( m \), and \( n \) are in AP, show that \( m = \frac{l + n}{2} \).

Since \( l \), \( m \), and \( n \) are in AP, the difference between consecutive terms is constant:
\[
m - l = n - m
\]

Adding \( l \) to both sides:
\[
2m = l + n
\]

Divide both sides by 2:
\[
m = \frac{l + n}{2}
\]

This shows that \( m \) is the arithmetic mean of \( l \) and \( n \).

### 7. (b) If the arithmetic mean between two numbers is 50 and the second number is 60, find the first number.

Let the first number be \( a \), and the second number is given as 60.

The arithmetic mean between two numbers is:
\[
\frac{a + 60}{2} = 50
\]

Multiply both sides by 2:
\[
a + 60 = 100
\]

Subtract 60 from both sides:
\[
a = 40
\]

So, the first number is 40.

### 7. (c) Two numbers are in the ratio of 2:3. If their arithmetic mean is 20, find the numbers.

Let the two numbers be \( 2x \) and \( 3x \).

The arithmetic mean of these two numbers is:
\[
\frac{2x + 3x}{2} = 20
\]

Simplify:
\[
\frac{5x}{2} = 20
\]

Multiply both sides by 2:
\[
5x = 40
\]

Divide by 5:
\[
x = 8
\]

So, the two numbers are:
\[
2x = 2(8) = 16 \quad \text{and} \quad 3x = 3(8) = 24
\]

Thus, the numbers are 16 and 24.

Do you have any questions, or would you like more details on any of these solutions?

Here are 5 related questions:

1. If \( 5, x, y, z, 20 \) are in AP, find \( x \), \( y \), and \( z \).
2. If the arithmetic mean between \( x-4 \) and \( 3x+2 \) is 12, find the value of \( x \).
3. In an AP, the 6th term is 30 and the 10th term is 50. Find the first term and common difference.
4. If \( x \), \( y \), and \( z \) are in AP and \( x + z = 16 \), find the value of \( y \).
5. Find the 15th term of an AP if the first term is 7 and the common difference is 3.

**Tip:** The sum of the first and last terms of an arithmetic sequence divided by 2 gives the arithmetic mean of the sequence.

Learn how to solve arithmetic progression (AP) problems step-by-step. Includes detailed solutions for finding terms in an AP sequence and determining arithmetic means. Suitable for high school students.

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