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.