To solve the problem step by step, we will follow the logical progression outlined in the video transcript.
### Step 1: Calculate the average of 35 consecutive natural numbers (N)
Let the first number in the sequence of 35 consecutive natural numbers be \( x \). The numbers are \( x, x+1, x+2, \ldots, x+34 \).
The average \( N \) of these numbers can be calculated as follows:
\[
N = \frac{x + (x + 1) + (x + 2) + \ldots + (x + 34)}{35}
\]
This simplifies to:
\[
N = \frac{35x + (0 + 1 + 2 + \ldots + 34)}{35}
\]
Using the formula for the sum of the first \( n \) natural numbers, \( \frac{n(n + 1)}{2} \), we find:
\[
0 + 1 + 2 + \ldots + 34 = \frac{34 \times 35}{2} = 595
\]
Thus, we have:
\[
N = \frac{35x + 595}{35} = x + \frac{595}{35} = x + 17
\]
### Step 2: Calculate the new average (M) after dropping and adding numbers
When we drop the first 10 numbers \( x, x+1, \ldots, x+9 \) and include the next 10 numbers \( x+35, x+36, \ldots, x+44 \), we need to find the new average \( M \).
The remaining numbers after dropping the first 10 are \( x+10, x+11, \ldots, x+34 \), which gives us 25 numbers. The sum of these numbers is:
\[
\text{Sum} = (x + 10) + (x + 11) + \ldots + (x + 34) = 25x + (10 + 11 + \ldots + 34)
\]
Calculating the sum \( 10 + 11 + \ldots + 34 \):
\[
10 + 11 + \ldots + 34 = \frac{(34 - 10 + 1)(10 + 34)}{2} = \frac{25 \times 44}{2} = 550
\]
Thus, the sum becomes:
\[
\text{Sum} = 25x + 550
\]
Now, we add the next 10 numbers:
\[
\text{New Sum} = 25x + 550 + (x + 35 + x + 36 + \ldots + x + 44) = 25x + 550 + (10x + 405) = 35x + 955
\]
The new average \( M \) is:
\[
M = \frac{35x + 955}{35} = x + \frac{955}{35} = x + 27.2857 \approx x + 28
\]
### Step 3: Relate M and N
From the problem, we know:
\[
M^2 - N^2 = 600
\]
Using the difference of squares:
\[
(M - N)(M + N) = 600
\]
Substituting \( M = N + 10 \) (since \( M \) is higher than \( N \)):
\[
(10)(M + N) = 600 \implies M + N = 60
\]
Now substituting \( N = x + 17 \) and \( M = x + 27 \):
\[
(x + 27) + (x + 17) = 60
\]
This simplifies to:
\[
2x + 44 = 60 \implies 2x = 16 \implies x = 8
\]
Thus:
\[
N = 8 + 17 = 25, \quad M = 8 + 27 = 35
\]
### Step 4: Calculate the average of \( 3M \) and \( 5N \)
Now we calculate:
\[
3M = 3 \times 35 = 105, \quad 5N = 5 \times 25 = 125
\]
The average of \( 3M \) and \( 5N \) is:
\[
\text{Average} = \frac{3M + 5N}{2} = \frac{105 + 125}{2} = \frac{230}{2} = 115
\]
### Final Answer
The average of \( 3M \) and \( 5N \) is \( \boxed{115} \).
