© All Rights reserved @ LearnWithDash
Step-by-Step Solution
Step 1: Understand the Problem
We have a set X of natural numbers from 1 to 17. A second set Y is formed by the linear transformation ax + b, where a and b are real numbers with a > 0. It is given that the mean of the elements of Y is 17, and the variance of the elements of Y is 216. We need to find the value of a + b.
Step 2: Calculate the Mean of X
The set X = {1, 2, 3, …, 17}. The mean of X is
$ \displaystyle \text{Mean}(X) = \frac{\sum_{x=1}^{17} x}{17}. $
We know
$ \displaystyle \sum_{x=1}^{17} x = 1 + 2 + \cdots + 17 = \frac{17 \times 18}{2} = 153.$
So,
$ \displaystyle \text{Mean}(X) = \frac{153}{17} = 9.$
Step 3: Express the Mean of Y
The elements of Y are of the form ax + b. Hence,
$ \displaystyle \text{Mean}(Y) = \frac{1}{17}\sum_{x=1}^{17} (ax + b). $
Given that Mean(Y) = 17, we write:
$ \displaystyle 17 = \frac{1}{17} \sum_{x=1}^{17} (ax + b). $
Simplify the summation:
$ \displaystyle \sum_{x=1}^{17} (ax + b) = a \sum_{x=1}^{17} x + b \sum_{x=1}^{17} 1 = a \cdot 153 + b \cdot 17. $
So,
$ \displaystyle 17 \;=\; \frac{a \cdot 153 + 17b}{17} = 9a + b. $
Thus, we obtain the linear equation:
$ \displaystyle 9a + b = 17. \quad (1) $
Step 4: Use the Variance of Y
The variance of a set S with n elements is
$ \displaystyle \text{Var}(S) = \frac{1}{n} \sum_{x \in S} (x - \text{mean}(S))^2. $
Alternatively,
$ \displaystyle \text{Var}(S) = \frac{1}{n} \sum_{x \in S} x^2 - \bigl(\text{mean}(S)\bigr)^2. $
Here, S corresponds to Y, and Var(Y) = 216. Also, Mean(Y) = 17. Hence,
$ \displaystyle 216
= \frac{1}{17} \sum_{x=1}^{17} (ax + b)^2 - 17^2. $
Rearrange to figure out the sum:
$ \displaystyle \frac{1}{17} \sum_{x=1}^{17} (ax + b)^2 = 216 + 17^2 = 216 + 289 = 505. $
Thus,
$ \displaystyle \sum_{x=1}^{17} (ax + b)^2 = 17 \times 505 = 8585. $
Step 5: Expand the Summation of Squares
We need:
$ \displaystyle \sum_{x=1}^{17} (ax + b)^2 = \sum_{x=1}^{17} \bigl(a^2 x^2 + 2ab\,x + b^2\bigr). $
This can be separated into:
$ \displaystyle a^2 \sum_{x=1}^{17} x^2
\;+\; 2ab \sum_{x=1}^{17} x
\;+\; b^2 \sum_{x=1}^{17} 1. $
We recall the standard formula
$ \sum_{x=1}^{n} x^2 = \frac{n(n+1)(2n+1)}{6}. $
For n = 17,
$ \displaystyle \sum_{x=1}^{17} x^2 = \frac{17 \times 18 \times 35}{6} = 1785, $
and as found before,
$ \displaystyle \sum_{x=1}^{17} x = 153, \quad \sum_{x=1}^{17} 1 = 17. $
Hence the expansion becomes:
$ \displaystyle \sum_{x=1}^{17} (ax + b)^2
= a^2 \cdot 1785 + 2ab \cdot 153 + b^2 \cdot 17. $
We know this sum must be 8585. Therefore,
$ \displaystyle 1785\,a^2 + 306\,a\,b + 17\,b^2 = 8585. \quad (2) $
Step 6: Solve the System of Equations
From step (3), we have:
$ \displaystyle b = 17 - 9a. $
Substitute b into equation (2) and solve for a. After finding a, we can find b. The detailed algebra yields:
$ \displaystyle a = 3, \quad b = -10. $
Step 7: Find a + b
With a = 3 and b = -10,
$ \displaystyle a + b = 3 + (-10) = -7. $
Final Answer
-7
