Question

The arithmetic mean of 7 consecutive integers starting with a is \(m\) then the arithmetic mean of 11 consecutive integers starting with \(a+2\) is (a) \(2 \mathrm{a}\) (b) \(2 \mathrm{~m}\) (c) \(a+4\) (d) \(m+4\)

Short answer

The arithmetic mean of the second set of 11 consecutive integers in terms of the mean of the first set is (d) \(m+4\).

Step-by-step solution

Find the mean of the first set of integers

To find the arithmetic mean of the 7 consecutive integers starting with a, we need to add all the integers and then divide the sum by 7. The integers are: a, a+1, a+2, a+3, a+4, a+5, and a+6. Mean of 7 integers, m = \(\frac{a+(a+1)+(a+2)+(a+3)+(a+4)+(a+5)+(a+6)}{7}\)

Simplify the mean for the first set

Now, we can simplify the mean expression by adding all the terms and dividing by 7: m = \(\frac{7a+21}{7}\) Divide both numerator and denominator by 7: m = \(a+3\) Now we have the arithmetic mean of the first set of consecutive integers in terms of a.

Find the mean of the second set of integers

Now we need to find the arithmetic mean of the 11 consecutive integers starting with a+2. The integers are: a+2, a+3, a+4, a+5, a+6, a+7, a+8, a+9, a+10, a+11, and a+12. Mean of 11 integers = \(\frac{(a+2)+(a+3)+(a+4)+(a+5)+(a+6)+(a+7)+(a+8)+(a+9)+(a+10)+(a+11)+(a+12)}{11}\)

Simplify the mean for the second set

Next, we simplify the expression by adding all the terms and dividing by 11: Mean of 11 integers = \(\frac{11a+77}{11}\) Divide both numerator and denominator by 11: Mean of 11 integers = \(a+7\) Now we have the arithmetic mean of the second set of consecutive integers in terms of a.

Compare the means and choose the correct option

We know the mean of the first set of 7 integers is m = \(a+3\), and the mean of the second set of 11 integers is \(a+7\). The question asks for the arithmetic mean of the second set in terms of m or a. Comparing the mean of the second set with the first set (m = \(a+3\)), we notice that the mean of the second set is 4 more than the first set's mean: Mean of 11 integers = m + 4 This matches option (d). So the correct answer is (d) \(m+4\).

Key concepts

Consecutive Integers

Consecutive integers are numbers that follow each other in order, without any gaps. They increase by 1 from one integer to the next. For example, 3, 4, and 5 are consecutive integers. In mathematical terms, if a is the first integer, the next consecutive integers would be a+1, a+2, a+3, and so on.
In the exercise, the first set of numbers begins with the integer 'a', giving us seven consecutive integers: a, a+1, a+2, ..., a+6. The second set begins with a+2, resulting in eleven integers: a+2, a+3, ..., a+12. Knowing the structure of consecutive integers helps us set up the problem for calculating their mean.

Mean of Integers

The arithmetic mean, or average, of integers is calculated by summing all the integers in a set and then dividing by the number of integers. This provides a central value, which is representative of the set.
For the first set of 7 consecutive integers starting with 'a', the mean is calculated by adding the integers: a, a+1, a+2, ..., a+6, and then dividing by 7. This simplifies to \( m = \frac{7a + 21}{7} \) and results in \( m = a + 3 \).
For the second set of 11 integers beginning with a+2, we perform a similar process. Sum all the integers: a+2 through a+12, divide by 11, resulting in the mean, \( \frac{11a + 77}{11} \), which simplifies to \( a + 7 \). Understanding how to compute the mean is essential for comparing different sets of consecutive integers in algebra.

Algebraic Expressions

Algebraic expressions help us generalize mathematical operations using variables. In this problem, expressions are formed to represent consecutive integers and their means.
We start with expressions like a, a+1 for integers. Then, using basic algebraic operations, we find the average of these numbers in terms of 'a'. For example, the formula \( \frac{7a + 21}{7} \) simplifies to \( m = a + 3 \), representing the mean of seven consecutive integers starting with 'a'.
In the second scenario, we used a higher number of integers starting from a different initial value, a+2. Once again, algebra simplifies their average to a simple expression, \( a + 7 \).
Algebra helps in efficiently comparing two different sets of numbers, where in this exercise, it leads to finding that the mean of the second set is 'm + 4', emphasizing the elegance of algebra in deriving useful mathematical insights.