Question

If \(a\) and \(b\) two \(A M s\) between \(c\) and \(d\), then \(a-c=\) (a) \(\frac{\mathrm{b}-\mathrm{c}}{2}\) (b) \(\mathrm{b}-\mathrm{c}\) (c) \(\frac{\mathrm{d}-\mathrm{c}}{2}\) (d) \(\frac{\mathrm{b}-\mathrm{c}}{3}\)

Short answer

#Answer# b: a-c=b-c

Step-by-step solution

Basic formula of Arithmetic Progression

To solve this, let's remember the basic formula of Arithmetic Progression (AP). For any AP, the nth term is given by: \(T_n = a + (n-1)d\) Where \(T_n\) is the nth term, \(a\) is the first term, \(d\) is the common difference, and \(n\) is the number of terms.

Find the common difference

In this case, we have a sequence of 4 terms as follows: c, a, b, d. Here, 'a' and 'b' are the two arithmetic means (AM) between 'c' and 'd'. Let's write the equations for each term using the formula: \(T_1 = a = c + (1-1)d = c\) \(T_2 = b = c + (2-1)d = c+d\) \(T_4 = d = c + (4-1)d = c+3d\)

Find the value of d

Now, let's subtract the first term from the second term to get the common difference: \(d=b-c\)

Express a-c in terms of b and c

The expression we want to find is a-c. Let's subtract c from both sides of the equation a=c+d: \(a-c=d\) Now, we found that \(d=b-c\). So, we can substitute this value for d in the equation for a-c: \(a-c=b-c\) Now, we can see that the correct answer to the exercise is (b).

Key concepts

Arithmetic Means

Arithmetic means, often abbreviated as AMs, are intermediate terms inserted into an arithmetic sequence such that the entire collection of numbers forms an unbroken progression. In simpler terms, arithmetic means maintain the uniform increase (or decrease) from one term to the next in a list of numbers.

For example, if you have two numbers, say 4 and 10, and wish to insert two arithmetic means between them, these would be 6 and 8, because each number increases by 2 to reach the next one: 4, 6, 8, 10. This consistent step-up is what defines an arithmetic progression, and the inserted numbers (6 and 8) are the arithmetic means.

In the exercise provided, the terms 'a' and 'b' are the arithmetic means between 'c' and 'd', thereby ensuring the sequence [c, a, b, d] ascends (or descends) by a constant amount

Common Difference

The common difference is the linchpin of an arithmetic sequence. It is denoted as 'd' and represents the fixed amount added to each term to arrive at the next term. If the common difference is positive, the sequence increases; if negative, the sequence decreases.

In an arithmetic sequence, to calculate the common difference, you can subtract the previous term from any term after the first. Mathematically, it is defined as:

\(d = T_{n} - T_{n-1}\) for any term \(T_{n}\) where \(n > 1\).

In the context of our exercise, we used the common difference to determine the relationship between the terms 'a', 'b', 'c', and 'd'. By understanding that 'a' and 'b' are means, we ascertained their positions in the sequence and used these positions to calculate their values based on the common difference.

Arithmetic Sequence

An arithmetic sequence is a list of numbers with a specific characteristic: each term after the first is arrived at by adding a constant value, known as the common difference, to the previous term.

The general form of an arithmetic sequence is given by the formula:

\[a, a+d, a+2d, a+3d, \.\.\.\] where 'a' is the first term and 'd' is the common difference. For any term in the sequence, the nth term can be calculated by using the formula:

\[T_n = a + (n-1)d\]

In essence, the common difference ties the sequence together, ensuring that the same interval is kept from one number to the next. This predictability is what makes arithmetic sequences particularly manageable and a fundamental concept in algebra and number theory.