Solving the Quadratic Equation and Understanding the Arithmetic Progression

The problem asks us to find the roots of a given quadratic equation, insert four terms between these roots to form an arithmetic progression (AP), and then find the sum of these four inserted terms.

Finding the Roots of the Quadratic Equation

The given quadratic equation is \(x^2 - 16x + 39 = 0\).

We can solve this equation by factoring or using the quadratic formula. Let's use factoring:

We need two numbers that multiply to 39 and add up to -16. These numbers are -3 and -13.

So, we can rewrite the equation as:

\((x - 3)(x - 13) = 0\)

Setting each factor to zero gives the roots:

• \(x - 3 = 0 \implies x = 3\)
• \(x - 13 = 0 \implies x = 13\)

The roots are 3 and 13. The problem states that the roots are m and n with m < n. Therefore, we have:

• \(m = 3\)
• \(n = 13\)

Forming the Arithmetic Progression (AP)

Four terms p, q, r, and s are inserted between m and n to form an AP. This means the sequence \(m, p, q, r, s, n\) is an arithmetic progression.

The terms of this AP are:

1. First term (\(a_1\)) = m = 3
2. Second term (\(a_2\)) = p
3. Third term (\(a_3\)) = q
4. Fourth term (\(a_4\)) = r
5. Fifth term (\(a_5\)) = s
6. Sixth term (\(a_6\)) = n = 13

This AP has a total of 6 terms.

Method 1: Finding Each Term and Summing Them

In an AP, the n-th term is given by the formula \(a_n = a_1 + (n-1)d\), where \(d\) is the common difference.

We know \(a_1 = 3\) and \(a_6 = 13\). Using the formula for \(a_6\):

\(a_6 = a_1 + (6-1)d\)

\(13 = 3 + 5d\)

\(13 - 3 = 5d\)

\(10 = 5d\)

\(d = \frac{10}{5} = 2\)

The common difference of the AP is 2.

Now we can find the values of p, q, r, and s:

• \(p = a_2 = a_1 + d = 3 + 2 = 5\)
• \(q = a_3 = a_1 + 2d = 3 + 2(2) = 3 + 4 = 7\)
• \(r = a_4 = a_1 + 3d = 3 + 3(2) = 3 + 6 = 9\)
• \(s = a_5 = a_1 + 4d = 3 + 4(2) = 3 + 8 = 11\)

The four inserted terms are 5, 7, 9, and 11.

Now, we find the sum \(p + q + r + s\):

\(p + q + r + s = 5 + 7 + 9 + 11\)

\(p + q + r + s = 12 + 20\)

\(p + q + r + s = 32\)

Method 2: Using Properties of AP

In any AP, the sum of terms equidistant from the beginning and the end is constant and equal to the sum of the first and last terms.

The AP is \(m, p, q, r, s, n\).

The sum of the first and last terms is \(m + n\).

The terms \(p\) and \(s\) are equidistant from the ends (second term from start, second term from end).

The terms \(q\) and \(r\) are equidistant from the ends (third term from start, third term from end).

So, we have:

• \(p + s = m + n\)
• \(q + r = m + n\)

The sum \(p + q + r + s\) can be written as \((p + s) + (q + r)\).

Substituting the relationships above:

\(p + q + r + s = (m + n) + (m + n) = 2(m + n)\)

For the quadratic equation \(x^2 - 16x + 39 = 0\), the sum of the roots (m + n) can be found using Vieta's formulas. The sum of roots of \(ax^2 + bx + c = 0\) is \(-b/a\).

Here, \(a=1\), \(b=-16\), \(c=39\).

Sum of roots \(m + n = -(-16)/1 = 16\).

Now, substitute the sum of roots into the expression for \(p + q + r + s\):

\(p + q + r + s = 2(m + n) = 2(16)\)

\(p + q + r + s = 32\)

Both methods yield the same result.

Conclusion

The value of \(p + q + r + s\) is 32.

Revision Table: Key Concepts

Additional Information: Arithmetic Means

When terms are inserted between two numbers to form an AP, these inserted terms are called Arithmetic Means.

If \(n\) terms are inserted between \(a\) and \(b\) to form an AP, the sequence is \(a, m_1, m_2, ..., m_n, b\). This is an AP with \(n+2\) terms.

• First term = a
• Last term = b (which is the (n+2)-th term)
• Common difference \(d = \frac{b - a}{n + 1}\)

In our problem, 4 terms (p, q, r, s) were inserted between 3 (m) and 13 (n). So, \(a=3\), \(b=13\), and \(n=4\). The number of terms in the AP is \(4+2=6\).

The common difference \(d = \frac{13 - 3}{4 + 1} = \frac{10}{5} = 2\), which matches our calculation.

The sum of the inserted terms (\(m_1 + m_2 + ... + m_n\)) can also be found using the formula \(n \times \text{Arithmetic Mean of a and b}\), where the Arithmetic Mean of a and b is \((a+b)/2\).

However, here \(n\) is the number of *inserted* terms (4), and the 'Arithmetic Mean' refers to the average of the *two endpoint numbers* (m and n). The sum of the inserted terms is \(n \times (\frac{m+n}{2})\) *if n is odd*. A more general approach for the sum of n arithmetic means inserted between a and b is \(n \times \frac{a+b}{2}\).

Sum of inserted terms \(p+q+r+s = 4 \times \frac{m+n}{2} = 4 \times \frac{3+13}{2} = 4 \times \frac{16}{2} = 4 \times 8 = 32\). This formula \(n \times \frac{a+b}{2}\) is the most efficient way to find the sum of the inserted means directly.