Question

Three consecutive positive integers a, b and c are such that b2
= 4(a + c). Find the value of c

Answer 1

Let the three consecutive positive integers be \( a \), \( b \), and \( c \). Since they are consecutive integers, we can express them as:

\[
a = n, \quad b = n + 1, \quad c = n + 2
\]

for some positive integer \( n \).

According to the problem, we have the equation:

\[
b^2 = 4(a + c)
\]

Substituting \( a \), \( b \), and \( c \) in terms of \( n \):

\[
(n + 1)^2 = 4(n + (n + 2))
\]

This simplifies to:

\[
(n + 1)^2 = 4(2n + 2)
\]

Calculating the right side:

\[
(n + 1)^2 = 8n + 8
\]

Expanding the left side:

\[
n^2 + 2n + 1 = 8n + 8
\]

Rearranging this equation gives:

\[
n^2 + 2n + 1 - 8n - 8 = 0
\]

Simplifying further:

\[
n^2 - 6n - 7 = 0
\]

Now, we can solve this quadratic equation using the quadratic formula:

\[
n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

In this case, \( a = 1 \), \( b = -6 \), and \( c = -7 \):

\[
n = \frac{6 \pm \sqrt{(-6)^2 - 4 \cdot 1 \cdot (-7)}}{2 \cdot 1}
\]

Calculating the discriminant:

\[
\sqrt{36 + 28} = \sqrt{64} = 8
\]

Thus:

\[
n = \frac{6 \pm 8}{2}
\]

Calculating the two possible values for \( n \):

\[
n = \frac{14}{2} = 7 \quad \text{or} \quad n = \frac{-2}{2} = -1
\]

Since \( n \) must be a positive integer, we take \( n = 7 \).

Now, substituting back to find \( c \):

\[
c = n + 2 = 7 + 2 = 9
\]

Thus, the value of \( c \) is:

\[
\boxed{9}
\]