Math Reasoning Exercises
The sum of three numbers in arithmetic progression is 30 and their product is 510. The largest of these three numbers is:
-
7
-
10
-
13
-
17
-
14
Try to solve it before checking the answer.
17
Let these numbers be \(a_1\), \(a_2\) and \(a_3\), and let \(d\) be the ratio of the progression. Since they are arithmetic progressions, it follows that:
a_2 = a_1 + d,
\]
a_3 = a_1 + 2d
\]
Hence,
\(
a_1 + (a_1 + d) + (a_1 + 2d) = 30 \hspace{2em} \boldsymbol{(1)}
\)
\(
a_1 (a_1 + d) (a_1 + 2d) = 510 \hspace{3em} \boldsymbol{(2)}
\)
From (1):
3 a_1 + 3d = 30 \Rightarrow a_1 + d = 10
\]
Then:
a_1 = 10 – d \hspace{3em} \boldsymbol{(4)}
\]
Replacing (4) in (2):
\begin{aligned}
&(10 – d) (10) (10 + d) = 510
\\[.5em]
&\hspace{3em}\Rightarrow
(10 – d) (10 + d) = 51
\\[.5em]
&\hspace{3em}\Rightarrow
100 – d^2 = 51
\\[.5em]
&\hspace{3em}\Rightarrow
d^2 = 49
\\[.5em]
&\hspace{3em}\Rightarrow
d = 7
\end{aligned}
\]
\begin{aligned}
(10 – d) (10) (10 + d) = 510
&\Rightarrow
(10 – d) (10 + d) = 51
\\[.5em]
&\Rightarrow
100 – d^2 = 51
\\[.5em]
&\Rightarrow
d^2 = 49
\\[.5em]
&\Rightarrow
d = 7
\end{aligned}
\]
Replacing \(d = 7\) in (4):
a_1 = 10 – 7 = 3
\]
Now, the largest of the three numbers is:
\begin{aligned}
a_3 &= a_1 + 2d
\\[.5em]
&= 3 + 2(7)
\\[.5em]
&= \boldsymbol{17}
\end{aligned}
\]