# If (1+3+5+...... +a ) + (1+3+5+ .......... +b) = (1+3+5......... +c), where each set of parentheses contains the sum of consecutive odd integers as shown such that a+b+c = 21, a>6. If G = Max{a,b,c} and L = Min{a,b,c}, then? The question has multiple ans

## A) G-L = 4 B) b-a = 2 C) G-L = 7 D) a-b = 2

Feb 4, 2018

$a = 7 \text{ "" "b=5" "" } c = 9$

The correct choices are A and D.

#### Explanation:

The sum of the first $n$ odd numbers is ${n}^{2}$.

$1 = 1$

$1 + 3 = 4$

$1 + 3 + 5 = 9$

$1 + 3 + 5 + 7 = 16$

...and so on.

A number's position within the set of odd numbers is $\frac{n + 1}{2}$

For example, 11 is the $\frac{11 + 1}{2} = 6$th odd number

Using these two facts, we can rewrite our expression as:

$\left(1 + 3 + 5 + \cdots + a\right) + \left(1 + 3 + 5 + \cdots + b\right) = \left(1 + 3 + 5 + \cdots + c\right)$

$\text{sum of first "(a+1)/2" odd numbers" + "sum of first "(b+1)/2" odd numbers" = "sum of first "(c+1)/2" odd numbers}$

${\left(\frac{a + 1}{2}\right)}^{2} + {\left(\frac{b + 1}{2}\right)}^{2} = {\left(\frac{c + 1}{2}\right)}^{2}$

Doing a little bit of algebra to this, we can simplify the expression:

${\left(a + 1\right)}^{2} / 4 + {\left(b + 1\right)}^{2} / 4 = {\left(c + 1\right)}^{2} / 4$

${\left(a + 1\right)}^{2} + {\left(b + 1\right)}^{2} = {\left(c + 1\right)}^{2}$

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Since $a$, $b$, and $c$ are all odd, $\left(a + 1\right)$, $\left(b + 1\right)$, and $\left(c + 1\right)$ must all be even.

We also know that $a + b + c = 21$, which means that:

$\left(a + 1\right) + \left(b + 1\right) + \left(c + 1\right)$
$= a + b + c + 3$
$= 21 + 3$
$= 24$

There's only one Pythagorean triple that adds up to 24:

${6}^{2} + {8}^{2} = {10}^{2}$

$6 + 8 + 10 = 24$

Since we know that $a > 6$, we also know that $a + 1 > 7$.

Therefore, we know that $a + 1$ must be $8$ since it can't be 6.

This means that $b + 1$ is 6 and $c + 1$ is 10.

Therefore, our three numbers are:

$a = 7 \text{ "" "b=5 " "" } c = 9$

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Using these numbers, we can figure out which of the 4 statements are true.

$G = \max \left\{a , b , c\right\} = \max \left\{7 , 5 , 9\right\} = 9$

$L = \min \left\{a , b , c\right\} = \min \left\{7 , 5 , 9\right\} = 5$

Statement A

$G - L = 9 - 5 = 4 \text{ "" "" "" }$ So, statement A is true.

Statement B

$b - a = 5 - 7 = - 2 \text{ "" "" }$ So, statement B is false.

Statement C

$G - L = 9 - 5 = 4 \text{ "" "" "" }$So, statement C is false

Statement D

$a - b = 7 - 5 = 2 \text{ "" "" "" }$ So, statement D is true.