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
A) G-L = 4
B) b-a = 2
C) G-L = 7
D) a-b = 2
1 Answer
The correct choices are A and D.
Explanation:
The sum of the first
#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
For example, 11 is the
#(11+1)/2 = 6# th odd number
Using these two facts, we can rewrite our expression as:
#(1+3+5+cdots+a) + (1+3+5+cdots+b) = (1+3+5+cdots+c)#
#"sum of first "(a+1)/2" odd numbers" + "sum of first "(b+1)/2" odd numbers" = "sum of first "(c+1)/2" odd numbers"#
#((a+1)/2)^2+((b+1)/2)^2 = ((c+1)/2)^2#
Doing a little bit of algebra to this, we can simplify the expression:
#(a+1)^2/4+(b+1)^2/4 = (c+1)^2/4#
#(a+1)^2+(b+1)^2 = (c+1)^2#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Since
We also know that
#(a+1)+(b+1)+(c+1)#
#= 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
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 " "" "b=5 " "" "c=9#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Using these numbers, we can figure out which of the 4 statements are true.
#G = max{a,b,c} = max{7,5,9} = 9#
#L = min{a,b,c} = min{7,5,9} = 5#
Statement A
#G - L = 9 - 5 = 4" "" "" "" "# So, statement A is true.
Statement B
#b - a = 5 - 7 = -2" "" "" "# So, statement B is false.
Statement C
#G - L = 9 - 5 = 4" "" "" "" "# So, statement C is false
Statement D
#a - b = 7 - 5 = 2" "" "" "" "# So, statement D is true.
