x and y are the first and last terms of a sequence of consecutive posi

Question

x and y are the first and last terms of a sequence of consecutive positive odd integers with a range of 22. If y^2-x^2 equals the square of the arithmetic mean of the sequence, what is the sum of x and y?

A. 76
B. 80
C. 84
D. 88
E. 92

A. 76
B. 80
C. 84
D. 88
E. 92

ScottTargetTestPrep · Accepted Answer

Since x and y are the first and last terms of a sequence of consecutive positive odd integers with a range of 22, we see that (y - x) = 22. We also know that the arithmetic mean of this evenly-spaced set is (x + y)/2. Since y^2 - x^2 equals the square of the arithmetic mean of the sequence, we have:

y^2 - x^2 =  ^2

(y + x)(y - x) = (x + y)^2/4

y - x = (x + y)/4

22 = (x + y)/4

88 = x + y

Answer: D

cbh · Answer

y - x = 22
y^2-x^2 = (y - x) ( y + x)
(y - x) ( y + x) = mean^2
(y - x) ( y + x) = ((y + x)/2)^2
lets substitute as y-x = 22 and y + x =2x + 22

22(2x+22) = ((2x + 22)/2)^2
simplify
44x + 484 = (x+ 11)^2
44x + 484 = x^2 + 22x + 121

everything to the right
x^2 - 22x - 363 = 0
(x -33) (x+11)
the sequence is a sequence of positive odd integers so x, first term is 33
then y 55 and sum of x + y = 88

septwibowo · Answer

- We are given set {x,x+2,x+4,....,x+22} which  x+22=y  (from the question).
- The number inside the set is 12 (you can write it down one by one from x to x+22).
- So, the mean of the set must be the median, which is the average of 6th and 7th number, or  (x+10)  and  (x+12) .

Thus, we can calculate :
-  y^2-x^2=(mean)^2 
- Substitute  y  with  x+22  and mean with the median, we get :  (x+22)^2-x^2=(\frac{(x+10)+(x+12)}{2})^2 
- Solve the equation, we get two roots of x,  x=-11  and  x=33 . Since the question tells that x is a positive, so we can choose 33.
- If  33=x , then  y=x+22=55 . So  x+y=88 .

D.

AbhimanyuDhar · Answer

x and y are the first and last terms of a sequence of consecutive positive odd integers with a range of 22. If y^2-x^2 equals the square of the arithmetic mean of the sequence, what is the sum of x and y?

A. 76
B. 80
C. 84
D. 88
E. 92

Given the range of the values of x and y is 22. 
that means y-x=22
so 
y^2 - x^2= {( x+y)/2}^2

because arithmetic mean of even spaced set is also equal to the mean of first and last term of the set.

so 
(y-x)(y+x)= {(x+y)(x+y)}/4
so
x+y= 4(y-x)
x+y=88

pappal · Answer

given y-x=22 ,range
(y^2- x^2)= ^2 (mean of an AP)
4(y+x)(y-x)=(x+y)^2
4(y-x)=(x+y)=4*22=88

Adit_ · Answer

Very straightforward arithmetic.

(y+x)(y-x) = (Mean)^2

(y+x)(22) = Mean^2

Now Mean = y+x/2

Thus, (y+x)(22) = (y+x/2)^2

22 = y+x/4
y+x = 88.

Answer:  Option D

x and y are the first and last terms of a sequence of consecutive posi

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x and y are the first and last terms of a sequence of consecutive posi

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