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The three-digit numbers C99, A6A, BC7, and B91 form an arithmetic sequence in this order. What is the value of A2 + B2 + C2 ?

 

I need a step by step or an explanation of how to do this problem, please. I don't understand how to solve this. Someone help.

 Feb 17, 2019
 #1
avatar+192 
+3

We can say

C99+d=A6A

A6A+d=BC7

BC7+d=B91
First equation means d = A6A-C99

Second gives us d=BC7-A6A

BC7-A6A=A6A-C99

2(A6A)=BC7+C99

From this we see that the ones digit on the right side must be 6 (9+7)

So digit A must be 8 or 3.

From the third equation we can see that d must have a ones digit of 4

Therefore A must be 3.

Knowing this, C99+d=363

We can guess that C is one or two by saying d is positive(Big assumption-_-)

if c is two then d=64

363+64=B27

B must be four then if the assumption is correct

Then 427+64=491

WE ARE CORRECT!!!!!!

A=3

B=4

C=2

32+42+22=96

Hope this helps

 Feb 18, 2019
 #2
avatar+32 
+1

Yes this helped a lot I think I understand how this problem works now Thank you soo much!

IDontKnowMath  Feb 18, 2019
 #3
avatar+192 
+4

Your're Welcome!

 Feb 18, 2019

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