+32 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.
+368 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
+32 Yes this helped a lot I think I understand how this problem works now Thank you soo much!
