The sum of the squares of three numbers is 138, while the sum of their products taken two at a time is 131. Their sum is:
A. 20
B. 30
C. 40
D. None of these
Option A. 20 is the right answer
Let suppose The sum of the squares of three numbers is 138 i.e
a2 + b2 + c2 = 138
And, the sum of their products taken two at a time is 131 i.e
(ab + bc + ca) = 131
(ab + bc + ca) = 131
Now, as we know that
(a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca)
(a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca)
By putting the values, we get
(a + b + c)2 = 138 + 2 x 131
(a + b + c)2 = 400
By taking under root on both sides, we get
√(a + b + c)2 = √400
(a + b + c) = 20
Hence, if the sum of the squares of three numbers is 138, while the sum of their products taken two at a time is 131. Thenm their sum is 20 as proved above.
which is our right answer:)