Find three consecutive numbers such that the sum of one-fourth the first and one-fifth the second is five less than one-seventh the third.
-10, -11 and -12
-12, -13 and -14
-14, -15 and -16

Question

Find three consecutive numbers such that the sum of one-fourth the first and one-fifth the second is five less than one-seventh the third.
-10, -11 and -12
-12, -13 and -14
-14, -15 and -16

anonymous · Answer

$$1/4a+1/5b=1/3c-5$$
a=a
b=a+1
c=a+1+1 

If i read right

anonymous · Answer

so would it be the 2nd one?

zepp · Answer

Okay, let the first term be a, second term be a+1, and third term be a+2, since they are consecutive
Then, write our equation. $\large \frac{1}{4}a+\frac{1}{5}(a+1)=\frac{1}{7}(a+2)$

anonymous · Answer

What zepp said. my 1/3 should be 1/7

zepp · Answer

Oh sorry, $\large \frac{1}{4}a+\frac{1}{5}(a+1)=\frac{1}{7}(a+2)-5$
Should be like that, I forgot the 5 less.

zepp · Answer

And solve for a.