1.

Three consecutive positive integers are taken such that the sum of the square of the first and the product of the other two is 154. Find the integers.​

Answer»

x,  x+1  and x+2

According to question

x^2+(x+1)(x+2)=154\\\\x^2+x(x+2)+1(x+2)=154\\\\x^2+x^2+2x+x+2=154\\\\2x^2+2x+x+2=154\\\\2x^2+3x+2=154\\\\2x^2+3x+2-154=0\\\\2x^2+3x-152=0\\\\2x^2+19x-16x-152=0\\\\x(2x+19)-8(2x+19)=0\\\\(2x+19)\:(x-8)\\\\\\2x+19=0\\\\2x=-19\\\\x=-\dfrac{19}{2}\:\bigg(x\neq\dfrac{-19}{2}\bigg)\\\\\\x-8=0\\\\x=8\:\bigg(x=8\bigg)

Here we get 'x' = 8

Now we put the value of 'x' we get,

First consecutive = x = 8

Second consecutive = x + 1 = 8 + 1 = 9

Third consecutive = x + 2 = 8 + 2 = 10

Therefore THREE consecutive POSITIVE INTEGERS are 8, 9 and 10


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