Finding numbers

Algebra Level 1

How many groups of three successive numbers have the property that the square of the middle number is greater by unity than the product of the other two numbers?

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4 8 Infinite 2

3 solutions

Anik Mandal
Jun 8, 2014

If the first of the sought-for numbers is x,then the equation we can set up looks like this,

$(x+1)^{2}$ =x(x+2)+1

Removing brackets we get the equation,

$x^{2}$ +2x+1= $x^{2}$ +2x+1

But we can't find x.This means that we have an identity and it holds true for all values of x.And so any three numbers taken in succession possess the property required.Indeed take any three numbers at random

2,3,4

and we see that

$3^{2}$ -2*4=1

Hence the answer is infinite.

I did it like this:

If a, b, c are 3 numbers

Arithmetic mean -

(a+c)/2 = b

Now, [(a+c)/2]^2 = 1 + ac

a^2 +b^2 + 2ab = 4 + 4ac

Hence, (a-b)^2 = 4

a-b = 2

Therefore, the answer is infinite.

Kartik Sharma - 7 years ago

Edwin Gray
Mar 28, 2019

Let the consecutive integers be n - 1, n, n +1. Then the square of the middle term minus the product of the other 2 terms is n^2 - (n^2 - 1) = 1.

Ram Meena
Jun 14, 2014

let select a number N then other numbers (N-1) and (N+1) their product(n+1)(n-1) = n^2 -1 always less than square of middle no. by 1 :)

Finding numbers

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