Thrilling Trio

Algebra Level 2

There are three non-zero numbers forming an increasing arithmetic progression such that their sum equals the product of the two smaller numbers amongst them. Find the smallest number of those numbers.


The answer is 3.

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3 solutions

Nihar Mahajan
Nov 9, 2016

Let those numbers forming AP be (a-r),a,(a+r) . According to the condition in the problem a(a-r)=(a-r)+a+(a+r) ==> a(a-r)=3a . But a is not 0, so a-r=3 which will always be present in all trios as the smallest number.

Furthermore, this property holds true for all increasing AP's whose smallest term is 3!

I find that really interesting :)

Calvin Lin Staff - 4 years, 7 months ago

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Thanks! :)

Nihar Mahajan - 4 years, 7 months ago

This is wrong as you wrote " 3! " which is not equal to 3

Arnav kumar Sinha - 1 year, 10 months ago

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Unexpected factorial!!

Mahdi Raza - 1 year ago
Mahdi Raza
Jun 4, 2020

Let the three numbers be: ( a d ) , ( a ) , ( a + d ) (a-d), (a), (a+d) with common difference d d . Given:

( a d ) + ( a ) + ( a + d ) = ( a d ) × a 0 = ( a d ) × a 3 a 0 = a × ( a d 3 ) \begin{aligned} (a-d) + (a) + (a+d) &= (a-d) \times a \\ 0 &= (a-d) \times a - 3a \\ 0 &= a \times (a-d-3) \end{aligned} a = d + 3 [ a 0 ] \implies {\color{#D61F06}{a = d+3}} \quad \quad [\because a \ne 0 ]

Smallest number is: ( a d ) = ( d + 3 d ) = 3 ({\color{#D61F06}{a}}-d) = ({\color{#D61F06}{d+3}} - d) = \boxed{3}

•First number: a a (smallest)

•Second number: b b

•Third number: a b a b ab-a-b

It follows ( a b a b ) b = b a \rightarrow (ab-a-b)-b = b-a

a = 3 \rightarrow \boxed{a=3}

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