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.
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Furthermore, this property holds true for all increasing AP's whose smallest term is 3!
I find that really interesting :)
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Thanks! :)
This is wrong as you wrote " 3! " which is not equal to 3
Let the three numbers be: ( a − d ) , ( a ) , ( a + d ) with common difference d . Given:
( a − d ) + ( a ) + ( a + d ) 0 0 = ( a − d ) × a = ( a − d ) × a − 3 a = a × ( a − d − 3 ) ⟹ a = d + 3 [ ∵ a = 0 ]
Smallest number is: ( a − d ) = ( d + 3 − d ) = 3
•First number: a (smallest)
•Second number: b
•Third number: a b − a − b
It follows → ( a b − a − b ) − b = b − a
→ a = 3
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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.