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Find 5 consecutive negative integers such that the product of the 3rd and 5th is 483.

 Apr 22, 2021

Best Answer 

 #1
avatar+524 
+6

Let's assume the integers are a-2, a-1, a, a+1, a+2.

Now, according to question 

 

\(a(a+2) =483\)

\(a^2+2a-483 = 0\)

\((a+23)(a-21)=0\)

 

∴ a = -23 or 21

Obviously, since a is negative,

⇒ a = -23

 

∴ 5 consecutive negative no.s are = -25, -24, -23, -22 and -21. 

 

~ Hope this helps :) 

 Apr 22, 2021
 #1
avatar+524 
+6
Best Answer

Let's assume the integers are a-2, a-1, a, a+1, a+2.

Now, according to question 

 

\(a(a+2) =483\)

\(a^2+2a-483 = 0\)

\((a+23)(a-21)=0\)

 

∴ a = -23 or 21

Obviously, since a is negative,

⇒ a = -23

 

∴ 5 consecutive negative no.s are = -25, -24, -23, -22 and -21. 

 

~ Hope this helps :) 

amygdaleon305 Apr 22, 2021

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