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