find sum of 3 consitive terms of an a,p and find their product of -120find three numvers
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Answer:
Let the number are x−y,x,x+y
Sum =−3
⇒x−y+3x+x+y=−3
⇒3x=−3
⇒x=−1
Now product =8
⇒(x−y)(x)(x+y)=8
Substituting x=−1
we get (−1−y)(−1)(−1+y)=8
(y
2
−1)=8
⇒y=±3
The no : are −4,−1,2 or 2,−1,−
Answered by
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Let the 3 consecutive numbers be a-d, a, and a+d respectively.
where a = first term of the AP and d=is their common difference.
Now given that sun of the three numbers is 6.
So we get =>
a + a-d + a + d = 6
=> 3a = 6
=>a = 2
Also, a(a-d)(a+d) = -9
Putting a = 2 we get =>
2(2-d)(2+d) = -9
By using (a+b)(a-b) = a²-b² we get =>
2(2² - d²) = -9
=> 2(4- d²) = -9
=> 8 -2d² = -17
=>-2d² = -25
=>d² = 25/2
=> d =√(25/2)
Hence the terms are =>
2 - √(25/2),2 and 2 + √(25/2) respectively.
Also we can write the terms as
2 - (5/√2) , 2 and 2 + (5/√2)
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