Sum of three consecutive terms in an AP is 15 and their product is 0. The sequence is
Both a and b
Since the three terms are in AP,
we can take the terms as a-d, a, a+d.
Sum of these terms= a - d + a +a + d = 3a
We have, 3a = 15
∴ a = 5.
Also given that their product is 0.
⇒(a−d)a(a+d)=a(a2−d2)=0
Since we already have a=5, a2−d2=0
⇒ a2=d2
∴a=±d or d=±a
⇒d=5,−5
For d=5, the sequence is 0,5,10.
For d=-5, the sequence is 10,5,0