The sumof three numbers in AP is 3 and their product is -35. Find the numbers
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given,
Sn= 3
S3= n/2[2a+(n-1)d]
3=3/2[2a+2d]
6=6a+6d
a+d=6....
a=6-d
....
product = (a-d)(a)(a+d)
-35=(a²-ad)6
-35=6a²-6ad
-35=6(36+d²-12d)-6(6d-d²)
-35=216+6d²-12d-36d+d²
-35=216+7d²-48d
7d²-48d+251=0.. hence, solve the eqn...
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