Question : The sum of three consecutive terms in an arithmetic progression is 6 and their product is −120. Find the three numbers.
Doubt by Pari
Solution :
Let the three consecutive terms of the AP be
a1=a-d,
a2=a,
a3=a+d
a1=a-d,
a2=a,
a3=a+d
ATQ
a1+a2+a3=6
a-d+a+a+d=6
3a=6
a-d+a+a+d=6
3a=6
a=6/3
a=2
Also
a1×a2×a3=-120
(a-d)×a×(a+d)=-120
(2-d)×2×(2+d)=-120 [∵a=2]
2(2²-d²)=-120
2(4-d²)=-120
4-d²=-120/2
4-d²=-60
(2-d)×2×(2+d)=-120 [∵a=2]
2(2²-d²)=-120
2(4-d²)=-120
4-d²=-120/2
4-d²=-60
-d²=-60-4
-d²=-64
d²=64
d=±√64
-d²=-64
d²=64
d=±√64
d=±8
When a=2 & d=8 then
a1=a-d=2-8=-6
a1=a-d=2-8=-6
a2=a=2
a3=a+d=2+8=10
When a=2 & d=-8 then
a1=a-d=2-(-8)=2+8=10
a2=a=2
a3=a+d=2+(-8)=2-8=-6
a3=a+d=2+(-8)=2-8=-6
Hence, the required three numbers are either -6,2,10 or 10,2,-6