Three numbers are in ap 0th their sum is 15 and product is 105 find them
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Let the AP be a-d,a,a+d
So
a-d+a+a+d=15
=>3a=15
a=5
Also
(a-d)(a)(a+d)=105
(a^2-ad)(a+d)=105
(25-5d)(5+d)=105
125+25d-25d-5d^2=105
125-105=5d^2
20=5d^2
d^2=20/5
d^2=4
d=root 4
d=2
So AP:- 5-2,5,5+2
AP:- 3,5,7
So
a-d+a+a+d=15
=>3a=15
a=5
Also
(a-d)(a)(a+d)=105
(a^2-ad)(a+d)=105
(25-5d)(5+d)=105
125+25d-25d-5d^2=105
125-105=5d^2
20=5d^2
d^2=20/5
d^2=4
d=root 4
d=2
So AP:- 5-2,5,5+2
AP:- 3,5,7
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