If sum of three number in A.P is 21 and their product is 231 Find the number
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Answered by
9
>>>>>>>>>>>>>(★)<<<<<<<<<<<<<<<
→let the three terms of AP be ,
(a-d),a,(a+d) then....
Sum:
a-d+a+a+d=21
3a=21
a=7
now,
product=231
(a^2-ad)(a+d)=231
(49-7d)(7+d)=231
343-49d+49d-7d^2=231
343-7d^2=231
7d^2=343-231
7d^2=112
d^2=16
d=4
now,
we have : a=7 and d=4
AP:(a-d)=3
a=7
(a+d)=11
>>>>>>>>>>>>>(★)<<<<<<<<<<<<<<<
→let the three terms of AP be ,
(a-d),a,(a+d) then....
Sum:
a-d+a+a+d=21
3a=21
a=7
now,
product=231
(a^2-ad)(a+d)=231
(49-7d)(7+d)=231
343-49d+49d-7d^2=231
343-7d^2=231
7d^2=343-231
7d^2=112
d^2=16
d=4
now,
we have : a=7 and d=4
AP:(a-d)=3
a=7
(a+d)=11
>>>>>>>>>>>>>(★)<<<<<<<<<<<<<<<
Answered by
1
let the three terms of AP be ,
(a-d),a,(a+d) then....
Sum:
a-d+a+a+d=21
3a=21
a=7
now,
product=231
(a^2-ad)(a+d)=231
(49-7d)(7+d)=231
343-49d+49d-7d^2=231
343-7d^2=231
7d^2=343-231
7d^2=112
d^2=16
d=4
now,
we have : a=7 and d=4
AP:(a-d)=3
a=7
(a+d)=11
>>>>>>>>>>>>>(★)<<<<<<<<<<<<<<<
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