Question

Sum of 3 numbers in AP is 37 & their product is 405. Find the number

Answer 1

LET THE THREE NUMBER BE a-d,a,a+d(we generally consider these)
then according to question
a-d+a+a+d=37
3a=37
a=12.3333....
we know that the terms are in AP
then ,
(a-d)a(a+d)=405
[tex]( a^{2} - d^{2} )a=450 [/tex]
$(151.29- d^{2} )12.3=405$
151.29-d^2=32.9
d^2=151.29-32.9
d^2=118.39
d=10.9(app)
then the numbers are
a-d=12.3-10.9=1.4
a=12.3
a+d=12.3+10.9=23.2
(these all values are approximate)