in a G.P sum of three consecutive terms is 43 and their product is 216 find the three numbers
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Answer:
Step-by-step explanation:Let 3 numbers in GP
r
a
⋅a⋅ar
now,
r
a
⋅a⋅ar=216
a
3
=216
a=6
r
a
⋅a/ar+ar⋅
r
a
=156
r
36
+36r+36=156
r
36
+36r=156−36
r
36
+36r=120
36[
r
1
+r]=120
36[1+
r
r
2
]=120
36r
2
−120r+36=0
90
2
−30r+9=0
3(3r
2
−10r+3)=0
3r
2
−10r+3=0
3r
2
−9r−r+3=0
3r(r−3)−1(r−3)=0
(3r−1)(r−3)=0
3r−1=0
r=
3
1
ifr=
3
1
&a=6
then,a
1
=
r
a
=
3
1
6
=18
a
2
=a=6
a
3
=ar
=6×
3
1
=2
ifr=3&a=6
then,
a
1
=
r
a
=
3
6
=2
a
2
=a=6
a
3
=ar=18
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