Terms of a G.P.
The first term of geometric progression
b
1
,
b
2
,
b
3
,
.
.
.
.
.
is 1.
Find the minimum value of
(
4
×
b
2
+
5
×
b
3
)
.
You can try my other Sequences And Series problems by clicking here : Part II and here : Part I.
The answer is -0.8.
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2 solutions
Alternative approach,
4 x + 5 x 2 = 5 ( x 2 + 5 4 x + 2 5 4 ) − 5 4
= 5 ( x + 5 2 ) 2 − 5 4
Minimum = − 5 4 = − 0 . 8
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Cool usage of the technique I hate the most!!
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I did the same thing. Quite didn't think of bringing messy algebra over here! Easy Minima ^_^
It is given that b 1=1. Now b 1,b 2,b 3 are in G.P. so b 2^2=b 1*b 3 therefore : b 2^2=b 3 min value of 4b 2+5b 3 = 5b 2^2+4b 2 which is equal to (-D/4a) where D is discriminant and 'a' is the co efficient of b 2^2
It is given that b 1 = 1 . Let the common ratio of the geometric progression be x , then b 2 = x and b 3 = x 2 . Let y = 4 b 2 + 5 b 3 , then:
y = 4 x + 5 x 2 ⇒ d x d y = 4 + 1 0 x
d x d y = 0 ⇒ x = − 0 . 4
Since d x 2 d 2 y > 0 ⇒ y m i n = 4 ( − 0 . 4 ) + 5 ( − 0 . 4 ) 2 = − 0 . 8