If 3 geometric means are inserted between 162 and 2, what is the fourth term of the resulting sequence?
Answers
ANSWER
Let G
1
,G
2
,G
3
be three geometric mean inserted between 2
4
1
&
9
4
then
4
9
,G
1
,G
2
,G
3
,
9
4
is a G.P with a=
4
9
&a
5
=
9
4
a
5
=
9
4
⇒ar
4
=
9
4
⇒
4
9
r
4
=
9
4
⇒r
4
=(
9
4
)
2
⇒r
4
=(
3
2
)
4
⇒r=
3
2
<1
So, G
1
& G
3
are the greatest and least mean respectively
Now ,
G
3
G
1
=
a
4
a
2
=
ar
3
ar
=
r
2
1
=(
2
3
)
2
Given: 3 geometric means are inserted between 162 and 2,
To Find: the fourth term of the resulting sequence
Solution:
3 geometric means are inserted between 162 and 2,
Hence total terms will be 5
a , ar , ar² , ar³ , ar⁴
a = 162
ar⁴ = 2
162r⁴ = 2
=> r⁴ = 2/162
=> r⁴ = 1/81
=> r = ± 1/3
Using r = 1/3
a , ar , ar² , ar³ , ar⁴
162 , 54 , 18 , 6 , 2
Hence 4th term of sequence is 6
Using r = -1/3
a , ar , ar² , ar³ , ar⁴
162 , -54 , 18 , -6 , 2
Hence 4th term of sequence is -6
So 4th term of sequence can be 6 or - 6
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