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Answer:
Three numbers a,b,ca,b,c in each type of sequence.
\implies\boxed{a,b,c\text{ (A.P)}\iff b=\dfrac{a+c}{2}}⟹
a,b,c (A.P)⟺b=
2
a+c
\implies\boxed{a,b,c\text{ (G.P)}\iff b^{2}=ac}⟹
a,b,c (G.P)⟺b
2
=ac
\implies\boxed{a,b,c\text{ (H.P)}\iff b=\dfrac{2ac}{a+c}}⟹
a,b,c (H.P)⟺b=
a+c
2ac
\red{\bigstar}★ Properties of logarithms.
\implies\log ab=\log a+\log b⟹logab=loga+logb
\implies\log m^{n}=n\log m⟹logm
n
=nlogm
\implies\log_{a}b=\dfrac{\log b}{\log a}⟹log
a
b=
loga
logb
\large\underline{\text{Solution}}
Solution
Since a,b,ca,b,c are in G.P
\implies b^{2}=ac⟹b
2
=ac
Taking log on both sides
\implies\log b^{2}=\log ac⟹logb
2
=logac
By the first and second properties of logarithms
\implies 2\log b=\log a+\log c⟹2logb=loga+logc
So
\implies \log b=\dfrac{\log a+\log c}{2}⟹logb=
2
loga+logc
Hence
\implies\log a,\log b,\log c\text{ (A.P)}⟹loga,logb,logc (A.P)
Dividing by \log xlogx
\implies\dfrac{\log a}{\log x},\dfrac{\log b}{\log x},\dfrac{\log c}{\log x}\text{ (A.P)}⟹
logx
loga
,
logx
logb
,
logx
logc
(A.P)
According to the definition of H.P
\implies\dfrac{\log x}{\log a},\dfrac{\log x}{\log b},\dfrac{\log x}{\log c}\text{ (H.P)}⟹
loga
logx
,
logb
logx
,
logc
logx
(H.P)
By the third property of logarithms
\implies\log_{a}x,\log_{b}x,\log_{c}x\text{ (H.P)}⟹log
a
x,log
b
x,log
c
x (H.P)
Hence proven.
\large\underline{\text{Proof of the concept No.1}}
Proof of the concept No.1
The common difference in an A.P is equal.
\implies d\text{(common difference)}⟹d(common difference)
\implies d=a-b,d=b-c⟹d=a−b,d=b−c
\implies a-b=b-c⟹a−b=b−c
\implies 2b=a+c⟹2b=a+c
\implies b=\dfrac{a+c}{2}\text{ (Arithmetic mean)}⟹b=
2
a+c
(Arithmetic mean)
The common ratio in a G.P is equal.
\implies r\text{(common ratio)}⟹r(common ratio)
\implies r=\dfrac{a}{b},r=\dfrac{b}{c}⟹r=
b
a
,r=
c
b
\implies \dfrac{a}{b}=\dfrac{b}{c}⟹
b
a
=
c
b
\implies b^{2}=ac\text{ (Geometric mean)}⟹b
2
=ac (Geometric mean)
An H.P consists of reciprocals of an A.P.
\implies\dfrac{1}{a},\dfrac{1}{b},\dfrac{1}{c}\text{ (A.P)}⟹
a
1
,
b
1
,
c
1
(A.P)
\implies \dfrac{2}{b}=\dfrac{1}{a}+\dfrac{1}{c}⟹
b
2
=
a
1
+
c
1
\implies \dfrac{1}{b}=\dfrac{a+c}{2ac}⟹
b
1
=
2ac
a+c
\implies b=\dfrac{2ac}{a+c}\text{ (Harmonic mean)}⟹b=
a+c
2ac
(Harmonic mean)
\large\underline{\text{Additional information}}
Additional information
We can notice that each result is the A.M, G.M, and H.M. Thereby, we can use the concept of means for finding a term between the two.
Answer:
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