if K^a, K^b, K^c are in G. P then a, b, c are in
Answers
Step-by-step explanation:
(i) If a, b, c are in G.P. show that 1/a, 1/b, 1/c are also in G.P.
From the question it is given that,
a,b,c are in G.P.
We know that, b
2
=ac
We have to show that, 1/a,1/b,1/c are also in G.P.
(1/b)
2
=(1/a)×(1/c)
(1/b
2
)=(1/ac)
By cross multiplication we get, ac=b
2
Hence it is proved that, 1/a,1/b,1/c are in G.P.
(ii) If K is any positive real number and K
a
,K
b
,K
c
are three consecutive terms of a G.P. prove that a,b,c are three consecutive terms of an A.P.
From the question it is given that,
K is any positive real number
K
a
,K
b
,K
c
are three consecutive terms of a G.P.
We know that, b
2
=ac (K
b
)
2
=k
a
×K
c
K
2b
=k
a+c
…[ from a
m
×a
n
=a
m+n
]
By comparing the left-hand side and the right-hand side we get,
2b=a+c
Therefore, a,b,c are three consecutive terms of an A.P.
(iii) If p,q,r are in A.P., show that p
th
,q
th
and r
th
terms of any G.P. are themselves in GP.
From the question, it is given that, p,q,r are in A.P.
So,2p=p+r
We have to show that
p
th
,q
th
and r
th
terms of any G.P. P
th
term in G.P.
=AR
p−1
Q
th
term in G.P.=AR
q−1
R
th
term in G⋅P.=AR
r−1
So, if (AR
q−1
)
2
=AR
p−1
×AR
r−1
A
2
R
2q−2
=A
2
R
p−1+r−1
A
2
R
2q−2
=A
2
R
p+r−2
R
2q−2
=R
p+r−2
By comparing left hand side and right hand side we get,
2p−2=p+r−2
2p=P+r