If pth,qth,rth,sth,terms of an A.P. are in G.P., then show that(p-q),(q-r),(r-s) are also in G.P.
Asked by | 3rd Jul, 2011, 03:37: PM
Expert Answer:
Let 1st term be a+ t and difference be t
We need to prove
(p-q)/(q-r) = (q-r)/(r-s)
So pth term = a + pt
Qth terem= a + qt
Rth term = a + rt
And sth term = a + st
Thery are in GP then
(a+st)/(a+rt) = (a+rt)/(a+qt) = (a+qt)/a+pt) ( common ratio)
Now if (a/b) = (c/d) then both = (a-c)/(b-d)
Using this we get
From (a+st)/(a+rt) = (a+rt)/(a+qt) both = (st-rt)/(rt-qt) = (s-r)/(r-q)
From (a+rt)/(a+qt) = (a+qt)/a+pt) both = (r-q)/(q-p) same as aboth
Hence
(s-r)/(r-q) = (r-q)/(q-p)
Or
(r-s)/(q-r) = (q-r)/(p-q) multiplying numerator and denominator of both sides by -1
Hence Proved
We need to prove
(p-q)/(q-r) = (q-r)/(r-s)
So pth term = a + pt
Qth terem= a + qt
Rth term = a + rt
And sth term = a + st
Thery are in GP then
(a+st)/(a+rt) = (a+rt)/(a+qt) = (a+qt)/a+pt) ( common ratio)
Now if (a/b) = (c/d) then both = (a-c)/(b-d)
Using this we get
From (a+st)/(a+rt) = (a+rt)/(a+qt) both = (st-rt)/(rt-qt) = (s-r)/(r-q)
From (a+rt)/(a+qt) = (a+qt)/a+pt) both = (r-q)/(q-p) same as aboth
Hence
(s-r)/(r-q) = (r-q)/(q-p)
Or
(r-s)/(q-r) = (q-r)/(p-q) multiplying numerator and denominator of both sides by -1
Hence Proved
Answered by | 4th Jul, 2011, 09:33: AM
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