If pth ,qth ,rth and sth terms of an A.P. are in G.P. then show that (p-q), (q-r), (r-s) are also in G.P.
 

Hi!
Here is the answer to your question.
 
Let a and d be the first term and common difference of the A.P.
ap, aq,ar, as are in G.P.
a + (q – 1)d = k[a + (p – 1)d], a + (r – 1)d = k [a + (q – 1)d], a + (s – 1)d = k [a + (r – 1)d]
[a + (q – 1)d] – [a + (p – 1)d] = k[a + (p – 1)d] – [a + (p – 1)d],
[a + (r – 1)d] – [a + (q – 1)d] = k[a + (q – 1)d] – [a + (q – 1)d] and
[a + (s – 1)d] – [a + (r – 1)d] = k[a + (r – 1)d] – [a + (r – 1)d]
 
d (p – q) = (k – 1) [a + (p – 1)d],
d(q – r) = (k – 1) [a + (q – 1)d],
d(r – s) = (k – 1) [a + (r – 1)d]
This shows that p – q, q – r and r – s are in G.P. 
 
Cheers!

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sir how did u get the 3rd step?

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vry confusing

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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 

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