Srry clicked post answer by mistake..... here is the answer:-
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pth term of an AP = a + (p-1)d
qth term of the AP = a + (q-1)d
rth term of the AP = a + (r-1)d
sth term of the AP = a + (s-1)d
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Since they are in GP, let x be the first term of GP and y be the common ratio.
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Therefore,
a+(p-1)d = x .................... 1
a+(q-1)d = xy .................. 2
a+(r-1)d = xy2................. 3
a+(s-1)d = xy3................. 4
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Hence,
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Subtracting 1 from 2:-
[a+(q-1)d] - {a+(p-1)d] = xy - x
(q-p)d = x(y-1) .................... 5
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Similarily,
By subtracting,
2 from 3 :- (r-q)d = xy(y-1) ............... 6
3 from 4:- (s-r)d = xy2 (y-1) ............. 7
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Hence
Dividing 6 by 5:-
(r-q) /(q-p)= xy/x = y ............... 8
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Similarily,
Dividing 7 by 6:-
(s-r) / (r-q) = y ............. 9
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From 8 and 9 :-
(r-q) / (q-p) = (s-r) / (r-q)
multiplying by (-1) throughout:-
(q-r) / (p-q) = (r-s) / (q-r)
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Hence (p-q) , (q-r) and (r-s) are in GP.