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 = xy^{2}................. 3

a+(s-1)d = xy^{3}................. 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 = xy**^{2} (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.