Answer :

Given,


pth, qth rth and sth terms of an AP are in GP .


Firstly we should find out pth, qth, rth and sth terms


Let a is the first term and d is the common difference of an AP


so, pth term = a + (p – 1)d


qth term = a + (q – 1)d


rth term = a + (r – 1)d


sth term = a + (s – 1)d


[a + (p – 1)d ], [a + (q – 1)d ], [a + (r – 1)d ], [a + (s – 1)d ] are in GP


so, Let first term of GP be α and common ratio is β


Then, [a + (p – 1)d ] = α


[a + (q – 1)d ] = αβ


[a + (r – 1)d ] = αβ2


[a + (s – 1)d ] = αβ3


now, here, it is clear that α, αβ, αβ2, αβ3 are in GP


NOTE: Using property of GP,we know that if a common term is multiplied with each number in a GP,series itself remains a GP


α(1 – β), αβ(1 – β), αβ2(1 – β) are in GP


Where the first term is α(1 – β), and the common ratio is β


so, α(1 – β) = [a + (p – 1)d] – [a + (q – 1)d ] = (p – q)


α(1 – β) = (p – q) ...... (1)


Similarly, αβ(1 – β) = αβ – αβ2 = [a + (q – 1)d ] – [a + (r – 1)d] = (q – r)


αβ(1 – β) = (q – r) …… (2)


And αβ2(1 – β) = αβ2 – αβ3 = [α + (r – 1)d] – [α + (s – 1)d] = (r – s)


αβ2(1 – β) = (r – s) …… (3)


From the above explanation, we got α(1 – β), αβ(1 – β), αβ2(1 – β) are in GP


From equations (1), (2) and (3),


(p – q), (q – r), (r – s) are in GP .


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