Let a be the first and d be the common difference of the given AP. Then,
`T_(p)=a+(p-1)d, T_(q)=a+(q-1)d`,
`T_(r)=a+(r-1)d, T_(s)=a+(s-1)d`.
It is given that `T_(p), T_(q), T_(r)` and `T_(s)` are in GP.
Let the first term of this GP be A and its common ratio be R. Then,
`T_(p)=A rArr a+(p-1)d=A` ...(i)
`T_(q)=AR rArr a+(q-1)d=AR` ...(ii)
`T_(r)=AR^(2) rArr a+(r-1)d=AR^(2)` ...(iii)
and `T_(s)=AR^(3) rArr a+(s-1)d=AR^(3)` ...(iv)
On subtracting (ii) from (i), we get
`(p-q)d=A(1-R)`. ...(v)
On subtracting (iii) from (ii), we get
`(q-r)d=AR(1-R)`. ...(vi)
On substracting (iv) from (iii), we get
`(r-s)d=AR^(2)(1-R)`. ...(vii)
Now, we have
`(q-r)^(2)d^(2)=A^(2)R^(2)(1-R)^(2)` and `(p-q)d xx (r-s)d=A^(2)R^(2)(1-R)^(2)`.
`:. (q-r)^(2)d^(2)=(p-q)d xx (r-s)d`
`rArr (q-r)^(2)=(p-q) xx (r-s)`
`rArr (p-q), (q-r)` and `(r-s)` are in GP.
Hence, `(p-q), (q-r), (r-s)` are in GP.
