If pth, qth , rth and sth terms of an AP are in GP then show that (p-q), (q-r), (r-s) are also in GP

Answer

N/a

Explanation

Let first term and common difference of A.P. be a and d respectively.
`therefore" In A.P."`
`T_(p)=a+(p-1)d`
`T_(q)=a+(q-1)d`
`T_(r)=a+(r-1)d`
`T_(s)=a+(s-1)d`
`"Now, "T_(p),T_(q),T_(r),T_(s)" are in G.P."`
`rArr" "(T_(q))/(T_(p))=(T_(r))/(T_(q))=(T_(s))/(T_(r))`
`rArr" "(T_(p))/(T_(q))=(T_(q))/(T_(r))=(T_(r))/(T_(s))`
`rArr" "(T_(p)-T_(q))/(T_(q)-T_(r))=(T_(q)-T_(r))/(T_(r)-T_(s))`
(From law of proportional)
`rArr" "(a+(p-1)d-a-(q-1)d)/(a+(q-1)d-a-(r-1)d)`
`(a+(q-1)d-a-(r-1)d)/(a+(r-1)d-a-(s-1)d)`
`rArr" "((p-q)d)/((q-r)d)=((q-r)d)/((r-s)d)`
`rArr" "(p-q)/(q-r)=(q-r)/(r-s)`
`rArr" "(q-r)^(2)=(p-q)(r-s)`
`rArr" "p-q,q-r,r-s" are in G.P."`