Let `a_1,a_2 ,a_3` …..be in A.P. and `a_p, a_q , a_r` be in G.P. Then `a_q : a_p` is equal to :

Options

A. `(q-p)/(r-p)`
B. `(r-q)/(q-p)`
C. `(q-p)/(r-q)`
D. none of thses

Answer

C

Explanation

It is given that `a_(p),a_(q),a_(r)` are in G.P.
`rArr" "(a_(q))/(q_(p))=(a_(r))/(a_(q))`
`rArr" "(a_(1)+(q-1)d)/(a_(1)+(p-1)d)=(a_(1)+(r-1)d)/(a_(1)+(q-1)d)`
`rArr" "((q-p)d)/(a_(1)+(p-1)d)=((r-q)d)/(a_(1)+(q-1)d)" "[("Subtracting 1 from"),("both sides")]`
`rArr" "(q-p)/(a_(p))=((r-q))/(a_(q))rArr(a_(p))/(a_(q))=(q-p)/(r-q)`