# AP AND GP QUESTIONS AND ANSWERS

Question 1 :

If the roots of the equation (q − r)x2 + (r − p)x + p − q = 0 are equal, then show that p, q and r are in AP.

Solution :

If the roots are real, then b2 - 4ac  =  0

here, a  =  q - r, b = r - p and c = p - q

(r - p)2  - 4 (q - r) (p - q)

r2 + p2 - 2rp - 4(pq - q2 - rp + rq)  =  0

r2 + p2 - 2rp - 4pq + 4q2 + 4rp - 4rq  =  0

r2 + p2 + 2rp - 4pq + 4q2 - 4rq  =  0

r2 + p2 + 4q+ 2rp - 4pq  - 4rq  =  0

r2 + p2 + (-2q)+ 2rp - 4pq  - 4rq  =  0

(r + p - 2q)2  =  0

r + p  =  2q

Hence p, q and r in A.P

Question 2 :

If a, b, c are respectively the pth, qth and rth terms of a GP, show that (q − r) log a + (r − p) log b + (p − q) log c = 0.

Solution :

t =  ARn-1

 pth term :tp  =  ARp-1  =  a qth term :tq  =  ARq-1  =  b rth term :tr  =  ARr-1  =  c

log a  =  log A + (p-1) log R  -----(1)

log b  =  log A + (q-1) log R  -----(2)

log c  =  log A + (r-1) log R  -----(3)

By applying the values of log a, log b and log c, we get

(q-r) log a + (r-p) log b + (p-q) log c

=  (q-r) (log A + (p-1) log R) + (r-p) (log A + (q-1) log R) + (p-q) (log A + (p-1) log R)

=  log A [(q - r) + (r - p) + (p - q)] + [(q-r) (p-1) + (r-p) (q-1) + (p-q) (p-1)] log R

=  log A (0) + (0 log R)

(q-r) log a + (r-p) log b + (p-q) log c =  0

Hence proved.

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