If the #p^(th), q^(th) and r^(th)# terms of an arithmetic progression are #P, Q, R# respectively, then #P(q-r) + Q(r-p) + R(p-q)# is equal to?

1 Answer
Nov 29, 2016

Answer:

#P(q-r)+Q(r-p)+R(p-q)=0#

Explanation:

The #n^"th"# term #a_n# of an arithmetic progression with common difference #d# and initial term #a_1# is given by

#a_n = a_1+(n-1)d#

thus, we can write

#P = a_1+(p-1)d#
#Q = a_1+(q-1)d#
#R = a_1+(r-1)d#

If we subtract #R# from #Q#, we find

#Q-R = a_1+(q-1)d - (a_1+(r-1)d) = (q-r)d#

Thus #q-r = (Q-R)/d#

Similarly, we have #r-p = (R-P)/d# and #p-q = (P-Q)/d#

Substituting these into the given expression, we get

#P(q-r)+Q(r-p)+R(p-q)#

#=(P(Q-R))/d+(Q(R-P))/d+(R(P-Q))/d#

#=(PQ-PR + QR-PQ + PR-QR)/d#

#=(PQ-PQ+PR-PR+QR-QR)/d#

#=0/d#

#=0#