Q. 314.3( 17 Votes )

# If pth

Answer :

Let a be the first term and d be the common difference of the given A.P.

pth term of AP = [ a + (p - 1)d] = a …………(i)

qth term of AP = [ a + (q - 1)d] = b …………(ii)

rth term of AP = [ a + (r - 1)d] = c …………(iii)

To show: (a – b)r + (b – c)p + (c – a)q = 0

Consider the L.H.S. = (a – b)r + (b – c)p + (c – a)q

= r[{a + (p - 1)d} - {a + (q - 1)d}]

+ p[{a + (q - 1)d} - {a + (r - 1)d}]

+ q[{a + (q - 1)d} - {a + (p - 1)d}]

= [r(p - 1- q + 1)d] + [p(q - 1- r + 1)d] + [q(r - 1- p + 1)d]

= d[p - q]r + d[q - r]p + d[r - p]q

= d[pr – qr + pq – pr + qr - pq]

= 0

Therefore, L.H.S. = RH.S.

Hence, proved.

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