sequence and series

Asked by tangocharliee | 15th Nov, 2009, 03:20: PM

Expert Answer:

Let a be the I term and d be the common difference of AP

So pth  , qth  ,rth     , and sth  terms of an A.P  will be a+(p-1)d , a+(q-1)d,a+(r-1)d ,a+(s-1)d

Now these terms are in GP so

a+(p-1)d /a+(q-1)d= a+(q-1)d/a+(r-1)d =a+(r-1)d /a+(s-1)d  = R

Applying compnendo and dividendo

R= (q-r)/(p-q)=(r-s)/(q-r) so (q-r)2=(r-s)(p-q) 

Therefore,(p-q) ,(q-r) ,(r-s) are  in G.P.


Answered by  | 30th Nov, 2009, 11:40: AM

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