A.P. and G.P.
Asked by tangocharliee
| 12th Nov, 2009,
05:22: 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,
11:36: AM
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