Maithreyi

• Apr 2nd, 2006

 

Answer is 8Take a,ar,a/r as the terms in G.P.a+ar+a/r=38a.ar.a/r=1728a^3=1728a=12Therefore 12+12r+12/r=38solve it u can get r=2/3 and 3/212/r will be the smallest term so, 12/(3/2)=8

Harsh

• Mar 11th, 2012

 

So lets pretend that were told there were, say, 3 terms.

Then we know the terms have the form a, ar and ar^2 for some a and r. We also have
a + ar + ar^2 = 38 and a(ar)(ar^2) = 1728
So we have two simultaneous equations in a and r. Simplifying them gives

a^3 r^3 = 1728
ar = cube root 1728 = 12. So set r = 12/a and put this into the first equation, we get
a + 12 + 144/a = 38
or a^2 - 26a + 144 = 0
so (a - 8) (a - 18) = 0
So we either have a=8, r=12/8 = 3/2 and the G.P. is 8, 12, 18; or a=18, r=12/18 = 2/3 and the G.P. is 18, 12, 8. In either case the smallest term is 8.