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Answer 1 · 2016-09-06T20:27:27

(a) The Common Ratio is #2/3#.

(b) The First Term is #9#.

Let the Infinite Geom. Seq. be #a,ar,ar^2,...,ar^(n-1),...#, where,

#a>0,&, r !in {0,1}#.

Since it is summable, #|r|<1, i.e., -1<,r<,1#.

But, #-1<,r<,0" makes the second term "#-ve#, we must have,

#0<,r<,1#.

Given that, #a+ar=a(1+r)=15...................(1)#.

Further given that, #S_oo=27 rArr a/(1-r)=27..................(2)#.

By #(2), a=27(1-r)#. Using this in #(1)#, we get,

#27(1-r)(1+r)=27(1-r^2)=15 rArr 1- r^2=15/27=5/9#

#rArr r^2=4/9 rArr r=+2/3 "[as, 0<,r<,1]"#.

Then, by #(1), a(1+2/3)=15 rArr a=9#.