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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 $(1), a(1+2/3)=15 rArr a=9$ .