The 1st term of a geometric progression exceeds the second term by 1 and the sum of the first three terms is $13/5$ .If there are positive as well as negative terms in the geometric progression, find the sum of the first five terms. ?

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Answer 1 · 2018-07-09T16:42:42

$41$

$a_1(r)^0=a_1(r)^1+1$

$a_1-a_1r=1$

$a_1(1-r)=1$

$a_1=1/(1-r)$

So we have:
$a_1(r)^0+a_1(r)^1+a_1(r)^2=13/5$

$1/(1-r)+r/(1-r)+r^2/(1-r)=13/5$

$(r^2+r+1)/(1-r)=13/5$

$5r^2+5r+5=13-13r$

$5r^2+18r-8=0$

$5r^2+20r-2r-8=0$

$5r(r+4)-2(r+4)=0$

$r=-4 or 2/5$

We know that there are positive and negative terms in this sequence so the ratio has to be negative, so therefore: $r=-4$

So therefore: $a_1=1/5$

Formula for geometric finite sum: $a_1((1-r^n)/(1-r))$

$S_5= 1/5((1-(-4)^5)/(1-(-4)))=41$

The 1st term of a geometric progression exceeds the second term by 1 and the sum of the first three terms is 13/513/5.If there are positive as well as negative terms in the geometric progression, find the sum of the first five terms. ?