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#