Question #a32a3

1 Answer
Oct 27, 2017

Let the given infinite Geometric series has first term #a# common ratio #r<1#

So sum of the infinite series #a/(1-r)=243......[1]#

And its sum up to 5th term #S_5=(a(1-r^5))/(1-r)=275......[2]#

Dividing [2] by [1] we get

#1-r^5=275/243#

#=>r^5=1-275/243=-32/243=(-2/3)^5#

#=>r=-2/3#

Inserting the value of #r=-2/3# in [1] we get

#a/(1+2/3)=243#

#=>a=243xx5/3=405#

Hence

1st term #t_1=a=405#

2nd term #t_2=axxr=405xx(-2/3)=-270#

3rd term #t_3=axxr^2=405xx(-2/3)^2=180#

4th term #t_4=ar^3=405xx(-2/3)^3=-120#