Question

Question #a32a3

Answer 1

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`