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The sum of an infinite geometric series is 15 and the sum of thesquares of these terms is 45. Find the series.

Answer» Let a be the first term and r be the common ratio.Then, the series is `a+ar+ar^(2)+...oo`.
Now, `(a+ar+ar^(2)+...oo)=15 rArr a/((1-r))=15` ...(i)
and `(a^(2)+a^(2)r^(2)+a^(2)r^(4)+...oo) rArr a^(2)/((1-r^(2)))=45`. ...(ii)
On squaring both sides of (i), we get `a^(2)/((1-r)^(2))=225` ...(iii)
On dividing (iii) by (ii), we get
`a^(2)/((1-r)^(2))xx((1-r^(2)))/a^(2)=225/45 rArr (1+r)/(1-r)=5`
`rArr 1+r=5-5r`
`rArr 6r=4 rArr r=2/3`.
Putting `r=2/3` in (i), we get
`a=15xx(1-2/3)=(15xx1/3)=5`.
Thus, `a=5` and `r=2/3`.
Hence, the required series is `(5+10/3+20/9+40/27+...oo)`.


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