Respuesta :
Answer:
r=3 where r is the common ratio
Step-by-step explanation:
The sum of geometric progression for n terms is given by:
[tex]S_{n}=a_{1} ( 1-r^{n})/ 1-r[/tex], where [tex]a_{1}[/tex] is the
first term of the series and r is the common ratio.
Now, according to the question, there are four terms and they form a geometric progression. So sum of four terms is given as [tex]S_{4}= 1+r[/tex] and [tex]a_{1}=1/10[/tex].
Also according to the above formula:
[tex]S_{4}=a_{1} ( 1-r^{4})/ 1-r[/tex]
Using the values as given in the question into the above equation we get:
[tex]1+r =1/10 ( 1-r^{4})/ 1-r[/tex]
[tex](1+r)(1-r)=1/10 ( 1-r^{4})[/tex]
[ Using formula [tex]a^{2} -b^{2} =( a+b)(a-b)[/tex] ]
[tex](1-r^{2})=1/10 ( 1-r^{2})(1+r^{2})[/tex]
[tex]10=(1+r^{2})[/tex]
[tex]10-1=(r^{2})[/tex]
[tex]\sqrt{9}=(r)[/tex]
[tex]r=3[/tex] which is the required answer.
