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Mathematics 14 Online
OpenStudy (osanseviero):

In a geometric series, a2=6 a5=48 Which is the first term, which is the constant and write the progression

OpenStudy (osanseviero):

I know that \[an=a1*r ^{n-1}\] What goes next?

OpenStudy (osanseviero):

6=a1*r^(n-1) ?

OpenStudy (osanseviero):

6=a1*r^(2-1), 6=a1*r

OpenStudy (anonymous):

As you said : \[a_n=a_1\times q^{n-1}\] where q is the constant of the serie, so we have : \[a_2=a_1\times q\\a_5=a_1\times q^4\] Is that true ?

OpenStudy (osanseviero):

Yep

OpenStudy (anonymous):

so we get : \[6=a_1\times q~~~~~~(1)\\48=a_1\times q^4~~~(2)\] Now, we can divide the equation (2) over (1) , what should we get ?

OpenStudy (osanseviero):

8=q^3, q=2!

OpenStudy (anonymous):

Good. Now the 1st term can be found easily, can't it ?

OpenStudy (osanseviero):

yepp :D

OpenStudy (osanseviero):

3 :) thanks

OpenStudy (osanseviero):

So an=3*2^n-1 ?

OpenStudy (anonymous):

Yes, it is ! And you are welcome !

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