The sum of the first 5 terms is 11 and the sum of the next 5

by **natasha.ch** » Fri Jun 19, 2009 8:12 am

In a geometric progression, the sum of the first 5 terms is 11 and the sum of the next 5 terms is $\frac{11}{32}$ What is the common ratio?

by **cheahtikwah** » Mon Jun 22, 2009 11:36 pm

Natasha, Try the following:

Find the sum to the first 5 terms; from here you get an expression for a in terms of r

Then find the sum to the first 10 terms, from here sum of first 10 terms - sum of first 5 terms = sum of last 5 terms we can get another expresion of a and r; substitute a in terms of r in the second equation, you only have r left, can solve for r already!

by **ilianko** » Tue Jun 23, 2009 3:27 am

$a_{n}=ar^{n-1}$

$\sum_{k=0}^{n}ar^{k}=\frac{a(r^{n+1}-1)}{r-1}$

$a_{5}=a_{0}r^{5}$

$X=\frac{a_{5}(r^{5+1}-1)}{r-1}=\frac{11}{32}$

$Y=\frac{a_{0}(r^{5+1}-1)}{r-1}=11$

$\frac{X}{Y}=\frac{a_{5}}{a_{0}}=r^{5}=\frac{1}{32}$

$r=\frac{1}{2}$

by **natasha.ch** » Sun Jun 28, 2009 7:32 am

Thanks a lot!!! Love you all :lol:

The sum of the first 5 terms is 11 and the sum of the next 5

The sum of the first 5 terms is 11 and the sum of the next 5

Geometric Progression

Progression problem Solved!!

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