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a geometric progression with 28 as its first term has the sum to infinity of 70 . find its common ratio .

 Jan 21, 2015

Best Answer 

 #3
avatar+26400 
+13

a geometric progression with 28 as its first term has the sum to infinity of 70 .

find its common ratio

$$\text{sum: }
\begin{array}{rlccl}
s &=& 28 & + & 28r+ 28r^2 + 28r^3+28r^4+28r^5 +\dots+ \\
r*s &=& && 28r +28r^2+28r^3+28r^4+28r^5+\dots+ \\
\hline
s-r*s &=& 28 &\\
&&&\\
\hline
&&&\\
\end{array}$$

$$r*s = s - 28 \\ \\
r = \dfrac{s - 28}{s} \quad | \quad s = 70\\\\
r = \dfrac{70 - 28}{70} \\\\
r = \frac{42}{70} \\\\
r = \frac{21}{35} \\\\
r = \frac{3}{5} \\\\
r = 0.6$$

 Jan 21, 2015
 #1
avatar+130536 
+13

We have

28 / (1 - r)  = 70    rearrange

28/70  = 1 - r

r = 1 - 28/70  = 14/35

 

Oops.....i made a slight error here....I forgot to subtract the 28/70 from the 1...the correct answer is ..

1 - 28/70  = 42/70 = 21/35 = 3/5  .....now, it matches heureka's solution!!!!

Thanks for calling my attention to that, heureka...!!!  DOH !!!!

 Jan 21, 2015
 #2
avatar+14 
0

thank you very much both of you

 Jan 21, 2015
 #3
avatar+26400 
+13
Best Answer

a geometric progression with 28 as its first term has the sum to infinity of 70 .

find its common ratio

$$\text{sum: }
\begin{array}{rlccl}
s &=& 28 & + & 28r+ 28r^2 + 28r^3+28r^4+28r^5 +\dots+ \\
r*s &=& && 28r +28r^2+28r^3+28r^4+28r^5+\dots+ \\
\hline
s-r*s &=& 28 &\\
&&&\\
\hline
&&&\\
\end{array}$$

$$r*s = s - 28 \\ \\
r = \dfrac{s - 28}{s} \quad | \quad s = 70\\\\
r = \dfrac{70 - 28}{70} \\\\
r = \frac{42}{70} \\\\
r = \frac{21}{35} \\\\
r = \frac{3}{5} \\\\
r = 0.6$$

heureka Jan 21, 2015
 #4
avatar+26400 
+5

Hi CPhill,

ditto!

 Jan 21, 2015

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