+0  
 
0
319
3
avatar

a5 = 7, a8 = 56a5 = 7, a8 = 56. Find a11a11.

Guest May 7, 2017
 #1
avatar
+1

That's impossible, from a5 = 7 we can conclude that a = 1.4.

 

1.4 * 8 =/= 56.

 

If a = 1.4 then a11a11 = 237.16.

Guest May 7, 2017
 #2
avatar+93683 
+2

Hi guest answerer, 

The question is not presented well but I think this is what is meant.

 

\(a_5 = 7, a_8 = 56\qquad Find \;\;a_{11}\)

 

First I want to know if it is an AP or a GP, I assume it is one or the other?

Melody  May 7, 2017
 #3
avatar+20025 
0

a5 = 7, a8 = 56. Find a11

 

\(\begin{array}{|lrcll|} \hline \text{AP:} &\mathbf{ t_z } &\mathbf{=}& \mathbf{ t_x\cdot (\frac{z-y}{x-y}) +t_y\cdot (\frac{x-z}{x-y}) } \\ \text{GP:} &\mathbf{ t_z } &\mathbf{=}& \mathbf{ t_x^{\frac{z-y}{x-y}} \cdot t_y^{\frac{x-z}{x-y}} } \\ \hline \end{array} \)

 

Let

\(\begin{array}{ll} x= 5 & t_x=t_5=7 \\ y=8 & t_y=t_8=56 \\ z=11 & t_z=t_{11}=? \\ \end{array}\)

 

\(\begin{array}{|lrcll|} \hline \text{AP:} &\mathbf{ t_z } &\mathbf{=}& \mathbf{ t_x\cdot (\frac{z-y}{x-y}) +t_y\cdot (\frac{x-z}{x-y}) } \\ &t_{11} &=& 7\cdot (\frac{11-8}{5-8}) + 56\cdot (\frac{5-11}{5-8}) \\ &t_{11} &=& 7\cdot (\frac{3}{-3}) + 56\cdot (\frac{-6}{-3}) \\ &t_{11} &=& 7\cdot (-1) + 56\cdot 2 \\ &t_{11} &=& -7 + 112 \\ & \mathbf{ t_{11} } & \mathbf{=} & \mathbf{105} \\ \hline \end{array} \)

 

\(\begin{array}{|lrcll|} \hline \text{GP:} &\mathbf{ t_z } &\mathbf{=}& \mathbf{ t_x^{\frac{z-y}{x-y}} \cdot t_y^{\frac{x-z}{x-y}} } \\ &t_{11} &=& 7^{\frac{11-8}{5-8}} \cdot 56^{\frac{5-11}{5-8}} \\ &t_{11} &=& 7^{-1} \cdot 56^{2} \\ &t_{11} &=& \frac{3136}{7} \\ & \mathbf{ t_{11} } & \mathbf{=} & \mathbf{448} \\ \hline \end{array} \)

 

laugh

heureka  May 8, 2017

7 Online Users

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.