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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.

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?

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

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

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

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