Question

What is the 8th term of the sequence below?2240, 1120, 560, 280, ..8.25

Answer 1

Given a sequence as shown below

`2240,\text{ 1120 , 560 , 280, }\ldots\ldots\ldots\ldots.\text{ 8.75}`
`\begin{gathered} Firstterm(T_1)\text{= 2240} \\ Secondterm(T_2)\text{= 1120} \\ Thirdterm(T_3)\text{= 560} \end{gathered}`

From the observation

`(T_2)/(T_1)\text{ =}(T_3)/(T_2)=\text{ }(1120)/(2240)\text{ =}(560)/(1120)\text{ = }\frac{1\text{ }}{2}`

For the 8th term

`\begin{gathered} 2240\text{ x (}(1)/(2))^(8-1) \\ 2240\text{ x (}(1)/(2))^7 \\ 2240\text{ x }(1)/(128) \\ 17.5 \end{gathered}`

Hence the 8th term of the sequence = 17.5