1 2 1 4 1 8 Solution
2 1 2 2 2 3.
1 2 1 4 1 8 solution. To make 1 6 dilution take 16 67 ml of the volume of the first solution of concentration a and dilute it to 100 ml to obtain a 6. 1 1 2 1 4 1 8 here first term a 1 common ratio r 1 2 1 1 2 r 1 number of terms to be added n 9 therefore s n a 1 r n 1 r. Again take 50 ml of the a 2 solution and dilute it to 100 ml to obtain a 4 solution. How do you express this series in summation notation.
And so you have the solutions a a 2 a 4 a 6 a 8. There are many different expressions that can be shown to be equivalent to the problem such as the form. Simple and best practice solution for 1 4 x 2 1 2 3x 1 1 8 x 1 equation. Then take 50 ml of the a 4 solution and dilute it to 100 ml to obtain a 8 solution.
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