Respuesta :
An arithmetic series is a series in which each word tends to increase with adding and subtracting of some k that's opposite to a simple geometric sequence in which each term tends to increase with a continual division / multiplying of k, and the further calculation can be defined as follows:
Given:
[tex]\bold{t_2=5}\\\\\bold{t_8=15}\\\\[/tex]
To find:
[tex]\bold{T_n=?}\\\\\bold{S_{10}=?}[/tex]
Solution:
[tex]\to \bold{t_2=5}\\\\ \bold{a+d=5} \\\\ \bold{a=5-d} ................(i)\\\\ \to \bold{t_8=15}\\\\ \bold{a+ 7d=15}.................(ii)\\\\[/tex]
Putting the equation (i) value into equation (ii):
[tex]\to \bold{5-d+7d=15}\\\\\to \bold{5+6d=15}\\\\\to \bold{6d=15-5}\\\\\to \bold{6d=10}\\\\\to \bold{d=\frac{10}{6}}\\\\\to \bold{d=\frac{5}{3}}\\\\[/tex]
Putting the value of (d) into equation (i):
[tex]\to \bold{a=5-\frac{5}{3}}\\\\\to \bold{a=\frac{15-5}{3}}\\\\\to \bold{a=\frac{10}{3}}\\\\[/tex]
Calculating the [tex]\bold{t_n :}[/tex]
[tex]\to \bold{t_n=a+(n-1)d}\\\\\to \bold{t_n=\frac{10}{3}+(n-1)\frac{5}{3}}\\\\\to \bold{t_n=\frac{10}{3}+\frac{5}{3}n-\frac{5}{3}}\\\\\to \bold{t_n=\frac{10}{3}-\frac{5}{3} +\frac{5}{3}n}\\\\\to \bold{t_n=\frac{10-5}{3} +\frac{5}{3}n}\\\\\to \bold{t_n=\frac{5}{3} +\frac{5}{3}n}\\\\ \to \bold{t_n=\frac{5}{3}(1+n)}\\\\[/tex]
Formula:
[tex]\bold{S_n = \frac{n}{2}[2a + (n - 1)d ] }\\[/tex]
Calculating the [tex]\bold{S_{10}} :[/tex]
[tex]\to \bold{S_{10} = \frac{10}{2}[2\frac{10}{3} + (10- 1)\frac{5}{3} ] }\\\\\to \bold{S_{10} = 5[\frac{20}{3} + (9)\frac{5}{3} ] }\\\\\to \bold{S_{10} = 5[\frac{20}{3} + 9 \times \frac{5}{3} ] }\\\\\to \bold{S_{10} = 5[\frac{20}{3} + 3\times 5 ] }\\\\\to \bold{S_{10} = 5[\frac{20}{3} + 15 ] }\\\\\to \bold{S_{10} = 5[\frac{20+ 45}{3} ] }\\\\\to \bold{S_{10} = 5[\frac{65}{3} ] }\\\\\to \bold{S_{10} = 5 \times \frac{65}{3} }\\\\\to \bold{S_{10} = \frac{325}{3} }\\\\\to \bold{S_{10} =108.33 }[/tex]
Therefore, the final answer is " [tex]\bold{\frac{5}{3}(1+n)\ and\ 108.33}[/tex]"
Learn more:
brainly.com/question/11853909
