Given
AP = 11/5 , 8/5 , 1 , 2/5 , ........
Find
Sum of first 16 terms
Explanation
As we know the formula of sum of an AP
[tex]s_n=\frac{n}{2}[a+(n-1)d][/tex]given a = 11/5
d = second term - first term
d = 8/5-11/5 = -3/5
n = 16
now, put values in formula
[tex]\begin{gathered} s_{16}=\frac{16}{2}[\frac{2\times11}{5}+(16-1)(-\frac{3}{5})] \\ s_{16}=8[\frac{22}{5}+15(-\frac{3}{5})] \\ s_{16}=8[\frac{22}{5}-9] \\ s_{16}=8[-\frac{23}{5}] \\ s_{16}=-\frac{184}{5} \\ \end{gathered}[/tex]Final Answer
Sum of first 16 terms = -184/5
