Question

Find the sum of the two middle most terms of the $AP:\frac{1}{3},\frac{2}{3},1,\dots \frac{20}{3}$

Answer 1

$a=\frac{1}{3}$

$d=\frac{2}{3}-\frac{1}{2}=\frac{1}{3}$

$T_n=a+(n-1)d$

$\frac{20}{3}=\frac{1}{3}+(n-1)\frac{1}{3}$

$20=1+(n-1)$

$20=1+n-1$

$20=n$

Middle terms $t=10,t=11$

$T_{10}=\frac{1}{3}+(10-1)\frac{1}{3}$

$T_{10}=\frac{1}{3}+9\left(\frac{1}{3}\right)$

$T_{10}=\frac{10}{3}$

$T_{11}=\frac{1}{3}+(11-1)\frac{1}{3}$

$T_{11}=\frac{1}{3}+10\left(\frac{1}{3}\right)$

$T_{11}=\frac{11}{3}$

$T_{10}+T_{11}=\frac{10}{3}+\frac{11}{3}=\frac{21}{3}=7$