\(a_1=2;\ r=-\dfrac{2}{3}\ therefore\ a_n=2\cdot\left(-\dfrac{2}{3}\right)^{n-1}\\\\a_1=2\\\\a_2=2\cdot\left(-\dfrac{2}{3}\right)=-\dfrac{4}{3}\\\\a_3=2\cdot\left(-\dfrac{2}{3}\right)^2=2\cdot\dfrac{4}{9}=\dfrac{8}{9}\\\\a_4=2\cdot\left(-\dfrac{2}{3}\right)^3=2\cdot\left(-\dfrac{8}{27}\right)=-\dfrac{16}{27}\\\\a_5=2\cdot\left(-\dfrac{2}{3}\right)^4=2\cdot\dfrac{16}{81}=\dfrac{32}{81}\\\\\vdots\)
