Question

plz slove this question ​

Answer 1

Answer:

$\bf t_n=\frac{5^{n-4}}{3^{n-3}} \\\\t_1=a=\frac{5^{1-4}}{3^{1-3}} =\frac{5^{-3}}{3^{-2}} =\frac{9}{125} \\\\t_2=a=\frac{5^{2-4}}{3^{2-3}} =\frac{5^{-2}}{3^{-1}} =\frac{3}{25}\\\\t_3=a=\frac{5^{3-4}}{3^{3-3}} =\frac{5^{-1}}{3^{0}} =\frac15\\\\\frac{t_2}{t_1} =\frac{\frac{3}{25} }{\frac{9}{125} } =\frac{3}{25} * \frac{125}{9}=5/3} \\\\Thus\ a=t_1=\frac{9}{125} \ and \ r=5/3$

Answer 2

Step-by-step explanation:

it's not the same

but you can get the idea from it