Question

s3=9
s9=81
an=?
a1=?
r=?​

Answer 1

$a_1+a_2+a_3 = 9 \Rightarrow a_1+(a_1+r)+(a_1+2r) = 9 \Rightarrow 3a_1+3r =9\Rightarrow\\ \Rightarrow a_1+r = 3\\ \\ a_1+a_2+...+a_9 = 81 \Rightarrow a_1+(a_1+r)+...+(a_1+8r) = 81 \Rightarrow \\\Rightarrow 9a_1+(1+2+3+...+8)r = 81 \Rightarrow 9a_1+\dfrac{8\cdot 9}{2}\cdot r =81 \Rightarrow\\ \Rightarrow a_1+4r = 9\\ \\ a_1+r = 3 \\ a_1+4r = 9\\ \\ \Rightarrow 3r = 6 \Rightarrow r = 2 \\ a_1+2 = 3 \Rightarrow a_1 = 1$

$\Rightarrow a_n = a_1+(n-1)r = 1+(n-1)\cdot 2 =1+2n-2 = 2n-1 \\ \\ \Rightarrow \boxed{a_n = 2n-1,\quad a_1 = 1,\quad r = 2}$