Question

Find the third term of a GP whose common ratio is 3 and the sum whose first seven term is 2186.

Answer 1

Given;
r=3;and S7=2186

S7=a(1 - 3(to the power 7)) / 1-3
2186=a(1 - 2187) / -2
2186 * -2=a(-2186) 
-4372=a(-2186)
-4372/-2186=a
    a=2.
T3=ar^2
    =2*(3)^2
    =18
therefore the third term is : 18