1. The problem states we have a geometric progression (GP) starting with 2, 6, 18, 54, ... and we want to find how many terms are added to get a sum of 19682. 2. Identify the first term $a = 2$ and the common ratio $r = \frac{6}{2} = 3$. 3. The sum of the first $n$ terms of a GP is given by the formula: $$ S_n = a \frac{r^n - 1}{r - 1} $$ 4. Substitute the known values: $$ 19682 = 2 \frac{3^n - 1}{3 - 1} = 2 \frac{3^n - 1}{2} $$ 5. Simplify the equation: $$ 19682 = 3^n - 1 $$ 6. Add 1 to both sides: $$ 19682 + 1 = 3^n $$ 7. So, $$ 19683 = 3^n $$ 8. Recognize that $19683 = 3^9$ because $3^9 = 19683$. 9. Therefore, $$ n = 9 $$ 10. The number of terms added to get the sum of 19682 is 9.