Find the sum of the infinitely decreasing geometric progression 9; -3; 1

By the condition of the problem, the first term b1 of the geometric progression is 9, the second term b2 is -3, the third term b1 is 1. Find the denominator q of this geometric progression using the ratio b2 = b1 * q.
q = b2 / b1 = -3/9 = -1/3.
Check if the relation b3 = b2 * q is satisfied:
1 = -3 * (- 1/3);
1 = 1.
Thus, this sequence is an exponential progression. Since the modulus of the denominator of this geometric progression is less than 1, this progression is infinitely decreasing, and to find the sum of all its members, we can use the formula for the sum of an infinitely decreasing geometric progression S = b1 / (1-q). In this case:
S = 9 / (1 – (- 1/3)) = 9 / (1+ 1/3) = 9 / (4/3) = 9 * 3/4 ​​= 27/4.

Answer: The sum of this infinitely decreasing geometric progression is 27/4.