Find the denominator of the infinitely decreasing geometric progression, the sum of which

Find the denominator of the infinitely decreasing geometric progression, the sum of which is 2.25 and the second term is 0.5.

Amount:

S = b1 / (1 – q), where q is the denominator, b1 is the first term.

b1 – unknown, b2 = 0.5;

q = b2 / b1 = 0.5 / b1 = 1 / 2b1;

Substitute the value of the sum and the expression for q:

2.25 = b1 / (1 – 1 / 2b1);

2.25 * (1 – 1 / 2b1) = b1;

2.25 – 2.25 / 2b1 = b1;

4.5b1 – 2.25 = 2b1 ^ 2;

2b1 ^ 2 – 4.5b1 + 2.25 = 0;

There are two possible roots – b11 and b12:

b11 = (4.5 – √ (4.5 ^ 2 – 4 * 2 * 2.25) / 4 =

= (4.5 – √ (20.25 – 18)) / 4 =

= (5 – √2.25) / 4 = (4.5 – 1.5) / 4 = 3/4;

b12 = (4.5 + 1.5) / 4 = 6/4 = 1.5;

q 11 = b11 / b2 = (1/2) / (3/4) = 4/6 = 2/3.

Verification:

S = (3/4) / (1 – 4/6) = (3/4) / (2/6) = 18/8 = 2.25;

q12 = b12 / b2 = 0.5 / 1.5 = 1/3.

Verification:

S = 1.5 / (1 – 1/3) = 1.5 / (2/3) = 4.5 / 2 = 2.25.

Answer:

1) b11 = 0.75, q11 = 2/3.

2) b12 = 1.5, q12 = 1/3.