Question 1064171
 A sequence of three real numbers forms an arithmetic progression with a first term of 9. If 2 is added to the second term and 20 is added to the third term, the three resulting numbers form a geometric progression. What is the smallest possible value for the third term of the geometric progression?
<pre>Adding 2 to the 2nd term of the AP makes the GP’s 2nd term: 9 + d + 2, or 11 + d 
Adding 20 to the 3rd term of the AP makes the GP’s 3rd term: 9 + 2d + 20, or 29 + 2d

Since this is now a GP, the common ratio is calculated as: {{{matrix(1,2, 2^(nd), term)/matrix(1,2, 1^(st), term) = matrix(1,2, 3^(rd), term)/matrix(1,2, 2^(nd), term)}}} =======> {{{(11 + d)/9 = (29 + 2d)/(11 + d)}}}
{{{(11 + d)^2 = 9(29 + 2d)}}} ------ Cross-multiplying
{{{121 + 22d + d^2 = 261 + 18d}}} 
{{{d^2 + 22d + 121 - 18d - 261 = 0}}}
{{{d^2 + 4d - 140 = 0}}}
(d - 10)(d + 14) = 0 
d, or common difference = 10, or – 14

As the SMALLER 3rd term is being sought, we use d = - 14
Therefore, the SMALLER 3rd term of the GP = {{{highlight_green(matrix(1,5, 29 + 2(- 14), or, 29 - 28, or, 1))}}}