Asequence 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?

The smallest possible value for the third term of the geometric progression is 1.

Step-by-step explanation:

Given : 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.

To find : What is the smallest possible value for the third term of the geometric progression?  

Solution :

The arithmetic progression is given by

First term a=9

Second term - 2 is added to second term i.e.

Third term - 20 is added to the third term i.e.

The geometric progression is given by

First term a=9

Second term -

Third term

r is the common ratio which is second term divided by first term,

So,

or  third term divided by second term,

So,

Equating both the r,

Cross multiply,

Solving by middle term split,

Substituting the value of d in the third term,

Third term

When d=-14,

Third term

When d=10,

Third term

Therefore, The smallest possible value for the third term of the geometric progression is 1.

alright well the answers to your question is

2xy/9 hope this have a nice day : )

step-by-step explanation: