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?

Answer: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 [tex]a,a+d,a+2d[/tex]First term a=9Second term - 2 is added to second term i.e. [tex]a+d+2=9+d+2=11+d[/tex]Third term - 20 is added to the third term i.e.[tex]a+2d+20=9+2d+20=29+2d[/tex]The geometric progression is given by [tex]a,ar,ar^2[/tex]First term a=9Second term -[tex]ar=11+d[/tex]Third term [tex]ar^2=29+2d[/tex]r is the common ratio which is second term divided by first term,So, [tex]r=\frac{ar}{a}=\frac{11+d}{9}[/tex]or  third term divided by second term, So, [tex]r=\frac{ar^2}{ar}=\frac{29+2d}{11+d}[/tex]Equating both the r, [tex]\frac{11+d}{9}=\frac{29+2d}{11+d}[/tex]Cross multiply, [tex](11+d)\times (11+d)=(29+2d)(9)[/tex] [tex]11^2+11d+11d+d^2=261+18d[/tex] [tex]121+22d+d^2-261-18d=0[/tex] [tex]d^2+4d-140=0[/tex]Solving by middle term split, [tex]d^2+14d-10d-140=0[/tex] [tex]d(d+14)-10(d+14)=0[/tex] [tex](d+14)(d-10)=0[/tex] [tex]d=-14,10[/tex]Substituting the value of d in the third term,Third term [tex]ar^2=29+2d[/tex]When d=-14,Third term [tex]ar^2=29+2(-14)=29-28=1[/tex]When d=10,Third term [tex]ar^2=29+2(10)=29+20=49[/tex]Therefore, The smallest possible value for the third term of the geometric progression is 1.