Question

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 1

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 $a,a+d,a+2d$

First term a=9

Second term - 2 is added to second term i.e. $a+d+2=9+d+2=11+d$

Third term - 20 is added to the third term i.e.$a+2d+20=9+2d+20=29+2d$

The geometric progression is given by $a,ar,ar^2$

First term a=9

Second term -$ar=11+d$

Third term $ar^2=29+2d$

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

So, $r=\frac{ar}{a}=\frac{11+d}{9}$

or  third term divided by second term,

So, $r=\frac{ar^2}{ar}=\frac{29+2d}{11+d}$

Equating both the r,

$\frac{11+d}{9}=\frac{29+2d}{11+d}$

Cross multiply,

$(11+d)\times (11+d)=(29+2d)(9)$

$11^2+11d+11d+d^2=261+18d$

$121+22d+d^2-261-18d=0$

$d^2+4d-140=0$

Solving by middle term split,

$d^2+14d-10d-140=0$

$d(d+14)-10(d+14)=0$

$(d+14)(d-10)=0$

$d=-14,10$

Substituting the value of d in the third term,

Third term $ar^2=29+2d$

When d=-14,

Third term $ar^2=29+2(-14)=29-28=1$

When d=10,

Third term $ar^2=29+2(10)=29+20=49$

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