Question

The sum of the first three terms of a geometric progression is 126. If 14, 36, and 4 are added to each of these terms, respectively, then the new numbers form an arithmetic progression. Find the sixth term of the geometric progression

Answer 1

Answer:

if r=4:

$\displaystyle a_6=6144$

if r=1/4:

$\displaystyle a_6=\frac{3}{32}$

Step-by-step explanation:

Geometric and Arithmetic Progressions

We define a geometric progression when each term $a_n$ is defined as the previous term $a_{n-1}$ times a constant called the common ratio. The iterative formula is

$\displaystyle a_n=a_1.r^{n-1}$

In an arithmetic progression, each term is found by adding a constant called common difference, to the previous term

$\displaystyle a_n=a_1+(n-1).r$

We are given the condition that the sum of the three first terms of a geometric progression is 126

$\displaystyle a_1+a_2+a_3=126$

Using the iterative formula, we have

$\displaystyle a_1+a_1.r+a_1.r^2=126$

Taking a common factor

$\displaystyle a_1(1+r+r^2)=126....[eq\ 1]$

We also know that if 14, 36, and 4 are added to each term, respectively, the new numbers form an arithmetic progression. It means they will have a common difference. The new numbers will be

$\displaystyle a_1'=a_1+14$

$\displaystyle a_2'=a_2+36$

$\displaystyle a_3'=a_3+4$

The common difference between term 2 and term 1 is

$\displaystyle a_2'-a_1'=a_2+36-a_1-14$

Using the iterative formula again

$\displaystyle a_2'-a_1'=a_1.r-a_1+22$

The common difference between term 3 and term 2 is

$\displaystyle a_3'-a_2'=a_3+4-a_2-36$

Using the iterative formula again

$\displaystyle a_3'-a_1'=a.r^2-a.r-32$

Both common differences must be equal

$\displaystyle a_1.r-a_1+22=a_1.r^2-a_1.r-32$

Rearranging

$\displaystyle 2a_1r-a_1r^2-a_1=-54$

Solving for $a_1$

$\displaystyle a_1=\frac{54}{1-2r+r^2}......[eq\ 2]$

Replacing in eq 1

$\displaystyle \frac{54(1+r+r^2)}{1-2r+r^2}=127$

Dividing by 18 and cross-multiplying

$\displaystyle 3+3r+3r^2=7-14r+7r^2$

Rearranging we have a second-degree equation

$\displaystyle 4r^2-17r+4=0$

Factoring

$\displaystyle (r-4)(4r-1)=0$

The solutions are

$\displaystyle r=4\ ,\ r=\frac{1}{4}$

If r=4, and using eq 2

$\displaystyle a_1=\frac{54}{1-8+16}=6$

Having $a_1$ and r, we compute $a_6$

$\displaystyle a_6=a_1.r^5=6.(4)^5$

$\displaystyle a_6=6144$

If we use the other solution r=1/4

$\displaystyle a_1=\frac{54}{1-\frac{1}{2}+\frac{1}{16}}=\frac{54}{\frac{9}{16}}$

$\displaystyle a_1=96$

The sixth term is

$\displaystyle a_6=96(\frac{1}{4})^5=\frac{96}{1024}$

$\displaystyle a_6=\frac{3}{32}$

Both solutions are feasible