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

if r=4:

`\displaystyle a_6=6144`

if r=1/4:

`\displaystyle a_6=(3)/(32)`

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=(54)/(1-2r+r^2)......[eq\ 2]`

Replacing in eq 1

`\displaystyle (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=(1)/(4)`

If r=4, and using eq 2

`\displaystyle a_1=(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=(54)/(1-(1)/(2)+(1)/(16))=(54)/((9)/(16))`

`\displaystyle a_1=96`

The sixth term is

`\displaystyle a_6=96((1)/(4))^5=(96)/(1024)`

`\displaystyle a_6=(3)/(32)`

Both solutions are feasible