Geometric, Arithmetic Progression

Question

I am not a student, not a homework. I got this in a test for job, hope someone can help with this

The first seven terms of an arithmetic progression are

$$7x+4, 9y, 7z+6, 4x+10, 3k, 3x+6, k+12.$$

The second, fifth and seventh terms of this progression form an infinite geometric progression. The sum of the terms of this geometric progression are?

Answer is $106$, I need to know HOW?

Given Options

a. 180
b. 126
c. 106
d. 162

Answer 1

$$7x+4,\ 9y,\ 7z+6,\ 4x+10,\ 3k,\ 3x+6,\ k+12$$

Let the common difference of the AP be $d$. Consider the first, fourth and sixth terms:

$$3d = (4x+10)-(7x+4) = 6-3x\\ 2d = (3x+6)-(4x+10) = -4 -x\\ 0 = 6d - 6d = 2(6-3x)-3(-4-x)\\ x= 8,\quad d= -6$$

So the first $7$ terms of the AP are

$$60,\ 9y,\ 7z+6,\ 42,\ 3k,\ 30,\ k+12$$ or $$60,\ 54,\ 48,\ 42,\ 36,\ 30,\ 24$$ The second, fifth and seventh terms form a GP with first term $54$ and common ratio $2/3$, and the infinite sum is $$\frac{54}{1-2/3} = 162$$

Answer 2

Concentrate on the $x$'s. We have that $(4x+10)-(7x+4)$ is three times the common difference, so the common difference is $2-x$.

So twice the common difference is $4-2x$. This is equal to $(3x+6)-(4x+10)$, which is $-x-4$. That gives $x=8$.

Now we know the arithmetic sequence, so can identify the second, fifth,and seventh terms, and solve the problem.

The sum of the infinite geometric progression with given first three terms is not $106$. It turns out to be $\frac{54}{1-\frac{2}{3}}$, which is $162$.

Answer 3

Ok This is what I have it doesnt really work out. (The problem was probably written by a committee.)Take the arithmetic progression. Say the difference is $a$, so you have

$$7x+4, 7x+4+a,7x+4+2a,7x+4+3a,7x+4+4a,7x+4+5a,7x+4+6a$$

That means that $$7x+4+3a=4x+10$$ and $$7x+4+5a=3x+6$$ Solving gives $x=8$ and $a=-6$ So the sequence is $$60, 54, 48,42, 36, 30, 24$$

So now $54+36+24=114$.