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
  • $\begingroup$ What kind of a job asks these questions ? $\endgroup$ Apr 30, 2016 at 17:24
  • $\begingroup$ Government Job, includes Math, Physics, English, Quant. Want to review where i did bad :) $\endgroup$ Apr 30, 2016 at 17:27
  • $\begingroup$ You have four unknowns $x,y,z,k$, thus you need 4 independent equations. Have you thought along this line ? How can you express that terms belong to an arithmetical (resp. geometrical) sequence ? $\endgroup$
    – Jean Marie
    Apr 30, 2016 at 17:32
  • $\begingroup$ @JeanMarie That's the complete question they gave, I tried to put in AP and GP equations but I am only stuck with more equations. Can anybody give complete answer, I am not a math student, my major is Electronics $\endgroup$ Apr 30, 2016 at 17:37
  • $\begingroup$ @angelaphilstine They phrased the question incorrectly. Of course three terms cannot literally "form an infinite geometric progression". It probably should have said "form a geometric progression; if this geometric progression is extended to an infinite geometric progression, then...". Although, considering the purported answer maybe they meant something completely different! Also, "the sum" is singular so the last word should have been "is". $\endgroup$
    – Erick Wong
    Apr 30, 2016 at 17:57

3 Answers 3


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

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 $$\frac{54}{1-2/3} = 162$$

  • $\begingroup$ That makes a lot of sense now, also how is the final answer 106? Thanks a lot peter $\endgroup$ Apr 30, 2016 at 17:45
  • $\begingroup$ Yes, one finds 114... Considering the neat way @peterwhy has used all the other conditions, it strictly means that the last condition is incompatible with the others... Strange test... $\endgroup$
    – Jean Marie
    Apr 30, 2016 at 17:53
  • $\begingroup$ But 114 he just added the 3 terms na? Won't there be any formula for sum of GP infinite terms. I know one formula for sum of n terms in GP S = a (1-r^n)/(1-r) Definitely Strange test, you should see my face on test site $\endgroup$ Apr 30, 2016 at 17:55
  • 1
    $\begingroup$ @angelaphilstine given the options, the answer is $\frac{54}{1-2/3} = 162$. $\endgroup$
    – peterwhy
    Apr 30, 2016 at 17:57
  • $\begingroup$ @peterwhy Can you explain how u got that formula, maybe i can raise an objection with them $\endgroup$ Apr 30, 2016 at 17:59

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 $162$.

  • $\begingroup$ Hey thanks andre :D $\endgroup$ Apr 30, 2016 at 18:00
  • $\begingroup$ You are welcome. There is really no substantially different way to tackle the problem, the three expressions in $x$ give us the only useful information. $\endgroup$ Apr 30, 2016 at 18:02
  • $\begingroup$ Yes! maybe i can ace it next time :) $\endgroup$ Apr 30, 2016 at 18:04
  • $\begingroup$ This sort of interview question has become quite faddish in the last few years. I am skeptical about how much useful information can be extracted from candidate performance, but unfortunately the candidate is forced to try to prepare for it. $\endgroup$ Apr 30, 2016 at 18:10

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 $54+36+24=114$.


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